Current Electricity: ICSE Class 10 Physics solutions

Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 8 Current Electricity: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed. 

Summary

Objects can have an electric charge, which can be positive or negative. This happens when tiny particles called electrons move between objects. Losing electrons makes an object positively charged; gaining them makes it negatively charged. The unit for charge is the coulomb. When charges flow through materials called conductors, this is electric current. Insulators don’t allow charge to flow easily. Current, measured in amperes, is the rate charge passes a point.

For current to flow, a potential difference, or voltage, is needed between two points. Voltage acts like a push for charges and is measured in volts. It represents work done to move a charge. Resistance opposes current flow and is measured in ohms. Ohm’s law states voltage equals current multiplied by resistance (V=IR). Conductors following this are ohmic; others are non-ohmic.

A conductor’s resistance depends on its material, length (longer means more resistance), thickness (thicker means less), and temperature. Materials have specific resistance or resistivity, indicating their resistance for a standard size. This helps choose materials: copper for wires (low resistivity) or nichrome for heaters (high resistivity). Superconductors show zero resistance at very low temperatures.

Electrical energy is carried by moving charges. It’s calculated by formulas like energy equals charge times voltage, or current squared times resistance times time. Its unit is the joule. Electrical power is the rate electrical energy is used. Power, measured in watts, can be found by power equals voltage times current. Household energy is often measured in kilowatt-hours (kWh). Appliances have power ratings, like ‘100W – 220V’, showing power used at a specific voltage. When current flows, it can also produce heat.

A cell provides an electromotive force (e.m.f.), its voltage when no current is drawn. When current flows, the terminal voltage is usually less than the e.m.f. due to the cell’s internal resistance. Resistors can be connected in series (end-to-end) or in parallel (side-by-side). In series, total resistance is the sum of individual resistances. In parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances. This affects current and voltage in a circuit.

Textbook solutions

Exercise (A)

MCQ

1. The charge on an electron is :

(a) + 1.6 × 10⁻¹⁹ C
(b) -1.6 x 10⁻²⁰ C
(c) + 1.6 × 10²⁰ C
(d) -1.6 × 10⁻¹⁹ C

Answer: (d) -1.6 × 10⁻¹⁹ C

2. The number of electrons in 1 C charge is :

(a) 6.25 × 10¹⁸
(b) 6.25 x 10¹⁹
(c) 6.25 x 10²⁵
(d) 1.6 x 10⁻¹⁹

Answer: (a) 6.25 × 10¹⁸

3. The current in a circuit is measured by ………… by connecting it in …………

(a) voltmeter, series
(b) voltmeter, parallel
(c) ammeter, series
(d) ammeter, parallel

Answer: (c) ammeter, series

4. If n electrons pass through a cross-section of a conductor in time t, then current in conductor is :

(a) I = ne/t
(b) I = net
(c) I = nt/e
(d) I = n/et

Answer: (a) I = ne/t

5. If W joule of work is done in bringing a test charge Q coulomb from infinity to the point P, then electric potential V at point P is:

(a) V = W Q
(b) V = Q/W
(c) V = W/Q
(d) none of the above

Answer: (c) V = W/Q

6. Potential at a point is said to be 1 volt, when ………… of work is done in bringing 1 coulomb charge from infinity to that point.

(a) 1 joule
(b) 5 joule
(c) 10 joule
(d) 1 erg

Answer: (a) 1 joule

7. The potential difference between two points in an electric circuit is measured by a ………… which is connected in ………… with the circuit.

(a) voltmeter, series
(b) voltmeter, parallel
(c) ammeter, series
(d) ammeter, parallel

Answer: (b) voltmeter, parallel

8. The obstruction offered to the flow of current by a wire is called :

(a) current
(b) potential
(c) resistance
(d) charge

Answer: (c) resistance

9. According to Ohm’s law :

(a) I ∝ V
(b) V/I = constant
(c) V = IR
(d) All of the above

Answer: (d) All of the above

10. 1 ohm is equal to :

(a) 1 volt
(b) Volt × ampere
(c) volt/ampere
(d) all of the above

Answer: (c) volt/ampere

11. The unit of conductance is :

(a) volt
(b) ohm
(c) ohm⁻¹
(d) ampere

Answer: (c) ohm⁻¹

12. Out of the following, non-ohmic resistor is :

(a) nichrome
(b) copper sulphate solution with copper electrodes and dil sulphuric acid.
(c) solar cell
(d) silver

Answer: (c) solar cell

13. The resistance of a conductor depends on :

(a) the temperature of the conductor
(b) the thickness of the conductor
(c) the length of the conductor
(d) all of the above

Answer: (d) all of the above

14. The specific resistance of a rod of aluminium as compared to that of a thin wire of aluminium is :

(a) Less
(b) More
(c) Same
(d) Depends upon the area of cross section and length of both.

Answer: (c) Same

15. If the length l of wire is increased to 3l by stretching, its resistance R increases to …………

(a) 3R
(b) 4R
(c) 6R
(d) 9R

Answer: (d) 9R

16. Which of the following is an ohmic resistor ?

(a) LED
(b) junction diode
(c) filament of bulb
(d) nichrome wire

Answer: (d) nichrome wire

17. For which of the following substances, resistance decreases with the increase in temperature :

(a) copper
(b) mercury
(c) carbon
(d) platinum

Answer: (c) carbon

18. Standard resistors are made of:

(a) Copper
(b) Aluminium
(c) Constantan
(d) Carbon

Answer: (c) Constantan

19. The properties for which nichrome wire is used in heating appliances is/are :

(a) Specific resistance is high
(b) Resistance decreases with an increase in temperature
(c) Resistance increases with an increase in temperature
(d) Both (a) and (c)

Answer: (d) Both (a) and (c)

20. A super conductor is a substance of :

(a) infinite conductance
(b) infinite resistance
(c) zero resistance
(d) both (a) and (c)

Answer: (d) both (a) and (c)

Very Short Answer Type Questions

1. Write an expression connecting the resistance of a wire and specific resistance of its material. State the meaning of the symbols used.

Answer: The expression connecting the resistance of a wire and its specific resistance is ρ = Ra/l.
The meaning of the symbols used are:
ρ is the specific resistance (or resistivity) of the material of the wire.
R is the resistance of the wire.
a is the area of cross section of the wire.
l is the length of the wire.

2. State the order of specific resistance of (i) a metal, (ii) a semiconductor, and (iii) an insulator.

Answer: The order of specific resistance is:
(i) For metals and alloys, it is very low, approximately 10⁻⁸ Ω m.
(ii) For semiconductors, it is low, approximately 10⁻⁵ Ω m.
(iii) For insulators, it is very high, approximately 10¹³ Ω m.

3. Name a substance of which the specific resistance remains almost unchanged by the increase in temperature.

Answer: For some alloys, e.g. constantan, manganin, the specific resistance remains practically unchanged or constant with change in temperature.

4. Name the material used for (i) filament of an electric bulb, and (ii) heating element of a room heater.

Answer:
(i) For the filament of an electric bulb, a tungsten wire is used.
(ii) For the heating element in appliances such as a heater, toaster, oven, etc., a nichrome wire is used.

5. A substance has zero resistance below 1 K. What is such a substance called?

Answer: Such substances are called superconductors. A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.

Short Answer Type Questions

1. Define the term current and state its S.I. unit.

Answer: Current is defined as the rate of flow of charge. The S.I. unit of current is ampere (symbol A).

2. Define the term electric potential. State its S.I. unit.

Answer: The potential at a point is defined as the amount of work done per unit charge in bringing a positive test charge from infinity to that point. The S.I. unit of potential is joule/coulomb (or JC⁻¹) which has been given the name volt (symbol V).

3. How is the electric potential difference between two points defined ? State its S.I. unit.

Answer: The potential difference (p.d.) between two points is equal to the work done per unit charge in moving a positive test charge from one point to the other. Potential difference (p.d.) is also expressed in volt (V).

4. Explain the statement ‘the potential difference between two points is 1 volt’.

Answer: The potential difference between two points is said to be 1 volt if the work done in moving 1 coulomb charge from one point to the other is 1 joule, i.e., 1 volt = 1 joule / 1 coulomb = 1 JC⁻¹.

5. (a) State whether the current is a scalar or vector ? What does the direction of current convey?
(b) State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey ?

Answer: (a) Current is a scalar quantity. By stating the direction of current in a conductor, we mean that the direction of motion of electrons is opposite to it.

(b) Potential is a scalar quantity. The potential is positive at a point in the vicinity of a positive charge since work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity, while it is negative at a point in the vicinity of a negative charge since the attractive force on the positive test charge due to the negative charge does work by itself (or work is obtained).

6. Define the term resistance. State its S.I. unit.

Answer: The obstruction offered to the flow of current by the conductor (or wire) is called its resistance. The S.I. unit of resistance is volt/ampere (or V A⁻¹) which is named ohm, denoted by the symbol Ω (omega).

7. (a) Name the particles which are responsible for the flow of current in a metallic wire.
(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part (a).
(c) What is the cause of resistance offered by the metallic wire in the flow of current through it ?

Answer: (a) In metals, the moving charges are the free electrons which constitute the current.

(b) When a metal is formed, its atoms lose electrons from its outer orbit. A metal (or conductor) has a large number of wandering electrons and an equal number of fixed positive ions. These electrons are called free electrons. When the ends of the metal wire are connected to a cell, i.e., when a potential difference is applied across the ends of the metal wire, the electrons inside it experience a force in the direction from the end at negative potential towards the end at positive potential of the conductor. As a result of this force, electrons begin to drift towards the positive end.

(c) During the movement of free electrons towards the positive end, they collide with the fixed positive ions and between themselves and lose some of their kinetic energy due to which their speed decreases. This process continues. As a result, the electrons do not move in bulk with a continuously increasing speed, but there is an overall drift of electrons towards the positive terminal. Thus, a metal wire offers some resistance to the flow of electrons through it. The resistance of a conductor depends on the number of collisions suffered by the electrons with the positive ions while moving from one end to the other end.

8. (a) Name and state the law which relates the potential difference and current in a conductor.
(b) What is the necessary condition for a conductor to obey the law named above in part (a) ?

Answer: (a) The law is Ohm’s law. According to Ohm’s law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant.

(b) Ohm’s law is obeyed only when the temperature of the conductor remains constant.

9. Give two differences between an ohmic and non-ohmic resistor.

Answer: Two differences between an ohmic and non-ohmic resistor are:

• An ohmic resistor obeys Ohm’s law, i.e., V/I is constant for all values of V or I, while a non-ohmic resistor does not obey Ohm’s law, i.e., V/I is not the same for all values of V or I.
• For an ohmic resistor, the graph for potential difference V versus current I is a straight line passing through the origin, while for a non-ohmic resistor, the graph for potential difference V versus current I is not a straight line, but is a curve which may not pass through the origin.

10. The figure shows I–V curves for two resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.

Answer: (b) is an ohmic resistor and (a) is a non-ohmic resistor.
Reason: For (b), the I-V graph is a straight line, while for (a), the graph is a curve.

11. How does the resistance of a wire depend on its length? Give a reason for your answer.

Answer: The resistance of a conductor is directly proportional to the length l of the conductor i.e., R ∝ l.

Reason: In a long conductor, the number of collisions of free electrons with the positive ions will be more as compared to a shorter one. Therefore, a longer conductor offers a proportionately more resistance.

12. How does the resistance of a metallic wire depend on its temperature ? Explain with reason.

Answer: The resistance of a conductor increases with an increase in its temperature.

Reason: With the increase in temperature of a conductor, the random motion of free electrons increases. As a result, the number of collisions of electrons with the positive ions increases. Hence, the resistance of conductor increases with an increase in its temperature.

13. Two wires, one of copper and the other of iron, are of the same length and same radius. Which will have more resistance ? Give reason.

Answer: The iron wire will have more resistance.

Reason: The specific resistance of iron (9.8 x 10⁻⁸ Ω m at 20°C) is more than that of copper (1.73 x 10⁻⁸ Ω m at 20°C).

14. Name three factors on which resistance of a given wire depends and state how it is affected by the factors stated by you.

Answer: Three factors on which the resistance of a given wire depends are its material, length, and thickness (or area of cross section). It also depends on temperature.
How it is affected:

• Material: Different materials have different concentration of free electrons; materials with lower concentration of free electrons offer higher resistance.
• Length (l): The resistance of a conductor is directly proportional to its length (R ∝ l). Longer wires offer more resistance.
• Thickness (area of cross section a): The resistance of a conductor is inversely proportional to its area of cross section (R ∝ 1/a). Thicker wires (larger area) offer less resistance.
• Temperature: For most conductors, the resistance increases with an increase in temperature.

15. Define the term specific resistance and state its S.I. unit.

Answer: Specific resistance of a material is the resistance of a wire of that material of unit length and unit area of cross section. In other words, the specific resistance of a material is the resistance between the opposite faces of a unit cube of that material. The S.I. unit of specific resistance is ohm × metre (or Ω m).

16. (a) Name two factors on which the specific resistance of a wire depends ?
(b) Two wires A and B are made of copper. The wire A is long and thin, while the wire B is short and thick. Which will have more specific resistance ?

Answer: (a) Two factors on which the specific resistance of a wire depends are:

• The material of the substance (it is its characteristic property).
• The temperature of the substance.

(b) Both wires A and B, being made of copper, will have the same specific resistance, as specific resistance is a characteristic property of the material and does not depend on the shape and size (length or thickness) of the conductor, assuming the temperature is the same.

17. How does the specific resistance of a semi-conductor change with the increase in temperature ?

Answer: The specific resistance of a semiconductor decreases with the increase in temperature.

18. How does (a) resistance, and (b) specific resistance of a wire depend on its (i) length, and (ii) radius ?

Answer: (a) Resistance of a wire:

(i) Length (l): Resistance is directly proportional to its length (R ∝ l).
(ii) Radius (r): Resistance is inversely proportional to the square of its radius (R ∝ 1/r²), since R ∝ 1/a and a = πr².

(b) Specific resistance of a wire: The specific resistance of the material of a conductor at a given temperature is constant and it does not depend on its length or radius.

19. (a) Name the material used for making connection wires. Give a reason for your answer.
(b) Why should a connection wire be thick ?

Answer: (a) Materials such as copper or aluminium are used for making connection wires.
Reason: Such wires should possess the least possible resistance, and copper and aluminium have very small specific resistance.

(b) Connection wires are made thick so that their resistance becomes even lower, as resistance is inversely proportional to the area of cross-section.

20. Name a material which is used for making a standard resistor. Give a reason for your answer.

Answer: Standard resistors are made from alloys such as manganin, constantan, etc.
Reason: These materials are used because their specific resistance is high and the effect of change in temperature on their resistance is negligible.

21. Name the material used for making a fuse wire. Give a reason.

Answer: A fuse wire is made from an alloy of lead and tin.

Reason: It is used because its melting point is low and its specific resistance is more than that of copper or aluminium. This ensures that the resistance of a short and thin fuse wire is more than that of the connecting wire. It permits current up to its safe limit to pass through it, but an excessive current melts it so that it blows off and the circuit is broken.

22. What is a superconductor? Give one example of it.

Answer: A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.
Example: Mercury below 4.2 K is a superconductor.

Long Answer Type Questions

1. State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.

Answer: According to Ohm’s law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant.

2. (a) Draw a V-I graph for a conductor obeying Ohm’s law.
(b) What does the slope of V-I graph for a conductor represent ?

Answer: (a) A V-I graph for a conductor obeying Ohm’s law is a straight line passing through the origin, with V on the Y-axis and I on the X-axis.

(b) The slope of the V-I graph (ΔV/ΔI) for a conductor represents the resistance R of the conductor.

3. Draw an I-V graph for a linear resistor. What does its slope represent ?

Answer: An I-V graph for a linear resistor is a straight line passing through the origin, with I on the Y-axis and V on the X-axis.

The slope of the I-V graph (ΔI/ΔV) is the reciprocal of the resistance of the conductor.

4. What is an ohmic resistor ? Give one example of an ohmic resistor. Draw a graph to show its current-voltage relationship. How is the resistance of the resistor determined from this graph ?

Answer: Ohmic resistors are conductors which obey Ohm’s law.

An example of an ohmic resistor is a metallic conductor such as silver or copper at a constant temperature.
The current-voltage relationship (V-I graph) for an ohmic resistor is a straight line passing through the origin when V is plotted against I.

The resistance of the resistor is determined by finding the slope of this V-I graph; the slope (ΔV/ΔI) gives the resistance R.

5. What are non-ohmic resistors ? Give one example and draw a graph to show its current-voltage relationship.

Answer: Non-ohmic resistors are conductors which do not obey Ohm’s law.

An example of a non-ohmic resistor is a junction diode or the filament of a bulb. The current-voltage relationship (V-I graph) for a non-ohmic resistor is not a straight line, but a curve. For example, for a junction diode, the V-I graph is a curve.

6. Draw a V-I graph for a conductor at two different temperatures. What conclusion do you draw from your graph for the variation of resistance of the conductor with temperature ?

Answer: A V-I graph for a conductor at two different temperatures T₁ and T₂ (where T₁ > T₂) will show two straight lines passing through the origin. The line corresponding to the higher temperature T₁ will be steeper than the line for the lower temperature T₂.

The conclusion drawn from the graph is that the resistance of the conductor is more at a higher temperature than at a lower temperature, meaning the resistance of the conductor increases with an increase in its temperature.

7. (a) How does the resistance of a wire depend on its radius ? Explain your answer.
(b) Two copper wires are of the same length, but one is thicker than the other. Which will have more resistance ?

Answer: (a) The resistance of a wire is inversely proportional to the square of its radius (R ∝ 1/r²). This is because resistance is inversely proportional to its area of cross-section (a), and for a wire with circular cross-section, a = πr².

Explanation: In a thick conductor (larger radius, thus larger area of cross-section), electrons get a larger area to flow as compared to a thin conductor. Therefore, a thick conductor offers less resistance.

(b) The thinner copper wire will have more resistance.

Numerical Questions

1. In a conductor, 6.25 × 10¹⁶ electrons flow from its end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10⁻¹⁹ C)

Answer:

Given:

Number of electrons (n) = 6.25 × 10¹⁶
Time (t) = 2 s
Charge of an electron (e) = 1.6 × 10⁻¹⁹ C

To find:

Current (I) = ?

Solution:

First, we need to find the total charge (Q) flowing through the conductor. The total charge is the product of the number of electrons and the charge of a single electron.

Q = n × e
=> Q = (6.25 × 10¹⁶) × (1.6 × 10⁻¹⁹) C
=> Q = (6.25 × 1.6) × (10¹⁶ × 10⁻¹⁹) C
=> Q = 10 × 10⁻³ C
=> Q = 10⁻² C or 0.01 C

Now, we can find the current (I), which is the rate of flow of charge.

I = Q / t
=> I = (10⁻² C) / (2 s)
=> I = 0.5 × 10⁻² A
=> I = 0.005 A

The current can also be expressed in milliamperes (mA). Since 1 A = 1000 mA:

I = 0.005 × 1000 mA
=> I = 5 mA

Therefore, the current flowing through the conductor is 5 mA.

2. A current of 3.2 mA flows through a conductor. If charge on an electron is – 1.6 × 10⁻¹⁹ coulomb, find the number of electrons that will pass each second through the cross section of that conductor.

Given:

Current (I) = 3.2 mA
Charge on an electron (e) = 1.6 × 10^-19 C (We use the magnitude of the charge for calculation)
Time (t) = 1 s

To find:

Number of electrons (n) = ?

Solution:

First, we convert the current from milliamperes (mA) to amperes (A).
We know, 1 mA = 10^-3 A
=> I = 3.2 × 10^-3 A

The formula for electric current is:
I = Q / t
where Q is the total charge flowing through the cross-section in time t.

The total charge (Q) is also given by the product of the number of electrons (n) and the charge of a single electron (e).
Q = n × e

By combining the two formulas, we get:
I = (n × e) / t

To find the number of electrons (n), we rearrange the formula:
n = (I × t) / e

Now, we substitute the given values into the formula:
=> n = (3.2 × 10^-3 A × 1 s) / (1.6 × 10^-19 C)
=> n = (3.2 / 1.6) × (10^-3 / 10^-19)
=> n = 2 × 10^(-3 – (-19))
=> n = 2 × 10^(-3 + 19)
=> n = 2 × 10^16

3. Find the potential difference required to flow a current of 300 mA in a wire of resistance 20 Ω.

Answer:
Given:

Current (I) = 300 mA
Resistance (R) = 20 Ω

To find:

Potential difference (V) = ?

Solution:

First, we need to convert the current from milliamperes (mA) to amperes (A), as the S.I. unit of current is Ampere.

We know that,
1 A = 1000 mA
So, 1 mA = 1/1000 A

Therefore,
I = 300 mA
=> I = 300 / 1000 A
=> I = 0.3 A

Now, according to Ohm’s Law, the potential difference (V) is the product of the current (I) and the resistance (R).
The formula is:
V = I × R

Substituting the values of I and R in the formula:
=> V = 0.3 × 20
=> V = 6 V

Thus, the potential difference required to flow a current of 300 mA in the wire is 6 Volts.

4. An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of the filament of the bulb while glowing.

Answer:

Given:

Current (I) = 1.2 A
Potential difference (V) = 6.0 V

To find:

Resistance (R) = ?

Solution:

According to Ohm’s Law, the relationship between potential difference (voltage), current, and resistance is given by the formula:

V = I * R

To find the resistance (R), we can rearrange the formula as:

R = V / I

Now, substituting the given values into the formula:

R = 6.0 / 1.2
=> R = 5 Ω

Therefore, the resistance of the filament of the bulb while glowing is 5 Ω.

5. A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb ? Will the resistance be more, same or less when the bulb is not glowing ?

Answer:

Given:

Voltage (V) = 12 volt
Current (I) = 2 A

To find:

Resistance (R) of the filament when the bulb is glowing = ?
Comparison of resistance when the bulb is not glowing.

Solution:

According to Ohm’s law, the relationship between voltage, current, and resistance is given by the formula:
V = I × R

To find the resistance (R) of the glowing filament, we can rearrange the formula:
=> R = V / I

Now, we substitute the given values into the formula:
=> R = 12 / 2
=> R = 6 Ω

Thus, the resistance of the filament of the bulb when it is glowing is 6 Ω.

Now, for the second part of the question:

The resistance of a conductor, like the filament in a bulb, depends on its temperature. For most metallic conductors, resistance increases as the temperature increases.

• When the bulb is glowing: The filament is very hot, causing its resistance to be high. The calculated resistance of 6 Ω is the resistance of the hot filament.
• When the bulb is not glowing: The filament is at room temperature, which is much colder. At this lower temperature, its resistance is also lower.

6. Calculate the current flowing through a wire of resistance 5 Ω connected to a battery of potential difference 3 V.

Given:

Resistance (R) = 5 Ω
Potential difference (V) = 3 V

To find:

Current (I) = ?

Solution:

According to Ohm’s law, the relationship between potential difference, current, and resistance is given by the formula:
V = I × R

Where,
V = Potential difference
I = Current
R = Resistance

To find the current (I), we can rearrange the formula as:
I = V / R

Substituting the given values into the formula:
=> I = 3 / 5
=> I = 0.6 A

Therefore, the current flowing through the wire is 0.6 A.

7. In an experiment of verification of Ohm’s law, following observations are obtained.

Draw a V-I graph and use this graph to find :

(a) the potential difference V when the current I is 0.5 A,
(b) the current I when the potential difference V is 0.75 V,
(c) the resistance in circuit.

Answer:

(a) To find the potential difference V when the current I is 0.5 A:

• On the V-I graph, locate the point corresponding to 0.5 A on the x-axis (Current axis).
• From this point, draw a vertical line upwards to meet the graph line at a point, let’s call it P.
• From point P, draw a horizontal line to the left to meet the y-axis (Potential Difference axis).
• The reading on the y-axis at this point gives the required potential difference.
• From the graph, this value is found to be 1.25 V.

Therefore, when the current I is 0.5 A, the potential difference V is 1.25 V.

(b) To find the current I when the potential difference V is 0.75 V:

• On the V-I graph, locate the point corresponding to 0.75 V on the y-axis (Potential Difference axis).
• From this point, draw a horizontal line to the right to meet the graph line at a point, let’s call it Q.
• From point Q, draw a vertical line downwards to meet the x-axis (Current axis).
• The reading on the x-axis at this point gives the required current.
• From the graph, this value is found to be 0.3 A.

Therefore, when the potential difference V is 0.75 V, the current I is 0.3 A.

(c) To find the resistance in the circuit:

The resistance (R) of the circuit is equal to the slope of the V-I graph.

Slope = (Change in Potential Difference) / (Change in Current) = ΔV / ΔI

To calculate the slope, we can choose any two points from the given data. Let’s take the points (I₁, V₁) = (0.2, 0.5) and (I₂, V₂) = (1.0, 2.5).

R = Slope = (V₂ – V₁) / (I₂ – I₁)
=> R = (2.5 – 0.5) / (1.0 – 0.2)
=> R = 2.0 / 0.8
=> R = 2.5 Ω

Therefore, the resistance in the circuit is 2.5 Ω.

8. Two wires of the same material and same length have radii 1 mm and 2 mm respectively. Compare : (i) their resistances, (ii) their specific resistance.

Answer:

Given:

There are two wires, let’s call them Wire 1 and Wire 2.

For Wire 1:
Radius (r₁) = 1 mm

For Wire 2:
Radius (r₂) = 2 mm

It is also given that the two wires are of the same material and have the same length.
This means:
Specific resistance of Wire 1 (ρ₁) = Specific resistance of Wire 2 (ρ₂)
Length of Wire 1 (L₁) = Length of Wire 2 (L₂)

To find:

(i) The comparison of their resistances (Ratio R₁ : R₂).
(ii) The comparison of their specific resistances (Ratio ρ₁ : ρ₂).

Solution:

(i) Comparison of their resistances

The formula for the resistance (R) of a wire is:
R = ρ * (L / A)
where,
ρ = specific resistance of the material
L = length of the wire
A = cross-sectional area of the wire

Since the wire is cylindrical, its cross-sectional area (A) is given by A = πr², where r is the radius.
So, the resistance formula becomes:
R = ρ * (L / (πr²))

For the first wire, the resistance (R₁) is:
R₁ = ρ₁ * (L₁ / (πr₁²)) —(1)

For the second wire, the resistance (R₂) is:
R₂ = ρ₂ * (L₂ / (πr₂²)) —(2)

To compare their resistances, we can find their ratio by dividing equation (1) by equation (2):
R₁ / R₂ = [ρ₁ * (L₁ / (πr₁²))] / [ρ₂ * (L₂ / (πr₂²))]

From the given information, we know that the material and length are the same, so:
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L

Substituting these into the ratio equation:
R₁ / R₂ = [ρ * (L / (πr₁²))] / [ρ * (L / (πr₂²))]

By cancelling the common terms (ρ, L, and π), we get:
R₁ / R₂ = (1 / r₁²) / (1 / r₂²)
=> R₁ / R₂ = r₂² / r₁²

Now, we substitute the given values of the radii:
r₁ = 1 mm
r₂ = 2 mm

=> R₁ / R₂ = (2)² / (1)²
=> R₁ / R₂ = 4 / 1

Therefore, the ratio of their resistances is R₁ : R₂ = 4 : 1.

(ii) Comparison of their specific resistance

Specific resistance (also known as resistivity, ρ) is a fundamental property of a material. It depends only on the nature of the material and its temperature, not on its physical dimensions like length or area.

The problem states that the two wires are made of the same material.

Therefore, their specific resistances must be equal.
ρ₁ = ρ₂

To compare them, we find their ratio:
ρ₁ / ρ₂ = 1 / 1

Therefore, the ratio of their specific resistances is ρ₁ : ρ₂ = 1 : 1.

9. A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance ?

Answer:

Given:

Initial Resistance (R₁) = 1 Ω
Let the initial length of the wire be L₁.
Let the initial cross-sectional area of the wire be A₁.
The wire is stretched such that its new length (L₂) is double its initial length.
=> New Length (L₂) = 2 * L₁

To find:

New Resistance (R₂) = ?

Solution:

When a wire is stretched, its material does not change, so its resistivity (ρ) remains constant. Also, the total volume (V) of the wire remains constant.

Since the volume of the wire remains constant:
Initial Volume (V₁) = Final Volume (V₂)
=> L₁ * A₁ = L₂ * A₂

Substitute the value of L₂ = 2 * L₁ into the equation:
=> L₁ * A₁ = (2 * L₁) * A₂
Dividing both sides by L₁:
=> A₁ = 2 * A₂
=> A₂ = A₁ / 2

This shows that when the length of the wire is doubled, its cross-sectional area is halved.

Now, we use the formula for resistance:
R = ρ * (L / A)

The initial resistance is:
R₁ = ρ * (L₁ / A₁) —(Equation 1)

The new resistance is:
R₂ = ρ * (L₂ / A₂) —(Equation 2)

Substitute the expressions for L₂ and A₂ in terms of L₁ and A₁ into Equation 2:
=> R₂ = ρ * ( (2 * L₁) / (A₁ / 2) )
=> R₂ = ρ * ( 2 * L₁ * (2 / A₁) )
=> R₂ = ρ * ( 4 * L₁ / A₁ )
=> R₂ = 4 * (ρ * L₁ / A₁)

From Equation 1, we know that R₁ = ρ * (L₁ / A₁). So, we can substitute R₁ into the expression for R₂:
=> R₂ = 4 * R₁

We are given the initial resistance, R₁ = 1 Ω.
=> R₂ = 4 * 1 Ω
=> R₂ = 4 Ω

Thus, the new resistance of the wire will be 4 Ω.

10. A wire of resistance 3 ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross section, what will be its new resistance ?

Answer:

Given:

Initial Resistance (R₁) = 3 ohm
Initial Length (L₁) = 10 cm
Final Length (L₂) = 30 cm

To find:

New Resistance (R₂) = ?

Solution:

The resistance of a wire is given by the formula:
R = ρL/A
where,
ρ = Resistivity of the material
L = Length of the wire
A = Area of cross-section

When the wire is stretched, its volume (V) and resistivity (ρ) remain constant.
Volume (V) = Area (A) × Length (L)

Since the volume is constant, the initial volume is equal to the final volume.
V₁ = V₂
=> A₁L₁ = A₂L₂
=> A₂ = A₁(L₁/L₂)

Now, let’s write the expressions for the initial and new resistance.
Initial Resistance, R₁ = ρL₁/A₁ —(1)
New Resistance, R₂ = ρL₂/A₂ —(2)

Dividing equation (2) by equation (1), we get:
R₂/R₁ = (ρL₂/A₂) / (ρL₁/A₁)
=> R₂/R₁ = (L₂/A₂) × (A₁/L₁)

Now, substitute the value of A₂ = A₁(L₁/L₂) into the equation:
=> R₂/R₁ = (L₂ / (A₁(L₁/L₂))) × (A₁/L₁)
=> R₂/R₁ = (L₂²/A₁L₁) × (A₁/L₁)
=> R₂/R₁ = L₂²/L₁²
=> R₂/R₁ = (L₂/L₁)²

Now, we can find the new resistance R₂:
=> R₂ = R₁ × (L₂/L₁)²

Substitute the given values:
=> R₂ = 3 × (30/10)²
=> R₂ = 3 × (3)²
=> R₂ = 3 × 9
=> R₂ = 27 ohm.

Thus, the new resistance of the wire will be 27 ohm.

11. A wire of resistance 9 ohm having length 30 cm is tripled on itself. What is its new resistance ?

Answer:

Given:

Original Resistance (R) = 9 ohm
Original Length (L) = 30 cm
The wire is tripled on itself.

To find:

New Resistance (R’) = ?

Solution:

The resistance of a wire is given by the formula:
R = ρ * (L / A)
Where,
ρ (rho) = Resistivity of the material of the wire
L = Length of the wire
A = Cross-sectional area of the wire

When the wire is “tripled on itself”, it is folded into three equal segments laid side by side. This changes its length and cross-sectional area.

• New Length (L’): The length of the wire becomes one-third of its original length.
  L’ = L / 3
• New Area (A’): Since the wire is now a bundle of three original strands, the new cross-sectional area becomes three times the original area.
  A’ = 3 * A
• Resistivity (ρ’): The material of the wire remains the same, so its resistivity does not change.
  ρ’ = ρ

Now, we can write the formula for the new resistance (R’):
R’ = ρ’ * (L’ / A’)

Substitute the new values of length and area:
=> R’ = ρ * ((L / 3) / (3A))
=> R’ = ρ * (L / (9 * A))
=> R’ = (1/9) * (ρ * L / A)

We know that the original resistance is R = ρ * L / A = 9 ohm.
Substituting the value of R into the equation for R’:
=> R’ = (1/9) * R
=> R’ = (1/9) * 9
=> R’ = 1 ohm.

Thus, the new resistance of the wire is 1 ohm.

12. What length of copper wire of specific resistance 1.7 × 10⁻⁸ Ωm and radius 1 mm is required so that its resistance is 2 Ω?

Answer:

Given:

Specific resistance (ρ) = 1.7 × 10⁻⁸ Ωm
Radius (r) = 1 mm
Resistance (R) = 2 Ω

To find:

Length of the wire (L) = ?

Solution:

First, we need to convert all the given quantities into the S.I. system of units. The radius is given in millimeters (mm), so we convert it to meters (m).

r = 1 mm
=> r = 1 × 10⁻³ m

Next, we calculate the cross-sectional area (A) of the wire. The cross-section of a wire is a circle.

Area (A) = πr²
=> A = 3.14 × (1 × 10⁻³)²
=> A = 3.14 × 10⁻⁶ m²

The formula for resistance (R) is given by:

R = ρ * (L / A)

To find the length (L), we rearrange the formula:

L = (R × A) / ρ

Now, we substitute the values into the rearranged formula:

L = (2 × 3.14 × 10⁻⁶) / (1.7 × 10⁻⁸)
=> L = (6.28 × 10⁻⁶) / (1.7 × 10⁻⁸)
=> L = (6.28 / 1.7) × 10⁻⁶⁺⁸
=> L = 3.694 × 10²
=> L = 369.4 m

Therefore, the required length of the copper wire is 369.4 m.

13. The filament of a bulb takes a current 100 mA when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 400 mA. Calculate the resistance of the filament in each case and account for the difference.

Given:

Case 1:
Potential Difference (V₁) = 0.2 V
Current (I₁) = 100 mA

Case 2:
Potential Difference (V₂) = 1.0 V
Current (I₂) = 400 mA

To find:

• Resistance of the filament in Case 1 (R₁).
• Resistance of the filament in Case 2 (R₂).
• The reason for the difference in resistance.

Solution:

First, we need to convert the current from milliamperes (mA) to amperes (A), since 1 A = 1000 mA.

For Case 1:
I₁ = 100 mA = 100 / 1000 A = 0.1 A

For Case 2:
I₂ = 400 mA = 400 / 1000 A = 0.4 A

Now, we can calculate the resistance in each case using Ohm’s Law.

Formula: Resistance (R) = Potential Difference (V) / Current (I)

(i) Calculation for Case 1:

R₁ = V₁ / I₁
=> R₁ = 0.2 V / 0.1 A
=> R₁ = 2 Ω

(ii) Calculation for Case 2:

R₂ = V₂ / I₂
=> R₂ = 1.0 V / 0.4 A
=> R₂ = 2.5 Ω

(iii) Accounting for the difference:

The resistance of the filament in Case 1 is 2 Ω, and in Case 2, it is 2.5 Ω. The resistance has increased.

This difference occurs because the resistance of the bulb’s filament is not constant; it depends on its temperature. When a larger potential difference (1.0 V) is applied, a larger current (400 mA) flows through the filament. This causes the filament to heat up significantly more than in the first case. For metallic conductors like the filament, resistance increases as the temperature increases. Therefore, the filament has a higher resistance when it is hotter (in Case 2).

Exercise (B)

MCQ

1. The emf of a cell depends on:

(a) the material of electrodes
(b) the shape of electrodes
(c) the distance between the electrodes
(d) the amount of electrolyte in the cell

Answer: (a) the material of electrodes

2. Emf of a cell is ……. the terminal voltage when cell is not in use, while …… the terminal voltage when cell is in use.

(a) greater than, smaller than
(b) smaller than, greater than
(c) equal to, greater than
(d) equal to, smaller than

Answer: (c) equal to, greater than

3. Current drawn from the cell:

(a) I = ɛ[R + r]
(b) I = ɛ / (R+r)
(c) I = ɛR
(d) I = E / r

Answer: (b) I = ɛ / (R+r)

4. In series combination of resistances :

(a) p.d. is the same across each resistance
(b) total resistance is reduced
(c) current is same in each resistance
(d) all of the above are true

Answer: (c) current is same in each resistance

5. In parallel combination of resistances :

(a) p.d. is the same across each resistance
(b) current is the same in each resistance
(c) total resistance is increased
(d) all of the above are true

Answer: (a) p.d. is the same across each resistance

6. For parallel combination of resistances, which of the statements are correct ?

(a) On I-V graph, the slope of line is more.
(b) The potential difference across each resistance is same.
(c) The current in a resistor is inversely proportional to the resistance.
(d) All of the above

Answer: (d) All of the above

7. Assertion (A) : The terminal voltage of a cell is always less than its e.m.f. Reason (R) : More the current drawn from the cell, the less is the terminal voltage.

(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false

Answer: (c) assertion is false but reason is true

Very Short Answer Type Questions

1. Write the expressions for the equivalent resistance R of three resistors R₁, R₂ and R₃ joined in (a) parallel, and (b) series.

Answer:

(a) For three resistors R₁, R₂, and R₃ joined in parallel, the equivalent resistance R (often denoted Rp) is given by the expression:
1/R = 1/R₁ + 1/R₂ + 1/R₃

(b) For three resistors R₁, R₂, and R₃ joined in series, the equivalent resistance R (often denoted Rs) is given by the expression:
R = R₁ + R₂ + R₃

2. State how are the two resistors joined with a battery in each of the following cases when :

(a) same current flows in each resistor,
(b) potential difference is same across each resistor,
(c) equivalent resistance is less than either of the two resistances, and
(d) equivalent resistance is more than either of the two resistances.

Answer: (a) When the same current flows in each resistor, the two resistors are joined in series with the battery.
(b) When the potential difference is the same across each resistor, the two resistors are joined in parallel with the battery.
(c) When the equivalent resistance is less than either of the two resistances, the two resistors are joined in parallel.
(d) When the equivalent resistance is more than either of the two resistances, the two resistors are joined in series.

3. Which of the following combinations have the same equivalent resistance between X and Y ?

Answer:

Given:
Four circuit combinations with resistors connected between points X and Y.

• Circuit (a): Two 2 Ω resistors connected in parallel.
• Circuit (b): Three resistors (1 Ω, 1 Ω, and 2 Ω) connected in parallel.
• Circuit (c): Two 1 Ω resistors connected in parallel.
• Circuit (d): A Wheatstone bridge arrangement with four 1 Ω resistors on the sides and one 1 Ω resistor in the middle.

To find:
Which of the given combinations have the same equivalent resistance.

Solution:
We will calculate the equivalent resistance (R_eq) for each circuit combination between points X and Y.

(i) For Circuit (a):
The two 2 Ω resistors are connected in parallel. For two resistors R₁ and R₂ in parallel, the equivalent resistance R_p is given by:
R_p = (R₁ × R₂) / (R₁ + R₂)

Here, R₁ = 2 Ω and R₂ = 2 Ω.
=> R_eq(a) = (2 × 2) / (2 + 2)
=> R_eq(a) = 4 / 4
=> R_eq(a) = 1 Ω

(ii) For Circuit (b):
The three resistors (1 Ω, 1 Ω, and 2 Ω) are connected in parallel. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances:
1 / R_p = 1 / R₁ + 1 / R₂ + 1 / R₃

Here, R₁ = 1 Ω, R₂ = 1 Ω, and R₃ = 2 Ω.
=> 1 / R_eq(b) = 1/1 + 1/1 + 1/2
=> 1 / R_eq(b) = (2 + 2 + 1) / 2
=> 1 / R_eq(b) = 5 / 2
=> R_eq(b) = 2 / 5 = 0.4 Ω

(iii) For Circuit (c):
The two 1 Ω resistors are connected in parallel.
Here, R₁ = 1 Ω and R₂ = 1 Ω.
=> R_eq(c) = (1 × 1) / (1 + 1)
=> R_eq(c) = 1 / 2
=> R_eq(c) = 0.5 Ω

(iv) For Circuit (d):
This arrangement is a Wheatstone bridge. For a balanced Wheatstone bridge, the ratio of resistances in the opposite arms is equal (R₁/R₂ = R₃/R₄). In this case, all resistors are 1 Ω.
Ratio of the upper arm resistances = 1 Ω / 1 Ω = 1
Ratio of the lower arm resistances = 1 Ω / 1 Ω = 1
Since the ratios are equal, the bridge is balanced. In a balanced Wheatstone bridge, no current flows through the central resistor, so it can be ignored.

The circuit simplifies to two parallel branches:

• Upper branch: Two 1 Ω resistors in series. R_upper = 1 + 1 = 2 Ω.
• Lower branch: Two 1 Ω resistors in series. R_lower = 1 + 1 = 2 Ω.

Now, these two branches are in parallel.
=> R_eq(d) = (R_upper × R_lower) / (R_upper + R_lower)
=> R_eq(d) = (2 × 2) / (2 + 2)
=> R_eq(d) = 4 / 4
=> R_eq(d) = 1 Ω

Conclusion:
By comparing the equivalent resistances of all four circuits:

• R_eq(a) = 1 Ω
• R_eq(b) = 0.4 Ω
• R_eq(c) = 0.5 Ω
• R_eq(d) = 1 Ω

The equivalent resistance of Circuit (a) is equal to the equivalent resistance of Circuit (d).
Therefore, combinations (I) and (IV) have the same equivalent resistance.

Short Answer Type Questions

1. State two differences between the e.m.f. and terminal voltage of a cell.

Answer: Two differences between the e.m.f. and terminal voltage of a cell are:

2. Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.

Answer: The internal resistance of a cell depends on the following factors:

(1) The surface area of the electrodes – larger the surface area of electrodes, less is the internal resistance.
(2) The distance between the electrodes – more the distance between the electrodes, greater is the internal resistance.
(3) The nature and concentration of the electrolyte – less ionic the electrolyte or higher the concentration of the electrolyte, greater is the internal resistance.
(4) The temperature of the electrolyte – higher the temperature of the electrolyte, less is the internal resistance.

Two factors on which the internal resistance of a cell depends are:

(i) The surface area of the electrodes: Larger the surface area of electrodes, less is the internal resistance.
(ii) The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

3. A cell of e.m.f. ɛ and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell, (c) the p.d. across the cell, and (d) voltage drop inside the cell.

Answer: (a) The total resistance of the circuit is R + r.
(b) The current drawn from the cell, I, is given by I = ɛ / (R + r).
(c) The p.d. across the cell (terminal voltage), V, is given by V = ɛ – Ir (or V = IR).
(d) The voltage drop inside the cell, v, is given by v = Ir.

4. A cell is used to send current to an external circuit. (a) How does the voltage across its terminals compare with its e.m.f. ? (b) Under what condition is the e.m.f. of the cell equal to its terminal voltage?

Answer: (a) When current is drawn from a cell, the terminal voltage V across its terminals is less than its e.m.f. ɛ. When the cell is not in use (i.e., no current is drawn), the terminal voltage is equal to the e.m.f.

\(b) The e.m.f. of the cell is equal to its terminal voltage when no current is drawn from the cell, i.e., when the cell is in an open circuit.

5. The V-I graph for a series combination and for a parallel combination of two resistors is shown in Fig. 8.41. Which of the two, A or B, represents the parallel combination ? Give a reason for your answer.

Answer: Line A represents the parallel combination.

Reason: The slope of the V-I graph (V on y-axis, I on x-axis) gives the resistance (R = V/I). Line A has a smaller slope than line B, which means line A represents a smaller resistance. In a parallel combination, the equivalent resistance is less than in a series combination (and also less than the smallest individual resistance). Therefore, line A, which indicates smaller resistance, represents the parallel combination.

Long Answer Type Questions

1. Explain the meaning of the terms e.m.f., terminal voltage, and internal resistance of a cell.

Answer:

• e.m.f. (electro-motive force): When no current is drawn from a cell, i.e., when the cell is in open circuit, the potential difference produced by the chemical reaction between the terminals of the cell is called its electro-motive force (or e.m.f.). The e.m.f. of a cell is defined as the energy spent (or the work done) per unit charge in taking a positive test charge around the complete circuit of the cell (i.e., in the circuit outside the cell as well as in the electrolyte inside the cell).
• Terminal voltage: When current is drawn from a cell, i.e., when the cell is in a closed circuit, the potential difference between the electrodes of the cell is known as its terminal voltage. The terminal voltage of a cell is defined as the work done per unit charge in carrying a positive test charge around the external circuit connected across the terminals of the cell.
• Internal resistance: When a cell is in a closed circuit, the current flows inside the cell through its electrolyte. The resistance offered by the electrolyte inside the cell, to the flow of current, is called the internal resistance of the cell. It is the internal resistance of the cell due to which its voltage drops on drawing current from it.

2. Explain why is the p.d. across the terminals of a cell more in an open circuit and reduced in a closed circuit.

Answer: In an open circuit, no current is drawn from the cell. The potential difference (p.d.) across the terminals of the cell is then equal to its electro-motive force (e.m.f.).
In a closed circuit, when current (I) is drawn from the cell, there is a voltage drop (v = Ir) across the internal resistance (r) of the cell itself due to the flow of current through the electrolyte. The terminal voltage (V) is then given by V = ɛ – Ir, where ɛ is the e.m.f. Thus, the p.d. across the terminals (terminal voltage) is reduced from the e.m.f. by an amount equal to the voltage drop inside the cell (Ir).

3. How would you connect two resistors in series ? Draw a diagram. Calculate the total equivalent resistance.

Answer: To connect two resistors in series, they are joined one after the other, such that one end of the first resistor is connected to a point (say, A), its other end is connected to one end of the second resistor, and the other end of the second resistor is connected to another point (say, C). The battery is then connected across points A and C. The same current flows through each resistor.

The total equivalent resistance (Rs) of two resistors R₁ and R₂ connected in series is the sum of their individual resistances:
Rs = R₁ + R₂

4. Show by a diagram how two resistors R₁ and R₂ are joined in parallel. Obtain an expression for the total resistance of the combination.

Answer: To join two resistors R₁ and R₂ in parallel, one end of each resistor is connected together at one common point (say, A), and the other end of each resistor is connected together at another common point (say, B). The battery is then connected between these two common points A and B. The potential difference across each resistor is the same.The expression for the total equivalent resistance (Rp) of two resistors R₁ and R₂ connected in parallel is given by:

1/Rp = 1/R₁ + 1/R₂

Alternatively, this can be written as:
Rp = (R₁R₂) / (R₁ + R₂)

Numericals

1. The diagram shows a cell of e.m.f. ɛ = 4 volt and internal resistance r = 2 ohm connected to an external resistance R = 8 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, and (ii) the key K is closed.

Answer:

Given:

e.m.f. of the cell (ɛ) = 4 V
Internal resistance (r) = 2 Ω
External resistance (R) = 8 Ω

To find:

The readings of the ammeter (I) and voltmeter (V) in two cases:
(i) When the key K is open.
(ii) When the key K is closed.

Solution:

(i) When the key K is open:

When the key K is open, the circuit is incomplete or open. Therefore, no current flows through the circuit.

Ammeter Reading:
Since no current is flowing in the circuit, the ammeter will show a reading of zero.
=> I = 0 A.

Voltmeter Reading:
The voltmeter measures the terminal voltage (V) across the cell. The relationship between terminal voltage, e.m.f., and internal resistance is given by:
V = ɛ – Ir

Since the current (I) is 0 A:
=> V = 4 – (0 × 2)
=> V = 4 – 0
=> V = 4 V.

When no current is drawn from the cell, the terminal voltage is equal to the e.m.f. of the cell.

(ii) When the key K is closed:

When the key K is closed, the circuit is complete, and current will flow through it.

Ammeter Reading:
First, we calculate the total resistance of the circuit, which is the sum of the external resistance and the internal resistance.
Total Resistance = R + r
=> Total Resistance = 8 Ω + 2 Ω
=> Total Resistance = 10 Ω

Now, we can find the current (I) flowing in the circuit using the formula:
I = ɛ / (R + r)
=> I = 4 V / (10 Ω)
=> I = 0.4 A.
The ammeter will read 0.4 A.

Voltmeter Reading:
The voltmeter measures the terminal voltage (V) across the cell.
V = ɛ – Ir

Using the value of current I = 0.4 A:
=> V = 4 – (0.4 × 2)
=> V = 4 – 0.8
=> V = 3.2 V.

The voltmeter will read 3.2 V.

2. A battery of e.m.f. 6.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 3 A, the voltmeter reads 5.4 V. Find the internal resistance of the battery.

Answer:

Given:

e.m.f. (E) = 6.0 V
Current (I) = 3 A
Terminal voltage (V) = 5.4 V

To find:

Internal resistance (r) = ?

Solution:

The relationship between the electromotive force (e.m.f.), terminal voltage, current, and internal resistance is given by the formula:
V = E – Ir

To find the internal resistance (r), we can rearrange the formula:
=> Ir = E – V
=> r = (E – V) / I

Now, we substitute the given values into the formula:
=> r = (6.0 V – 5.4 V) / 3 A
=> r = 0.6 V / 3 A
=> r = 0.2 Ω

Therefore, the internal resistance of the battery is 0.2 Ω.

3. A cell of e.m.f. 1.8 V and internal resistance 2 Ω is connected in series with an ammeter of resistance 0.7 Ω and a resistor of 4.5 Ω as shown in Fig. 8.43.

(a) What would be the reading of the ammeter ?

Answer:

Given:

e.m.f. of the cell (ε) = 1.8 V
Internal resistance of the cell (r) = 2 Ω
Resistance of the ammeter (R_A) = 0.7 Ω
Resistance of the resistor (R) = 4.5 Ω

To find:

Reading of the ammeter (I) = ?

Solution:

Since the cell, ammeter, and resistor are all connected in series, the total resistance of the circuit is the sum of their individual resistances.

Total resistance (R_total) = Internal resistance + Ammeter resistance + Resistor
=> R_total = r + R_A + R
=> R_total = 2 Ω + 0.7 Ω + 4.5 Ω
=> R_total = 7.2 Ω

The current flowing through the circuit is given by Ohm’s law, which is the total e.m.f. divided by the total resistance. The ammeter will read this current.

Current (I) = Total e.m.f. / Total resistance
=> I = ε / R_total
=> I = 1.8 V / 7.2 Ω
=> I = 0.25 A

Therefore, the reading of the ammeter would be 0.25 A.

(b) What is the potential difference across the terminals of the cell ?

Answer:

Given:

e.m.f. of the cell (ε) = 1.8 V
Internal resistance of the cell (r) = 2 Ω
Current in the circuit (I) = 0.25 A (calculated in part a)

To find:

Potential difference across the terminals of the cell (V) = ?

Solution:

The potential difference (or terminal voltage) across the terminals of a cell when it is supplying current is given by the formula:

V = ε – I * r

This accounts for the potential drop across the internal resistance of the cell.

V = 1.8 V – (0.25 A × 2 Ω)
=> V = 1.8 V – 0.5 V
=> V = 1.3 V

Therefore, the potential difference across the terminals of the cell is 1.3 V.

4. The music system draws a current of 400 mA when connected to a 12 V battery. (a) What is the resistance of the music system ? (b) The music system if left playing for several hours and finally the battery voltage drops and the music system stops playing when the current drops to 320 mA. At what battery voltage does the music system stop playing ?

(a) What is the resistance of the music system ?

Answer:

Given:

Voltage (V) = 12 V
Current (I) = 400 mA

To find:

Resistance (R) = ?

Solution:

First, we convert the current from milliamperes (mA) to amperes (A).
We know that 1 A = 1000 mA.

=> I = 400 / 1000 A
=> I = 0.4 A

According to Ohm’s Law, the relationship between voltage, current, and resistance is given by the formula:
V = I × R

To find the resistance (R), we can rearrange the formula as:
R = V / I

Now, we substitute the given values into the formula:
=> R = 12 V / 0.4 A
=> R = 30 Ω

Therefore, the resistance of the music system is 30 Ω.

(b) The music system if left playing for several hours and finally the battery voltage drops and the music system stops playing when the current drops to 320 mA. At what battery voltage does the music system stop playing ?

Answer:

Given:

Resistance (R) = 30 Ω (from part a)
New Current (I’) = 320 mA

To find:

New Battery Voltage (V’) = ?

Solution:

We assume the resistance of the music system remains constant.
First, we convert the new current from milliamperes (mA) to amperes (A).

=> I’ = 320 / 1000 A
=> I’ = 0.32 A

Using Ohm’s Law again:
V’ = I’ × R

Now, we substitute the known values:
=> V’ = 0.32 A × 30 Ω
=> V’ = 9.6 V

Therefore, the music system stops playing when the battery voltage drops to 9.6 V.

5. A cell of e.m.f. ɛ and internal resistance r sends a current of 1.0 A when it is connected to an external resistance of 1.9 Ω. But it sends a current of 0.5 A when it is connected to an external resistance of 3.9 Ω. Calculate the values of ɛ and r.

Answer:

Given:

The information can be summarized in the following table:

To find:

E.M.F. (ɛ) = ?
Internal resistance (r) = ?

Solution:

The relationship between e.m.f. (ɛ), internal resistance (r), external resistance (R), and current (I) is given by the formula:
ɛ = I(R + r)

For Case 1:
We have, I₁ = 1.0 A and R₁ = 1.9 Ω.
Substituting these values into the formula:
=> ɛ = 1.0(1.9 + r)
=> ɛ = 1.9 + r —(i)

For Case 2:
We have, I₂ = 0.5 A and R₂ = 3.9 Ω.
Substituting these values into the formula:
=> ɛ = 0.5(3.9 + r)
=> ɛ = 1.95 + 0.5r —(ii)

Now, we have two equations for ɛ. By equating equation (i) and equation (ii), we can solve for r:
=> 1.9 + r = 1.95 + 0.5r
=> r – 0.5r = 1.95 – 1.9
=> 0.5r = 0.05
=> r = 0.05 / 0.5
=> r = 0.1 Ω

Now, substitute the value of r back into equation (i) to find ɛ:
=> ɛ = 1.9 + r
=> ɛ = 1.9 + 0.1
=> ɛ = 2.0 V

Thus, the e.m.f. of the cell is 2.0 V and the internal resistance is 0.1 Ω.

6. Two resistors having resistance 8 Ω and 12 Ω are connected in parallel. Find their equivalent resistance.

Answer:

Given:

Resistance of the first resistor, R₁ = 8 Ω
Resistance of the second resistor, R₂ = 12 Ω

To find:

The equivalent resistance of the parallel combination, R_p.

Solution:

For two resistances R₁ and R₂ connected in parallel, the equivalent resistance R_p is given by the formula:

R_p = (R₁ × R₂) / (R₁ + R₂)

Substituting the given values into the formula:

=> R_p = (8 × 12) / (8 + 12)
=> R_p = 96 / 20
=> R_p = 4.8 Ω

Therefore, the equivalent resistance of the parallel combination is 4.8 Ω.

7. Four resistors each of resistance 5 Ω are connected in parallel. What is the effective resistance ?

Answer:

Given:

Number of resistors, n = 4
Resistance of each resistor, R = 5 Ω

To find:

The effective resistance of the parallel combination, R_p.

Solution:

When n equal resistances, each of value R, are connected in parallel, the equivalent resistance R_p is given by the formula (8.34):

R_p = R / n

Substituting the given values into the formula:

=> R_p = 5 / 4
=> R_p = 1.25 Ω

Therefore, the effective resistance of the parallel combination is 1.25 Ω.

8. You have three resistors of values 2 Ω, 3 Ω, and 5 Ω. How will you join them so that the total resistance is less than 1 Ω? Draw a diagram and find the total resistance.

Answer: Given:

Resistance of the first resistor, R₁ = 2 Ω
Resistance of the second resistor, R₂ = 3 Ω
Resistance of the third resistor, R₃ = 5 Ω

To Find:

The arrangement of resistors that results in a total equivalent resistance of less than 1 Ω.

Solution:

Reasoning:

• Series Combination: The equivalent resistance is the sum of individual resistances (R_s = R₁ + R₂ + R₃). This always results in a total resistance that is greater than the largest individual resistance. In this case, R_s = 2 + 3 + 5 = 10 Ω, which is not less than 1 Ω.
• Parallel Combination: In the parallel combination, the equivalent resistance R_p is less than even the smallest resistance connected.

The smallest resistance we have is 2 Ω. Therefore, connecting all three resistors in parallel is the only configuration that guarantees a total resistance less than 2 Ω, and it is the method we must use to try and achieve a resistance less than 1 Ω.

Calculation:

We will connect the three resistors in parallel. The equivalent resistance R_p is calculated as:

1 / R_p = 1 / R₁ + 1 / R₂ + 1 / R₃

Substituting the given values:

=> 1 / R_p = 1/2 + 1/3 + 1/5

To add these fractions, we find a common denominator, which is 30.

=> 1 / R_p = (15/30) + (10/30) + (6/30)
=> 1 / R_p = (15 + 10 + 6) / 30
=> 1 / R_p = 31 / 30

Now, we find R_p by taking the reciprocal:

=> R_p = 30 / 31 Ω

Converting this to a decimal:

=> R_p ≈ 0.97 Ω

Since 0.97 Ω is less than 1 Ω, this arrangement is correct.

9. Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.

Answer: Given:

Three resistors, R₁, R₂, and R₃, each with a resistance of 2 Ω.
The desired total resistance is 3 Ω.

To Find:

The circuit arrangement that provides a total resistance of 3 Ω.

Solution:

We need to find a combination of series and parallel connections to achieve the target resistance of 3 Ω. L

Step 1: Calculate the equivalent resistance of the two resistors in parallel (R_p).

Using the formula for two parallel resistors:
R_p = (R₁ × R₂) / (R₁ + R₂)

=> R_p = (2 × 2) / (2 + 2)
=> R_p = 4 / 4
=> R_p = 1 Ω

Step 2: Calculate the total resistance by adding the series resistor.

The parallel combination (R_p = 1 Ω) is in series with the third resistor (R₃ = 2 Ω). The total resistance (R_total) is the sum:
R_total = R_p + R₃

=> R_total = 1 Ω + 2 Ω
=> R_total = 3 Ω

This matches the desired total resistance.

10. Calculate the equivalent resistance between the points A and B in the figure if each resistance is 2.0 Ω.

Answer: Answer:

Given:

A circuit diagram with four resistors connected between points A and B.
The resistance of each resistor is 2.0 Ω.

To find:

The equivalent resistance between points A and B (R_AB).

Solution:

This problem involves a combination of resistors in series and parallel. We will solve it by simplifying the circuit in parts.

Step 1: Calculate the equivalent resistance of the initial series combination.

The first two resistors from point A are connected in series. Let their equivalent resistance be R_s.
According to the formula for series combination (R_s = R₁ + R₂):

=> R_s = 2.0 Ω + 2.0 Ω
=> R_s = 4.0 Ω

Step 2: Calculate the equivalent resistance of the parallel combination.

The next two resistors are connected in parallel. Let their equivalent resistance be R_p.
According to the formula for two parallel resistors (R_p = (R₁ × R₂) / (R₁ + R₂)):

=> R_p = (2.0 × 2.0) / (2.0 + 2.0)
=> R_p = 4.0 / 4.0
=> R_p = 1.0 Ω

Step 3: Calculate the total equivalent resistance between A and B.

The circuit has now been simplified to the series combination (R_s) connected in series with the parallel combination (R_p). The total equivalent resistance R_AB is the sum of these two parts.

=> R_AB = R_s + R_p
=> R_AB = 4.0 Ω + 1.0 Ω
=> R_AB = 5.0 Ω

Therefore, the equivalent resistance between the points A and B is 5.0 Ω.

11. A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.

Answer: Given:

Resistance of each individual resistor = 2 Ω
Number of resistors in each series set = 3
Number of similar sets connected in parallel = 4

To find:

The total equivalent resistance of the entire combination.

Solution:

This problem involves a combination of resistors in both series and parallel. The strategy is to first find the resistance of each series branch (set) and then find the equivalent resistance of these branches connected in parallel.

Step 1: Calculate the resistance of one series set.

Each set consists of three 2 Ω resistors connected in series. According to the formula for series combination (Equation 8.30 on page 19), the equivalent resistance of one set (R_set) is the sum of the individual resistances.

R_set = R₁ + R₂ + R₃
=> R_set = 2 Ω + 2 Ω + 2 Ω
=> R_set = 6 Ω

So, the resistance of each of the four similar sets is 6 Ω.

Step 2: Calculate the total resistance of the four sets connected in parallel.

We now have four identical sets, each with a resistance of 6 Ω, connected in parallel. According to the formula for parallel combination of ‘n’ equal resistances (Equation 8.34 on page 20), the total equivalent resistance (R_total) is given by:

R_total = R_set / n

Here, R_set = 6 Ω and n = 4.

=> R_total = 6 / 4 Ω
=> R_total = 1.5 Ω

Therefore, the total resistance of the combination is 1.5 Ω.

12. In the circuit shown below, calculate the value of x if the equivalent resistance between the points A and B is 4 Ω.

Answer: Answer:

Given:

The circuit diagram shows two parallel branches connected between points A and B.
Equivalent resistance between A and B, R_eq = 4 Ω

In the upper branch, two resistors are connected in series:
R_upper1 = 8 Ω
R_upper2 = 4 Ω

In the lower branch, two resistors are connected in series:
R_lower1 = 5 Ω
R_lower2 = x Ω

To find:

The value of x.

Solution:

First, we calculate the equivalent resistance of the upper branch. Since the 8 Ω and 4 Ω resistors are in series, their equivalent resistance (let’s call it R_upper) is their sum.

R_upper = R_upper1 + R_upper2
=> R_upper = 8 + 4
=> R_upper = 12 Ω

Next, we calculate the equivalent resistance of the lower branch. Since the 5 Ω and x Ω resistors are in series, their equivalent resistance (let’s call it R_lower) is their sum.

R_lower = R_lower1 + R_lower2
=> R_lower = (5 + x) Ω

Now, the upper branch (R_upper) and the lower branch (R_lower) are connected in parallel between points A and B. The total equivalent resistance (R_eq) is given by the formula for parallel resistors:

1 / R_eq = 1 / R_upper + 1 / R_lower

We are given that R_eq = 4 Ω. Substituting the values into the equation:

1 / 4 = 1 / 12 + 1 / (5 + x)

To solve for x, we first isolate the term containing x:

1 / (5 + x) = 1 / 4 – 1 / 12

Find a common denominator for the right side (which is 12):

1 / (5 + x) = (3 / 12) – (1 / 12)
=> 1 / (5 + x) = (3 – 1) / 12
=> 1 / (5 + x) = 2 / 12

Simplify the fraction on the right side:

1 / (5 + x) = 1 / 6

By inverting both sides (or cross-multiplying), we get:

5 + x = 6

Now, solve for x:

x = 6 – 5
=> x = 1 Ω

Therefore, the value of x is 1 Ω.

13. Calculate the effective resistance between the points A and B in the circuit shown.

Answer: To find the effective resistance between points A and B, we need to simplify the given circuit by identifying series and parallel combinations of resistors.

Let’s label the junctions in the circuit for clarity. Let the junction after the first 1 Ω resistor from A be C, and the junction before the last 1 Ω resistor to B be F.

The circuit between points C and F consists of three parallel branches:

1. The upper branch: It has three 1 Ω resistors connected in series.
2. The middle branch: It has a single 2 Ω resistor.
3. The lower branch: It has three 2 Ω resistors connected in series.

Resistance of the upper and lower branches.

For resistors in series, the equivalent resistance is the sum of the individual resistances (R_s = R₁ + R₂ + R₃ + …).

Resistance of the upper branch (R_upper):
=> R_upper = 1 Ω + 1 Ω + 1 Ω
=> R_upper = 3 Ω

Resistance of the lower branch (R_lower):
=> R_lower = 2 Ω + 2 Ω + 2 Ω
=> R_lower = 6 Ω

Equivalent resistance of the parallel combination between C and F.

Now, the circuit between C and F has three resistors in parallel: R_upper = 3 Ω, R_middle = 2 Ω, and R_lower = 6 Ω.

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances (1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + …).

Let the equivalent resistance between C and F be R_CF.
=> 1/R_CF = 1/R_upper + 1/R_middle + 1/R_lower
=> 1/R_CF = 1/3 + 1/2 + 1/6

To add these fractions, we find a common denominator, which is 6.
=> 1/R_CF = 2/6 + 3/6 + 1/6
=> 1/R_CF = (2 + 3 + 1) / 6
=> 1/R_CF = 6/6
=> 1/R_CF = 1
=> R_CF = 1 Ω

Total effective resistance between A and B.

The entire circuit is now simplified to three resistors connected in series:

• The 1 Ω resistor between A and C.
• The equivalent resistance R_CF = 1 Ω.
• The 1 Ω resistor between F and B.

The total effective resistance between A and B (R_AB) is the sum of these series resistances.
=> R_AB = R_AC + R_CF + R_FB
=> R_AB = 1 Ω + 1 Ω + 1 Ω
=> R_AB = 3 Ω

Thus, the effective resistance between the points A and B is 3 Ω.

14. A uniform wire with a resistance of 27 Ω is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.

Answer:

Given:

Resistance of the uniform wire, R = 27 Ω
Number of equal pieces = 3

To find:

The equivalent resistance of the parallel combination, Rₚ = ?

Solution:

First, we need to find the resistance of each of the three equal pieces. Since the wire is uniform, its resistance is directly proportional to its length. When the wire is divided into three equal pieces, the resistance of each piece will be one-third of the total resistance.

Let the resistance of each piece be R’.

R’ = Total Resistance / Number of pieces
=> R’ = 27 Ω / 3
=> R’ = 9 Ω

So, we have three resistors, each with a resistance of 9 Ω. Let’s call them R₁, R₂, and R₃.
R₁ = 9 Ω
R₂ = 9 Ω
R₃ = 9 Ω

Now, these three resistors are connected in parallel. The formula for the equivalent resistance (Rₚ) of a parallel combination is:
1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

Substituting the values:
=> 1/Rₚ = 1/9 + 1/9 + 1/9
=> 1/Rₚ = (1 + 1 + 1) / 9
=> 1/Rₚ = 3 / 9
=> 1/Rₚ = 1 / 3

To find Rₚ, we take the reciprocal:
=> Rₚ = 3 Ω

Thus, the equivalent resistance of the parallel combination is 3 Ω.

15. A circuit consists of a resistor of 1 ohm in series with a parallel arrangement of resistors of 12 ohm and 6 ohm. Calculate the total resistance of the circuit. Draw a diagram of the arrangement.

Answer:

Given:

A resistor R₁ = 1 Ω is connected in series.

Two resistors R₂ = 12 Ω and R₃ = 6 Ω are connected in parallel with each other.

To find:

The total resistance of the circuit (R_total).

Solution:

First, we will find the equivalent resistance (R_p) of the parallel combination of resistors R₂ and R₃.

The formula for the equivalent resistance of a parallel combination is:
1/R_p = 1/R₂ + 1/R₃

Substituting the given values:
1/R_p = 1/12 + 1/6

To add the fractions, we find a common denominator, which is 12.
1/R_p = (1 + 2) / 12
1/R_p = 3 / 12
1/R_p = 1 / 4

Therefore, the equivalent resistance of the parallel part is:
R_p = 4 Ω

Now, this equivalent resistance R_p is in series with the resistor R₁. The total resistance of the circuit is the sum of the resistances in series.

The formula for the equivalent resistance of a series combination is:
R_total = R₁ + R_p

Substituting the values:
R_total = 1 Ω + 4 Ω
R_total = 5 Ω

Thus, the total resistance of the circuit is 5 Ω.

16. Calculate the effective resistance between the points A and B in the network shown below.

Answer:

Given:

A circuit diagram with the following resistors:

• A 2 Ω resistor connected to a junction.
• Three resistors of 12 Ω, 6 Ω, and 4 Ω connected in parallel.
• A 5 Ω resistor connected after the parallel combination.

To find:

The effective resistance between points A and B (R_AB).

Solution:

First, we will find the equivalent resistance of the parallel combination. The resistors of 12 Ω, 6 Ω, and 4 Ω are connected in parallel. Let the equivalent resistance of this combination be Rp.

Using the formula for resistors in parallel:
1 / Rp = 1 / R1 + 1 / R2 + 1 / R3
=> 1 / Rp = 1 / 12 + 1 / 6 + 1 / 4

To add these fractions, we find the least common multiple (LCM) of the denominators 12, 6, and 4, which is 12.
=> 1 / Rp = 1/12 + 2/12 + 3/12
=> 1 / Rp = (1 + 2 + 3) / 12
=> 1 / Rp = 6 / 12
=> 1 / Rp = 1 / 2

Therefore, the equivalent resistance of the parallel part is:
Rp = 2 Ω

Now, the circuit is simplified to three resistors connected in series: the initial 2 Ω resistor, the equivalent resistance of the parallel part (Rp = 2 Ω), and the final 5 Ω resistor.

The total effective resistance between points A and B (R_AB) is the sum of these resistances in series.

Using the formula for resistors in series:
R_AB = R_a + Rp + R_b
=> R_AB = 2 Ω + 2 Ω + 5 Ω
=> R_AB = 9 Ω

Thus, the effective resistance between the points A and B is 9 Ω.

17. Calculate the equivalent resistance between the points A and B in the figure.

Answer: Given:

The circuit diagram shows a network of five resistors connected between points A and B.
Let’s name the resistors:
R₁ = 3 Ω
R₂ = 2 Ω
R₃ = 6 Ω
R₄ = 4 Ω
R₅ = 30 Ω

To find:

The equivalent resistance between points A and B (R_AB).

Solution:

The given arrangement of resistors is a Wheatstone bridge.

First, we check the condition for a balanced bridge, which is:
R₁ / R₃ = R₂ / R₄

Substituting the given values:
3 / 6 = 1/2
2 / 4 = 1/2

Since R₁ / R₃ = R₂ / R₄, the bridge is balanced.

When the bridge is balanced, no current flows through the resistor connected between the two branches (R₅ = 30 Ω). Therefore, this resistor can be removed from the circuit for calculation.

The simplified circuit consists of two parallel branches:

1. The upper branch with resistors R₁ (3 Ω) and R₂ (2 Ω) in series.
2. The lower branch with resistors R₃ (6 Ω) and R₄ (4 Ω) in series.

Step 1: Calculate the equivalent resistance of the upper branch (R_upper).

Since resistors R₁ and R₂ are in series, their equivalent resistance is the sum of the individual resistances.
R_upper = R₁ + R₂
=> R_upper = 3 Ω + 2 Ω
=> R_upper = 5 Ω

Step 2: Calculate the equivalent resistance of the lower branch (R_lower).

Since resistors R₃ and R₄ are in series, their equivalent resistance is the sum of the individual resistances.
R_lower = R₃ + R₄
=> R_lower = 6 Ω + 4 Ω
=> R_lower = 10 Ω

Step 3: Calculate the total equivalent resistance between A and B (R_AB).

The upper branch (R_upper) and the lower branch (R_lower) are connected in parallel. The equivalent resistance (R_AB) is given by the formula for parallel combination:
1 / R_AB = 1 / R_upper + 1 / R_lower
=> 1 / R_AB = 1/5 + 1/10

To add the fractions, we find a common denominator, which is 10.
=> 1 / R_AB = (2 + 1) / 10
=> 1 / R_AB = 3 / 10

Therefore, the equivalent resistance is:
=> R_AB = 10 / 3 Ω or 3.33 Ω.

18. In the network shown in the figure, calculate the equivalent resistance between the points (a) A and B, and (b) A and C.

Answer: Given:
The circuit shows four resistors, each of resistance 2 Ω, connected to form a square ABCD.
Resistance in arm AB = 2 Ω
Resistance in arm BC = 2 Ω
Resistance in arm CD = 2 Ω
Resistance in arm DA = 2 Ω

To find:
Equivalent resistance between points A and B.

Solution:
To find the equivalent resistance between points A and B, we can see that the current entering at A will divide into two paths to reach B.
Path 1: The direct path through the arm AB.
Path 2: The path through the arms AD, DC, and CB.

First, we consider Path 2. The resistors in the arms AD, DC, and CB are connected in series. The equivalent resistance of this branch (Rs) is:
Rs = Resistance (AD) + Resistance (DC) + Resistance (CB)
=> Rs = 2 Ω + 2 Ω + 2 Ω
=> Rs = 6 Ω

Now, this series combination of resistance 6 Ω is connected in parallel with the resistor in arm AB (2 Ω).
If the equivalent resistance between A and B is Rp, then:
1 / Rp = 1 / Rs + 1 / R_AB
=> 1 / Rp = 1 / 6 + 1 / 2
=> 1 / Rp = (1 + 3) / 6
=> 1 / Rp = 4 / 6
=> 1 / Rp = 2 / 3
=> Rp = 3 / 2 Ω
=> Rp = 1.5 Ω

The equivalent resistance between points A and B is 1.5 Ω.

(b) Find the equivalent resistance between the points A and C.

Answer:

Given:
The same circuit with four resistors, each of resistance 2 Ω, connected to form a square ABCD.

To find:
Equivalent resistance between points A and C.

Solution:
To find the equivalent resistance between points A and C, we can see that the current entering at A will divide into two paths to reach C.
Path 1: The path through the arms AD and DC.
Path 2: The path through the arms AB and BC.

In Path 1, the resistors in arms AD and DC are connected in series. The equivalent resistance of this branch (Rs1) is:
Rs1 = Resistance (AD) + Resistance (DC)
=> Rs1 = 2 Ω + 2 Ω
=> Rs1 = 4 Ω

In Path 2, the resistors in arms AB and BC are connected in series. The equivalent resistance of this branch (Rs2) is:
Rs2 = Resistance (AB) + Resistance (BC)
=> Rs2 = 2 Ω + 2 Ω
=> Rs2 = 4 Ω

Now, these two series combinations (Rs1 and Rs2) are connected in parallel between the points A and C.
If the equivalent resistance between A and C is Rp, then:
1 / Rp = 1 / Rs1 + 1 / Rs2
=> 1 / Rp = 1 / 4 + 1 / 4
=> 1 / Rp = 2 / 4
=> 1 / Rp = 1 / 2
=> Rp = 2 Ω

The equivalent resistance between points A and C is 2 Ω.

19. Five resistors, each of 3 Ω, are connected as shown in the figure. Calculate the resistance (a) between the points P and Q, and (b) between the points X and Y.

Answer:

Given:
A combination of resistors is connected between points P and Q.

• Two resistors of 3 Ω each are connected in series in the upper branch.
• One resistor of 3 Ω is connected in the lower branch.
• These two branches are connected in parallel between points P and Q.

To find:
The equivalent resistance between points P and Q (R_PQ).

Solution:
First, we calculate the equivalent resistance of the upper branch (R_s) where two 3 Ω resistors are connected in series.
R_s = 3 Ω + 3 Ω = 6 Ω

Now, this series combination (R_s) is connected in parallel with the lower branch resistor (3 Ω). The equivalent resistance between P and Q (R_PQ) can be calculated as follows:
1 / R_PQ = 1 / R_s + 1 / 3
=> 1 / R_PQ = 1 / 6 + 1 / 3
To find the sum, we take the least common multiple (LCM) of the denominators, which is 6.
=> 1 / R_PQ = (1 + 2) / 6
=> 1 / R_PQ = 3 / 6
=> 1 / R_PQ = 1 / 2
=> R_PQ = 2 Ω

Thus, the resistance between the points P and Q is 2 Ω.

(b) Calculate the resistance between the points X and Y.

Answer:

Given:

• Resistance between X and P (R_XP) = 3 Ω
• Equivalent resistance between P and Q (R_PQ) = 2 Ω (as calculated in part (a))
• Resistance between Q and Y (R_QY) = 3 Ω

To find:
The equivalent resistance between points X and Y (R_XY).

Solution:
The entire circuit from X to Y consists of the resistor R_XP, the equivalent resistance R_PQ, and the resistor R_QY, all connected in series.

The total equivalent resistance between X and Y is the sum of these individual resistances.
R_XY = R_XP + R_PQ + R_QY
=> R_XY = 3 Ω + 2 Ω + 3 Ω
=> R_XY = 8 Ω

Thus, the resistance between the points X and Y is 8 Ω.

20. Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

Answer: (a) When the resistors are connected in series

Given:
Resistance of first resistor, R₁ = 4.0 Ω
Resistance of second resistor, R₂ = 6.0 Ω
Potential difference of the battery, V = 6.0 V

To find:
(i) The circuit diagram.
(ii) The current through the battery, I.

Solution:
(i) The circuit diagram for the series combination is as follows:

(ii) First, we calculate the total equivalent resistance of the circuit.
In a series combination, the equivalent resistance (Rₛ) is the sum of the individual resistances.
Rₛ = R₁ + R₂
=> Rₛ = 4.0 Ω + 6.0 Ω
=> Rₛ = 10.0 Ω

Now, we calculate the current flowing through the battery using Ohm’s law.
Current (I) = Potential Difference (V) / Equivalent Resistance (Rₛ)
=> I = 6.0 V / 10.0 Ω
=> I = 0.6 A

Therefore, the current through the battery in the series circuit is 0.6 A.

(b) When the resistors are connected in parallel

Given:
Resistance of first resistor, R₁ = 4.0 Ω
Resistance of second resistor, R₂ = 6.0 Ω
Potential difference of the battery, V = 6.0 V

To find:
(i) The circuit diagram.
(ii) The current through the battery, I.

Solution:
(i) The circuit diagram for the parallel combination is as follows:

(ii) First, we calculate the total equivalent resistance of the circuit.
In a parallel combination, the reciprocal of the equivalent resistance (Rₚ) is the sum of the reciprocals of the individual resistances.
1 / Rₚ = 1 / R₁ + 1 / R₂
=> 1 / Rₚ = 1 / 4.0 + 1 / 6.0
To add the fractions, we find a common denominator, which is 12.0.
=> 1 / Rₚ = (3 + 2) / 12.0
=> 1 / Rₚ = 5 / 12.0
=> Rₚ = 12.0 / 5 Ω
=> Rₚ = 2.4 Ω

Now, we calculate the current flowing through the battery using Ohm’s law.
Current (I) = Potential Difference (V) / Equivalent Resistance (Rₚ)
=> I = 6.0 V / 2.4 Ω
=> I = 2.5 A

Therefore, the current through the battery in the parallel circuit is 2.5 A.

21. A resistor of 6 Ω is connected in series with another resistor of 4 Ω. A potential difference of 20 V is applied across the combination. Calculate : (a) the current in the circuit, and (b) the potential difference across the 6 Ω resistor.

Answer: Given:

Resistance of the first resistor, R₁ = 6 Ω
Resistance of the second resistor, R₂ = 4 Ω
Total potential difference applied, V = 20 V
The resistors are connected in series.

To find:

The current in the circuit, I = ?

Solution:

When resistors are connected in series, the total equivalent resistance (Rs) is the sum of the individual resistances.
Rs = R₁ + R₂
=> Rs = 6 Ω + 4 Ω
=> Rs = 10 Ω

According to Ohm’s Law, the current in the circuit is given by:
I = V / Rs
=> I = 20 V / 10 Ω
=> I = 2 A

Therefore, the current in the circuit is 2 A.

(b) Find the potential difference across the 6 Ω resistor.

Answer:

Given:

Current in the circuit, I = 2 A (calculated in part a)
Resistance of the resistor, R₁ = 6 Ω

To find:

Potential difference across the 6 Ω resistor, V₁ = ?

Solution:

In a series combination, the current flowing through each resistor is the same as the total current in the circuit.
So, the current through the 6 Ω resistor is I = 2 A.

Using Ohm’s Law to find the potential difference (V₁) across the 6 Ω resistor:
V₁ = I × R₁
=> V₁ = 2 A × 6 Ω
=> V₁ = 12 V

Therefore, the potential difference across the 6 Ω resistor is 12 V.

22. Two resistors of resistance 2 Ω and 3 Ω are connected in parallel to a cell to draw current 0.5 A from the cell. (a) Draw a labelled diagram of the arrangement. (b) Calculate the current in each resistor.

Answer:

(a)

(b) Given:

Resistance of the first resistor, R₁ = 2 Ω
Resistance of the second resistor, R₂ = 3 Ω
Total current drawn from the cell, I = 0.5 A
The resistors are connected in parallel.

To find:

Current through the first resistor, I₁ = ?
Current through the second resistor, I₂ = ?

Solution:

First, we calculate the equivalent resistance (R_p) of the parallel combination. For two resistors in parallel, the formula is:
R_p = (R₁ × R₂) / (R₁ + R₂)

=> R_p = (2 × 3) / (2 + 3)
=> R_p = 6 / 5
=> R_p = 1.2 Ω

Now, we can find the potential difference (V) across the parallel combination using Ohm’s law. Since the potential difference is the same across both resistors, let’s call it V.
V = I × R_p
=> V = 0.5 A × 1.2 Ω
=> V = 0.6 V

The potential difference across both the 2 Ω resistor and the 3 Ω resistor is 0.6 V.

Now, we can calculate the current through each resistor using Ohm’s law (I = V / R).

Current through the first resistor (R₁):
I₁ = V / R₁
=> I₁ = 0.6 V / 2 Ω
=> I₁ = 0.3 A

Current through the second resistor (R₂):
I₂ = V / R₂
=> I₂ = 0.6 V / 3 Ω
=> I₂ = 0.2 A

Thus, the current through the 2 Ω resistor is 0.3 A, and the current through the 3 Ω resistor is 0.2 A.

23. Calculate the current flowing through each of the resistors A and B in the circuit shown.

Answer:

Given:

Resistance of resistor A (R_A) = 1 Ω
Resistance of resistor B (R_B) = 2 Ω
Potential difference of the battery (V) = 2 V

The resistors A and B are connected in a parallel combination across the battery.

To find:

Current flowing through resistor A (I_A) = ?
Current flowing through resistor B (I_B) = ?

Solution:

In a parallel combination of resistors, the potential difference (voltage) across each resistor is the same and is equal to the voltage supplied by the battery.

Therefore,
Potential difference across resistor A (V_A) = 2 V
Potential difference across resistor B (V_B) = 2 V

According to Ohm’s Law, the relationship between potential difference (V), current (I), and resistance (R) is given by:
V = IR
or
I = V / R

First, we will calculate the current flowing through resistor A (I_A).

I_A = V_A / R_A
=> I_A = 2 V / 1 Ω
=> I_A = 2 A

Next, we will calculate the current flowing through resistor B (I_B).

I_B = V_B / R_B
=> I_B = 2 V / 2 Ω
=> I_B = 1 A

Thus, the current flowing through resistor A is 2 A and the current flowing through resistor B is 1 A.

24. In the figure, calculate : (a) the total resistance of the circuit, (b) the value of R, and (c) the current flowing in R.

Answer:

Given:

Potential difference across the parallel combination (from voltmeter reading), V = 4 V
Total current flowing through the circuit (from ammeter reading), I = 0.4 A
Resistance of one parallel branch, R₁ = 20 Ω

(a) Find the total resistance of the circuit.

To find:

Total resistance, R_total = ?

Solution:

According to Ohm’s law, the total resistance of the circuit is the ratio of the potential difference across the circuit to the total current flowing through it.

R_total = V / I
=> R_total = 4 V / 0.4 A
=> R_total = 10 Ω

Thus, the total resistance of the circuit is 10 Ω.

(b) Find the value of R.

To find:

Resistance, R = ?

Solution:

The resistors R₁ (20 Ω) and R are connected in parallel. For a parallel combination, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.

1 / R_total = 1 / R₁ + 1 / R
Substituting the known values:
=> 1 / 10 = 1 / 20 + 1 / R
=> 1 / R = 1 / 10 – 1 / 20
To subtract the fractions, we find a common denominator, which is 20.
=> 1 / R = (2 / 20) – (1 / 20)
=> 1 / R = (2 – 1) / 20
=> 1 / R = 1 / 20
=> R = 20 Ω

Thus, the value of R is 20 Ω.

(c) Find the current flowing in R.

To find:

Current flowing in R, I_R = ?

Solution:

In a parallel circuit, the potential difference across each branch is the same as the total potential difference across the combination. So, the potential difference across resistor R is 4 V.

Applying Ohm’s law to the resistor R:
I_R = V / R
Substituting the values V = 4 V and R = 20 Ω (from part b):
=> I_R = 4 V / 20 Ω
=> I_R = 0.2 A

Thus, the current flowing in R is 0.2 A.

25. A particular resistance wire has a resistance of 3.0 ohm per metre. Find : (a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel. (b) The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that the resistance of battery is negligible). (c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.

Answer:

(a) Given:

Resistance per metre of the wire = 3.0 Ω/m
Length of each wire (l) = 1.5 m
Number of wires = 3
Arrangement = Parallel

To find:

The total (equivalent) resistance of the parallel combination (Rp).

Solution:

First, we calculate the resistance of a single 1.5 m long wire.
Resistance of one wire (R) = Resistance per metre × length
=> R = 3.0 Ω/m × 1.5 m
=> R = 4.5 Ω

Now, three such wires (R₁, R₂, R₃) are connected in parallel, where R₁ = R₂ = R₃ = 4.5 Ω.
For a parallel combination, the equivalent resistance (Rp) is given by the formula:
1/Rp = 1/R₁ + 1/R₂ + 1/R₃
=> 1/Rp = 1/4.5 + 1/4.5 + 1/4.5
=> 1/Rp = 3 / 4.5
=> Rp = 4.5 / 3
=> Rp = 1.5 Ω

The total resistance is 1.5 Ω.

(b) Given:

The three wires are connected in parallel.
Current in each wire (I_each) = 2.0 A
Resistance of each wire (R) = 4.5 Ω (from part a)
Internal resistance of the battery is negligible.

To find:

The potential difference of the battery (V).

Solution:

In a parallel circuit, the potential difference across each branch is the same and is equal to the potential difference of the battery.
Using Ohm’s Law (V = I × R) for any one of the wires:
V = I_each × R
=> V = 2.0 A × 4.5 Ω
=> V = 9.0 V

The potential difference of the battery is 9.0 V.

(c) Given:

Initial wire has a resistance of 3.0 Ω per metre.
Length of the new wire (l’) = 5 m
The new wire is made of the same material, but its area of cross-section (a’) is twice the original area (a).
i.e., a’ = 2a.

To find:

The resistance of the new wire (R’).

Solution:

The resistance (R) of a wire is given by the formula:
R = ρ (l/a)
where ρ is the specific resistance (resistivity), l is the length, and a is the area of cross-section.

For the original wire, we are given that its resistance is 3.0 Ω for a length of 1 m.
Let the original area be ‘a’.
R₁ = 3.0 Ω for l₁ = 1 m.
So, 3.0 = ρ (1/a)
=> ρ/a = 3.0 Ω

Now, for the new wire:
Length (l₂) = 5 m
Area (a₂) = 2a
The resistance of the new wire (R₂) is:
R₂ = ρ (l₂/a₂)
=> R₂ = ρ (5 / 2a)
=> R₂ = (5/2) × (ρ/a)
Substituting the value of (ρ/a) from the original wire:
=> R₂ = (5/2) × 3.0
=> R₂ = 2.5 × 3.0
=> R₂ = 7.5 Ω

The resistance of the new wire is 7.5 Ω.

26. A cell supplies a current of 1.2 A through two resistors each of 2 Ω connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate : (i) the internal resistance, and (ii) e.m.f. of the cell.

Answer:

Given:

Resistance of each resistor, R_each = 2 Ω

Case 1: Resistors in Parallel

Current, I₁ = 1.2 A
Two resistors of 2 Ω are connected in parallel.

Case 2: Resistors in Series

Current, I₂ = 0.4 A
Two resistors of 2 Ω are connected in series.

Let the internal resistance of the cell be ‘r’ and its e.m.f. be ‘ε’.

To find:

(i) The internal resistance, r = ?
(ii) The e.m.f. of the cell, ε = ?

Solution:

First, we need to find the equivalent external resistance in both cases.

For Case 1 (Parallel Combination):

The equivalent resistance (R_p) for two resistors connected in parallel is given by:
1 / R_p = 1 / R_each + 1 / R_each
=> 1 / R_p = 1 / 2 + 1 / 2
=> 1 / R_p = 2 / 2 = 1
=> R_p = 1 Ω

Using the formula for a cell, ε = I (R + r):
=> ε = I₁ (R_p + r)
=> ε = 1.2 (1 + r) ………. (i)

For Case 2 (Series Combination):

The equivalent resistance (R_s) for two resistors connected in series is given by:
R_s = R_each + R_each
=> R_s = 2 + 2
=> R_s = 4 Ω

Using the formula for a cell, ε = I (R + r):
=> ε = I₂ (R_s + r)
=> ε = 0.4 (4 + r) ………. (ii)

(i) To find the internal resistance (r):

Since the e.m.f. (ε) of the cell is the same in both cases, we can equate equations (i) and (ii):
From eqns. (i) and (ii),
1.2 (1 + r) = 0.4 (4 + r)
=> 1.2 + 1.2r = 1.6 + 0.4r
=> 1.2r – 0.4r = 1.6 – 1.2
=> 0.8r = 0.4
=> r = 0.4 / 0.8
=> r = 0.5 Ω

The internal resistance of the cell is 0.5 Ω.

(ii) To find the e.m.f. (ε):

Substitute the value of r = 0.5 Ω into equation (i):
ε = 1.2 (1 + r)
=> ε = 1.2 (1 + 0.5)
=> ε = 1.2 (1.5)
=> ε = 1.8 V

The e.m.f. of the cell is 1.8 V.

27. A battery of e.m.f. 16 V and internal resistance 2 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find : (a) the current through the battery, (b) the p.d. between the terminals of the battery, (c) the current in 3 Ω resistor, (d) the current in 6 Ω resistor.

Answer:

Given:

e.m.f. of the battery, ε = 16 V
Internal resistance, r = 2 Ω
Resistance 1, R₁ = 3 Ω
Resistance 2, R₂ = 6 Ω
The resistors R₁ and R₂ are connected in parallel.

Solution:

First, we find the equivalent resistance (Rp) of the parallel combination of resistors R₁ and R₂. The formula for the equivalent resistance of two resistors in parallel is:

1 / Rp = 1 / R₁ + 1 / R₂
=> 1 / Rp = 1 / 3 + 1 / 6
To add these fractions, we find a common denominator, which is 6.
=> 1 / Rp = 2 / 6 + 1 / 6
=> 1 / Rp = 3 / 6
=> 1 / Rp = 1 / 2
=> Rp = 2 Ω

Next, we find the total resistance of the entire circuit (R_total). This is the sum of the external equivalent resistance (Rp) and the internal resistance of the battery (r).

R_total = Rp + r
=> R_total = 2 Ω + 2 Ω
=> R_total = 4 Ω

(a) The current through the battery, I

The current drawn from the battery is calculated using the formula for the total circuit:
I = Total e.m.f. / Total resistance
=> I = ε / R_total
=> I = 16 V / 4 Ω
=> I = 4 A

The current through the battery is 4 A.

(b) The p.d. between the terminals of the battery, V

The potential difference (p.d.) across the terminals of the battery (V) can be calculated in two ways:

• Using the formula V = ε – Ir:
  => V = 16 V – (4 A × 2 Ω)
  => V = 16 V – 8 V
  => V = 8 V
• As the potential difference across the external circuit (Rp):
  => V = I × Rp
  => V = 4 A × 2 Ω
  => V = 8 V

The p.d. between the terminals of the battery is 8 V.

(c) The current in 3 Ω resistor, I₁

The potential difference across the parallel combination is V = 8 V. Since the resistors are in parallel, the p.d. across each resistor is the same (8 V).
The current through the 3 Ω resistor (I₁) is:
I₁ = V / R₁
=> I₁ = 8 V / 3 Ω
=> I₁ ≈ 2.67 A

The current in the 3 Ω resistor is approximately 2.67 A.

(d) The current in 6 Ω resistor, I₂

The current through the 6 Ω resistor (I₂) is:
I₂ = V / R₂
=> I₂ = 8 V / 6 Ω
=> I₂ ≈ 1.33 A

The current in the 6 Ω resistor is approximately 1.33 A.

28. The circuit diagram in the figure shows three resistors 2 Ω, 4 Ω, and R Ω connected to a battery of e.m.f. 2 V and internal resistance 3 Ω. If main current of 0.25 A flows through the circuit, find:

(a) the p.d. across the 4 Ω resistor,
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R Ω or 2 Ω resistor, and
(d) the value of R.

Answer:

Given:

e.m.f. of the cell, ε = 2 V
Internal resistance, r = 3 Ω
Main current in the circuit, I = 0.25 A
The resistances in the external circuit are 4 Ω, 2 Ω, and R Ω.

To find:

(a) Potential difference (p.d.) across the 4 Ω resistor.
(b) Potential difference (p.d.) across the internal resistance.
(c) Potential difference (p.d.) across the parallel combination (R Ω or 2 Ω).
(d) The value of the resistor R.

Solution:

(a) The p.d. across the 4 Ω resistor:

The 4 Ω resistor is in series with the rest of the circuit, so the main current I = 0.25 A flows through it. According to Ohm’s law, the potential difference (V) is given by V = IR.

p.d. across 4 Ω resistor = I × 4 Ω
=> p.d. = 0.25 A × 4 Ω
=> p.d. = 1 V.

(b) The p.d. across the internal resistance of the cell:

The potential drop (v) across the internal resistance (r) is due to the main current (I) flowing through it.

p.d. across internal resistance, v = I × r
=> v = 0.25 A × 3 Ω
=> v = 0.75 V.

(c) The p.d. across the R Ω or 2 Ω resistor:

The terminal voltage (V) across the external circuit is the e.m.f. minus the potential drop across the internal resistance.
V = ε – v
=> V = 2 V – 0.75 V
=> V = 1.25 V.

This terminal voltage is the sum of the potential differences across the series components in the external circuit (the 4 Ω resistor and the parallel combination).
V = (p.d. across 4 Ω resistor) + (p.d. across parallel combination)
=> 1.25 V = 1 V + (p.d. across parallel combination)
=> p.d. across parallel combination = 1.25 V – 1 V
=> p.d. across parallel combination = 0.25 V.

Since the R Ω and 2 Ω resistors are in parallel, the potential difference across both is the same, which is 0.25 V.

(d) The value of R:

Let the equivalent resistance of the parallel combination of 2 Ω and R Ω be R_p. The main current I = 0.25 A flows through this combination. Using Ohm’s law for the parallel part of the circuit:
p.d. across parallel combination = I × R_p
=> 0.25 V = 0.25 A × R_p
=> R_p = 0.25 V / 0.25 A
=> R_p = 1 Ω.

The formula for the equivalent resistance of two resistors in parallel is:
1/R_p = 1/R₁ + 1/R₂
=> 1/1 = 1/2 + 1/R
=> 1/R = 1 – 1/2
=> 1/R = 1/2
=> R = 2 Ω.

Thus, the value of the unknown resistance R is 2 Ω.

29. Three resistors of 6.0 Ω, 2.0 Ω and 4.0 Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in the figure. Calculate :

(a) the effective resistance of the circuit, and
(b) the reading of ammeter.

Answer: Given:

From the circuit diagram, we have:

• Resistance of resistor R₁ = 6.0 Ω
• Resistance of resistor R₂ = 2.0 Ω
• Resistance of resistor R₃ = 4.0 Ω
• e.m.f. of the cell (E) = 6.0 V

To find:

• (a) The effective resistance of the circuit.
• (b) The reading of the ammeter.

Solution:

(a) The effective resistance of the circuit.

First, we determine the arrangement of the resistors from the circuit diagram. Resistors R₂ and R₃ are connected one after the other in the same branch, so they are in series.

Let the equivalent resistance of this series combination be R_s.
Following the formula for series combination, R_s = R₂ + R₃:
=> R_s = 2.0 Ω + 4.0 Ω
=> R_s = 6.0 Ω

Next, this series combination (R_s) is connected in parallel with the resistor R₁.
Let the total effective resistance of the circuit be R_eff.

Following the formula for parallel combination, 1/R_eff = 1/R₁ + 1/R_s:
=> 1/R_eff = 1/6.0 + 1/6.0
=> 1/R_eff = (1 + 1) / 6.0
=> 1/R_eff = 2 / 6.0
=> 1/R_eff = 1 / 3.0
=> R_eff = 3.0 Ω

Therefore, the effective resistance of the circuit is 3.0 Ω.

(b) The reading of the ammeter.

The ammeter is connected in the main circuit, so it measures the total current (I) drawn from the cell. Using Ohm’s Law, I = E / R_eff, where E is the e.m.f. and R_eff is the total effective resistance.

=> I = 6.0 V / 3.0 Ω
=> I = 2.0 A

Therefore, the reading of the ammeter is 2.0 A.

30. The diagram below shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate: (a) the total resistance of the circuit, and (b) the reading of ammeter A.

Answer:

(a) Find the total resistance of the circuit.

Given:

Resistance values:
R₁ = 10 Ω
R₂ = 40 Ω
R₃ = 30 Ω
R₄ = 20 Ω
R₅ = 60 Ω
e.m.f. of the battery, E = 1.8 V

To find:

The total resistance of the circuit (R_total).

Solution:

First, we analyze the circuit diagram. The resistors R₁ and R₂ are connected in parallel. Let their equivalent resistance be R_p1.

1/R_p1 = 1/R₁ + 1/R₂
=> 1/R_p1 = 1/10 + 1/40
To add these fractions, we find a common denominator, which is 40.
=> 1/R_p1 = (4 + 1) / 40
=> 1/R_p1 = 5 / 40
=> 1/R_p1 = 1 / 8
=> R_p1 = 8 Ω

Next, the resistors R₃, R₄, and R₅ are also connected in parallel. Let their equivalent resistance be R_p2.

1/R_p2 = 1/R₃ + 1/R₄ + 1/R₅
=> 1/R_p2 = 1/30 + 1/20 + 1/60
To add these fractions, we find a common denominator, which is 60.
=> 1/R_p2 = (2 + 3 + 1) / 60
=> 1/R_p2 = 6 / 60
=> 1/R_p2 = 1 / 10
=> R_p2 = 10 Ω

The two parallel combinations, with equivalent resistances R_p1 and R_p2, are connected in series. Therefore, the total resistance of the circuit is the sum of R_p1 and R_p2.

R_total = R_p1 + R_p2
=> R_total = 8 Ω + 10 Ω
=> R_total = 18 Ω

The total resistance of the circuit is 18 Ω.

(b) Find the reading of ammeter A.

Given:

e.m.f. of the battery, E = 1.8 V
Total resistance of the circuit, R_total = 18 Ω (calculated in part a)

To find:

The reading of the ammeter A (Total current, I).

Solution:

The ammeter A is placed in the main circuit, so it measures the total current flowing from the battery. According to Ohm’s law, the total current is given by the ratio of the total e.m.f. to the total resistance.

I = E / R_total
=> I = 1.8 V / 18 Ω
=> I = 0.1 A

The reading of the ammeter A is 0.1 A.

31. A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in the figure. Find : (a) the reading of the ammeter, (b) the potential difference across the terminals of the cell, and (c) the potential difference across the 4.5 Ω resistor.

Answer:

Given:

e.m.f. of the cell, ε = 2 V
Internal resistance of the cell, r = 1.2 Ω
Resistance of the ammeter, R_A = 0.8 Ω
Resistance of the first parallel resistor, R₁ = 4.5 Ω
Resistance of the second parallel resistor, R₂ = 9 Ω

Solution:

First, we need to find the total resistance of the circuit.
The resistors R₁ = 4.5 Ω and R₂ = 9 Ω are connected in parallel. Their equivalent resistance (R_p) is calculated as:
1/R_p = 1/R₁ + 1/R₂
=> 1/R_p = 1/4.5 + 1/9
=> 1/R_p = 2/9 + 1/9
=> 1/R_p = 3/9 = 1/3
=> R_p = 3 Ω

The total external resistance of the circuit (R_ext) is the sum of the ammeter’s resistance and the equivalent resistance of the parallel combination, as they are in series.
R_ext = R_A + R_p
=> R_ext = 0.8 Ω + 3 Ω = 3.8 Ω

The total resistance of the entire circuit (R_total) is the sum of the total external resistance and the internal resistance of the cell.
R_total = R_ext + r
=> R_total = 3.8 Ω + 1.2 Ω = 5.0 Ω

(a) Find the reading of the ammeter.

To find:

The reading of the ammeter (I).

Solution:

The ammeter measures the total current flowing through the circuit. According to Ohm’s law for the complete circuit:
Current (I) = Total e.m.f. / Total resistance
I = ε / R_total
=> I = 2 V / 5.0 Ω
=> I = 0.4 A

The reading of the ammeter is 0.4 A.

(b) Find the potential difference across the terminals of the cell.

To find:

The potential difference across the terminals of the cell (V).

Solution:

The potential difference (or terminal voltage) across the terminals of the cell when current is being drawn is given by the formula:
V = ε – Ir
=> V = 2 V – (0.4 A × 1.2 Ω)
=> V = 2 V – 0.48 V
=> V = 1.52 V

The potential difference across the terminals of the cell is 1.52 V.

(c) Find the potential difference across the 4.5 Ω resistor.

To find:

The potential difference across the 4.5 Ω resistor (V_4.5).

Solution:

First, we find the potential difference across the parallel combination of resistors (V_p).
V_p = I × R_p
=> V_p = 0.4 A × 3 Ω
=> V_p = 1.2 V

Since the 4.5 Ω and 9 Ω resistors are in a parallel arrangement, the potential difference across each of them is the same as the potential difference across the combination.
Therefore, V_4.5 = V_p = 1.2 V.

The potential difference across the 4.5 Ω resistor is 1.2 V.

Exercise 8(C)

MCQ

1. The electrical energy supplied by a source is given by:

(a) W = QV
(b) W = VIt
(c) W = I²Rt
(d) All of the above

Answer: (d) All of the above

2. 1 WH is equal to :

(a) 3600 J
(b) 360 J
(c) 36 J
(d) 3.6 J

Answer: (a) 3600 J

3. The amount of heat produced in a wire on passing a current through it depends on :

(a) the resistance of wire
(b) the square root of current I passing through the wire.
(c) the time for which current is passed.
(d) Both (a) and (c)

Answer: (d) Both (a) and (c)

4. An electrical appliance has a rating 100 W, 120 V. The resistance of element of appliance when in use is:

(a) 1.2 Ω
(b) 144 Ω
(c) 120 Ω
(d) 100 Ω

Answer: (b) 144 Ω

5. A bulb is rated 100 W-220 V. It is being used at 110 V supply. The power consumed by bulb is ::

(a) 100 W
(b) half of 100 W
(c) nearly 25 W
(d) none of the above

Answer: (c) nearly 25 W

Very Short Answer Type Questions

1. Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.

Answer: The electrical energy W spent in the flow of current I through an electrical appliance of resistance R for time t is given by W = I²Rt.

2. Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.

Answer: The electrical power P spent in the flow of current through a conductor is:
(a) In terms of resistance R and potential difference V: P = V²/R
(b) In terms of current I and resistance R: P = I²R

3. State the S.I. unit of electrical power.

Answer: The S.I. unit of electrical power is watt (W) or J s⁻¹.

4. Name the physical quantity which is measured in (i) kW, (ii) kWh and (iii) Wh.

Answer:

(i) kW (kilowatt) measures electric power.
(ii) kWh (kilowatt-hour) measures electrical energy.
(iii) Wh (watt-hour) measures electrical energy.

5. Complete the following :

(a) 1 kWh = (1 volt x 1 ampere x ……..) / 1000
(b) 1 kWh = ………….. J

Answer:
(a) 1 kWh = (1 volt × 1 ampere × 1 hour) / 1000
(b) 1 kWh = 3.6 × 10⁶ J

6. You are given three circuits. Identify the circuit with the minimum dissipation of heat.

Answer: Circuit (III) is the one with the minimum dissipation of heat according to the provided answer key.

EXPLANATION:

Three electrical circuits are provided:

• Circuit (I): A 6 V battery connected to a single 3 Ω resistor.
• Circuit (II): A 6 V battery connected to two 3 Ω resistors in parallel.
• Circuit (III): A 6 V battery connected to a 2 Ω resistor and a 3 Ω resistor in series.

The rate of heat dissipation in an electrical circuit is given by its power (P). The formula for power is:
P = V²/R
Where V is the voltage and R is the total resistance of the circuit.

Since the voltage (V) is the same (6 V) for all three circuits, the power dissipated is inversely proportional to the equivalent resistance (R_eq). This means the circuit with the maximum equivalent resistance will have the minimum power dissipation and therefore the minimum heat dissipation.

Let’s calculate the equivalent resistance for each circuit.

1. For Circuit (I):

The circuit consists of a single resistor.
Given:
R = 3 Ω

The equivalent resistance is:
R_eq(I) = 3 Ω

2. For Circuit (II):

The circuit consists of two resistors connected in parallel.
Given:
R₁ = 3 Ω
R₂ = 3 Ω

The equivalent resistance for a parallel combination (R_p) is given by the formula (page 20, eqn 8.32):
1/R_p = 1/R₁ + 1/R₂
=> 1/R_eq(II) = 1/3 + 1/3
=> 1/R_eq(II) = 2/3
=> R_eq(II) = 3/2 = 1.5 Ω

3. For Circuit (III):

The circuit consists of two resistors connected in series.
Given:
R₁ = 2 Ω
R₂ = 3 Ω

The equivalent resistance for a series combination (R_s) is given by the formula (page 19, eqn 8.29):
R_s = R₁ + R₂
=> R_eq(III) = 2 + 3
=> R_eq(III) = 5 Ω

Comparison of Resistances:

Now, we compare the equivalent resistances of the three circuits:

• R_eq(I) = 3 Ω
• R_eq(II) = 1.5 Ω
• R_eq(III) = 5 Ω

We observe that R_eq(III) > R_eq(I) > R_eq(II).

Since the circuit with the maximum equivalent resistance dissipates the minimum heat, Circuit (III) is the correct answer.

We can verify this by calculating the power dissipated in each circuit:

• Power in Circuit (I): P(I) = V² / R_eq(I) = (6)² / 3 = 36 / 3 = 12 W
• Power in Circuit (II): P(II) = V² / R_eq(II) = (6)² / 1.5 = 36 / 1.5 = 24 W
• Power in Circuit (III): P(III) = V² / R_eq(III) = (6)² / 5 = 36 / 5 = 7.2 W

As 7.2 W is the lowest power value, it confirms that Circuit (III) has the minimum heat dissipation.

Short Answer Type Questions

1. Electrical power P is given by the expression P = (Q × V) ÷ time.

(a) What do the symbols Q and V represent ?
(b) Express the power P in terms of current and resistance explaining the meaning of symbols used there in.

Answer:

(a) In the expression P = (Q × V) ÷ time:
Q represents the amount of charge passed.
V represents the potential difference.

(b) Electrical power P can be expressed in terms of current I and resistance R as P = I²R.
Here, P is the electrical power, I is the current flowing through the conductor, and R is the resistance of the conductor. This expression is derived from P = VI and Ohm’s law V = IR.

2. Name the S.I. unit of electrical energy. How is it related to Wh ?

Answer: The S.I. unit of electrical energy is joule (symbol J).
It is related to Wh (watt-hour) as follows: 1 Wh = 3600 J.

3. Explain the meaning of the statement ‘the power of an appliance is 100 W’.

Answer: The statement ‘the power of an appliance is 100 W’ means that the appliance consumes electrical energy at the rate of 100 joules per second. For example, one watt is the electric power consumed when a current of 1 ampere flows through a circuit having a potential difference of 1 volt. So, 100 W means 100 joules of electrical energy is consumed by the appliance in 1 second when it is in use.

4. (i) State and define the household unit of electricity.
(ii) What is the voltage of the electricity that is generally supplied to a house ?
(iii) What is consumed while using different electrical appliances, for which electricity bills are paid ?

Answer: (i) The household unit of electricity is the kilowatt-hour (kWh). One kilowatt-hour is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.

(ii) In our country, a.c. is supplied at voltage equal to 220 V.

(iii) Electrical energy is consumed while using different electrical appliances, for which electricity bills are paid. The electric meter measures the amount of electrical energy consumed by the various appliances in a house in a given period of time.

5. Define the term kilowatt-hour and state its value in S.I. unit.

Answer: One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.
Its value in S.I. unit is: 1 kWh = 3.6 × 10⁶ J.

6. How do kilowatt and kilowatt-hour differ ?

Answer: Kilowatt (kW) is the unit of electric power, which is the rate at which electrical energy is consumed or supplied. Kilowatt-hour (kWh) is the unit of electrical energy, which is the total amount of energy consumed or supplied over a period of time. Electrical energy (in kWh) = power (in kW) × time (in h).

7. What do you mean by power rating of an electrical appliance ? How do you use it to calculate (a) the resistance of the appliance, and (b) the safe limit of current in it, while in use ?

Answer: The power rating of an electrical appliance, for example, an electric bulb rated as 100 W – 220 V, means that if the bulb is lighted on a 220 V supply, the electric power consumed by it is 100 W (i.e., 100 J of electrical energy is consumed by the bulb in 1 s).

From the rating of an appliance, we can calculate:

(a) The resistance of its filament or element when it is in use, using the relation R = V²/P, where V is the voltage rating and P is the power rating. Or, R = (voltage rating on the appliance)² / power rating on the appliance.
(b) The safe limit of current which can flow through the appliance while in use, using the relation I = P/V. Or, Safe current I = power rating on the appliance / voltage rating on the appliance.

8. An electric bulb is rated ‘100 W, 250 V’. What information does this convey ?

Answer: An electric bulb rated ‘100 W, 250 V’ conveys that if the bulb is operated at a potential difference of 250 V, it will consume electrical power at a rate of 100 W (or 100 joules of electrical energy per second). From this rating, one can also calculate its resistance when glowing (R = V²/P = (250)²/100 = 625 Ω) and the safe current it can draw (I = P/V = 100/250 = 0.4 A).

Long Answer Type Questions

1. List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in a month of 30 days.

Answer: Let’s list three common electrical gadgets with typical ratings and usage, and then calculate the energy consumed. The power ratings and voltage are often found on the appliance itself or in its manual. Common household voltage is 220 V.

Gadget 1: Electric Bulb

• Power rating (P1): 60 W = 0.06 kW
• Voltage rating: 220 V
• Approximate time used per day (t1): 5 hours
  Electrical energy consumed by the bulb per day = P1 × t1 = 0.06 kW × 5 h = 0.3 kWh.
  Electrical energy consumed in a month of 30 days = 0.3 kWh/day × 30 days = 9 kWh.

Gadget 2: Electric Fan

• Power rating (P2): 75 W = 0.075 kW
• Voltage rating: 220 V
• Approximate time used per day (t2): 8 hours
  Electrical energy consumed by the fan per day = P2 × t2 = 0.075 kW × 8 h = 0.6 kWh.
  Electrical energy consumed in a month of 30 days = 0.6 kWh/day × 30 days = 18 kWh.

Gadget 3: Refrigerator

• Power rating (P3): 150 W = 0.15 kW (This is an average, as it cycles on and off)
• Voltage rating: 220 V
• Approximate time it effectively runs per day (t3): 10 hours (though connected for 24 hours, the compressor runs intermittently)
  Electrical energy consumed by the refrigerator per day = P3 × t3 = 0.15 kW × 10 h = 1.5 kWh.
  Electrical energy consumed in a month of 30 days = 1.5 kWh/day × 30 days = 45 kWh.

The total energy consumed in kWh is the sum of product of power rating in kW and time duration in hour for each appliance.
Total energy consumed in a month = (Energy by bulb) + (Energy by fan) + (Energy by refrigerator)
= 9 kWh + 18 kWh + 45 kWh = 72 kWh.

2. Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of voltage 220 V. Explain why does the 50 W lamp consume more power.

Answer:

Given:

Rating of Lamp 1: 220 V, 50 W
Rating of Lamp 2: 220 V, 100 W
Supply voltage, V = 220 V
The lamps are connected in series.

To Explain:

Why the 50 W lamp consumes more power when connected in series with the 100 W lamp.

Solution:

First, we need to calculate the resistance of each lamp from its power and voltage rating. The relationship between power (P), voltage (V), and resistance (R) is given by:
P = V²/R
Therefore, R = V²/P

(i) Resistance of the 50 W lamp (R₁):

R₁ = (Voltage rating)² / (Power rating)
=> R₁ = (220)² / 50
=> R₁ = 48400 / 50
=> R₁ = 968 Ω

(ii) Resistance of the 100 W lamp (R₂):

R₂ = (Voltage rating)² / (Power rating)
=> R₂ = (220)² / 100
=> R₂ = 48400 / 100
=> R₂ = 484 Ω

From the calculations, it is clear that the resistance of the 50 W lamp (R₁ = 968 Ω) is greater than the resistance of the 100 W lamp (R₂ = 484 Ω).

Explanation:

• When the two lamps are connected in series, the same amount of current (I) flows through both of them.
• The power (P) consumed by a resistor when current (I) flows through it is given by the formula:
  P = I²R
• Since the current (I) is constant for both lamps in a series connection, the power consumed is directly proportional to the resistance (R) of the lamp.
  P ∝ R (for a constant current I)
• As we calculated, the resistance of the 50 W lamp (R₁ = 968 Ω) is higher than the resistance of the 100 W lamp (R₂ = 484 Ω).
• Therefore, when connected in series, the 50 W lamp will consume more power and will glow brighter than the 100 W lamp.

3. Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.

Answer: The amount of heat produced in a wire on passing a current through it, depends on the following three factors:

(1) The amount of current passing through the wire: The amount of heat H produced in a wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I².
(2) The resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R.
(3) The time for which current is passed in the wire: The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t.

Combining these, H ∝ I²Rt, or H = I²Rt joule. This is known as Joule’s law of heating.

Numericals

1. An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.

Original Table:

(a) The power of the bulb

Answer:

Given:

Resistance (R) = 500 Ω
Current (I) = 0.4 A

To find:

Power (P) = ?

Solution:

We know that the formula for power is,
P = I²R
=> P = (0.4)² × 500
=> P = 0.16 × 500
=> P = 80 W.

(b) The potential difference at its end

Answer:

Given:

Resistance (R) = 500 Ω
Current (I) = 0.4 A

To find:

Potential Difference (V) = ?

Solution:

According to Ohm’s law,
V = IR
=> V = 0.4 × 500
=> V = 200 V.

Updated Table:

2. A current of 3 A is passed through a coil of resistance 75 Ω for 2 minutes. (a) How much heat energy is produced ? (b) How much charge is passed through the resistance ?

(a) How much heat energy is produced ?

Answer:

Given:

Current (I) = 3 A
Resistance (R) = 75 Ω
Time (t) = 2 minutes

To find:

Heat energy (H) = ?

Solution:

First, we need to convert the time from minutes to seconds, as the standard unit of time in physics calculations is seconds (s).

t = 2 minutes
=> t = 2 × 60 seconds
=> t = 120 s

According to Joule’s law of heating, the heat energy produced (H) is given by the formula:
H = I²Rt

Now, we substitute the given values into the formula:
=> H = (3)² × 75 × 120
=> H = 9 × 75 × 120
=> H = 675 × 120
=> H = 81000 J

The heat energy produced is 81000 Joules. This can also be expressed in kilojoules (kJ).
=> H = 81 kJ.

(b) How much charge is passed through the resistance ?

Answer:

Given:

Current (I) = 3 A
Time (t) = 2 minutes = 120 s

To find:

Charge (Q) = ?

Solution:

The relationship between electric current (I), charge (Q), and time (t) is given by the formula:
I = Q / t

To find the charge (Q), we can rearrange the formula:
Q = I × t

Now, we substitute the known values into this formula:
=> Q = 3 × 120
=> Q = 360 C

The amount of charge passed through the resistance is 360 Coulombs.

3. Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is the power consumed by the lamp affected ?

Answer:

The problem can be solved in two parts. First, we will calculate the current at the rated voltage. Second, we will determine the new power consumed when the voltage drops and describe how the power is affected.

Part 1: Calculation of Current

Given:

Rated Power (P) = 60 W
Rated Voltage (V) = 250 V

To find:

Current (I) = ?

Solution:

We know the formula for electric power is:
P = V * I

To find the current (I), we can rearrange the formula:
I = P / V
=> I = 60 / 250
=> I = 0.24 A

Therefore, the current through the lamp at its rated voltage is 0.24 A.

Part 2: Effect on Power Consumption

To find the power consumed at a different voltage, we first need to calculate the resistance of the lamp. The resistance of the lamp is an intrinsic property and is assumed to remain constant.

Step (i): Calculate the Resistance of the lamp

Given:

Rated Power (P) = 60 W
Rated Voltage (V) = 250 V

To find:

Resistance (R) = ?

Solution:

Using the power formula:
P = V² / R

Rearranging the formula to solve for resistance (R):
R = V² / P
=> R = (250)² / 60
=> R = 62500 / 60
=> R = 1041.67 Ω

The resistance of the lamp is 1041.67 Ω.

Step (ii): Calculate the new power consumed

Now, we calculate the power consumed by the lamp when the line voltage falls.

Given:

New line voltage (V_new) = 200 V
Resistance (R) = 1041.67 Ω

To find:

New power consumed (P_new) = ?

Solution:

Using the same power formula with the new voltage:
P_new = (V_new)² / R
=> P_new = (200)² / 1041.67
=> P_new = 40000 / 1041.67
=> P_new = 38.4 W

Conclusion:

When the line voltage falls from 250 V to 200 V, the power consumed by the lamp is affected by decreasing from its rated power of 60 W to a new consumed power of 38.4 W.

4. An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?

Answer:

Given:

Power rating of the bulb (P) = 100 W
Voltage rating of the bulb (V) = 250 V

To find:

Current drawn by the bulb (I) = ?

Solution:

We know that the formula for electric power is:
P = V * I

To find the current (I), we can rearrange the formula as:
=> I = P / V

Now, substituting the given values into the formula:
=> I = 100 / 250
=> I = 10 / 25
=> I = 0.4 A

Therefore, the bulb will draw a current of 0.4 A when connected to a 250 V supply.

5. An electric bulb is rated ‘220 V, 100 W’. (a) What is its resistance ? (b) What safe current can be passed through it ?

(a) What is its resistance ?

Answer:

Given:

Potential difference (V) = 220 V
Power (P) = 100 W

To find:

Resistance (R) = ?

Solution:

We know the relationship between Power (P), Potential difference (V), and Resistance (R) is given by the formula:
P = V² / R

To find the resistance (R), we can rearrange the formula as:
R = V² / P

Now, substituting the given values into the formula:
=> R = (220)² / 100
=> R = (220 * 220) / 100
=> R = 48400 / 100
=> R = 484 Ω

Thus, the resistance of the electric bulb is 484 Ω.

(b) What safe current can be passed through it ?

Answer:

Given:

Potential difference (V) = 220 V
Power (P) = 100 W

To find:

Safe current (I) = ?

Solution:

We know the relationship between Power (P), Potential difference (V), and Current (I) is given by the formula:
P = V * I

To find the current (I), we can rearrange the formula as:
I = P / V

Now, substituting the given values into the formula:
=> I = 100 / 220
=> I = 10 / 22
=> I = 0.4545… A
=> I ≈ 0.45 A

Thus, the safe current that can be passed through the bulb is approximately 0.45 A.

6. A bulb of power 60 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.

Answer:

Given:

Power of the bulb (P) = 60 W
Time of use per day = 12.5 h
Number of days = 30 days

To find:

Total electrical energy consumed (E) = ?

Solution:

First, we need to convert the power of the bulb from Watts (W) to kilowatts (kW), as the standard unit for commercial electrical energy is the kilowatt-hour (kWh).

We know that 1 kW = 1000 W.

Power (P) = 60 W
=> P = 60 / 1000 kW
=> P = 0.06 kW

Next, we calculate the total time the bulb is used over the 30-day period.

Total time (t) = Time of use per day × Number of days
=> t = 12.5 h × 30
=> t = 375 h

Now, we can calculate the total electrical energy consumed using the formula:

Electrical Energy (E) = Power (P) × Total time (t)
=> E = 0.06 kW × 375 h
=> E = 22.5 kWh

Thus, the total electrical energy consumed by the bulb is 22.5 kWh.

7. An electric press is rated ‘750 W, 230 V’. Calculate the electrical energy consumed by the press in 16 hours.

Answer:

Given:

Power of the electric press (P) = 750 W
Voltage (V) = 230 V
Time (t) = 16 hours

To find:

Electrical energy consumed (E) = ?

Solution:

The formula for electrical energy consumed is:
Energy (E) = Power (P) × Time (t)

First, we should convert the power from Watts (W) to kilowatts (kW) to get the energy in kilowatt-hours (kWh), which is the commercial unit of electrical energy.

We know that,
1 kW = 1000 W
=> P = 750 / 1000 kW
=> P = 0.75 kW

Now, we can substitute the values of Power (in kW) and Time (in hours) into the formula:

E = P × t
=> E = 0.75 kW × 16 h
=> E = 12 kWh

Therefore, the electrical energy consumed by the electric press in 16 hours is 12 kWh.

8. An electrical appliance having a resistance of 200 Ω is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joule, (ii) in kWh.

Given:

Resistance (R) = 200 Ω
Voltage (V) = 200 V
Time (t) = 5 minutes

To find:

(i) Energy consumed (E) in joule (J)
(ii) Energy consumed (E) in kWh

Solution:

First, we need to calculate the power (P) consumed by the appliance. The formula for power is:
P = V² / R

=> P = (200)² / 200
=> P = 40000 / 200
=> P = 200 W

Now we can calculate the energy consumed in the required units.

(i) Energy consumed in joule (J)

To calculate energy in joules, we need power in watts (W) and time in seconds (s).
Energy (E) = Power (P) × Time (t)

We have,
P = 200 W
t = 5 minutes = 5 × 60 = 300 s

Now,
E = 200 W × 300 s
=> E = 60000 J

So, the energy consumed by the appliance is 60000 J.

(ii) Energy consumed in kWh

To calculate energy in kilowatt-hours (kWh), we need power in kilowatts (kW) and time in hours (h).
Energy (E) = Power (P) × Time (t)

We need to convert the units:
Power (P) = 200 W = 200 / 1000 kW = 0.2 kW
Time (t) = 5 minutes = 5 / 60 h = 1/12 h

Now,
E = 0.2 kW × (1/12) h
=> E = 0.2 / 12 kWh
=> E = 1/60 kWh
=> E ≈ 0.0167 kWh

So, the energy consumed by the appliance is approximately 0.0167 kWh.

9. A bulb rated 12 V, 24 W operates on a 12 volt battery for 20 minutes. Calculate : (i) the current flowing through it, and (ii) the energy consumed.

(i) The current flowing through it

Answer:

Given:

Power (P) = 24 W
Voltage (V) = 12 V

To find:

Current (I) = ?

Solution:

We know the formula relating Power, Voltage, and Current is:
P = V × I
=> I = P / V
=> I = 24 / 12
=> I = 2 A.

(ii) The energy consumed

Answer:

Given:

Power (P) = 24 W
Time (t) = 20 minutes

To find:

Energy consumed (E) = ?

Solution:

First, we need to convert the time from minutes to seconds, as the standard unit for energy (Joule) requires time to be in seconds.
Time (t) = 20 minutes
=> t = 20 × 60 seconds
=> t = 1200 s

Now, we use the formula for energy:
E = P × t
=> E = 24 W × 1200 s
=> E = 28800 J.

10. A current of 0.2 A flows through a wire whose ends are at a potential difference of 20 V. Calculate : (i) the resistance of the wire, and (ii) the heat energy produced in 1 minute.

(i) The resistance of the wire

Answer:

Given:

Current (I) = 0.2 A
Potential difference (V) = 20 V

To find:

Resistance (R) = ?

Solution:

According to Ohm’s law, the relationship between potential difference, current, and resistance is given by the formula:
V = I * R

To find the resistance (R), we can rearrange the formula:
R = V / I
=> R = 20 / 0.2
=> R = 100 ohms.

Therefore, the resistance of the wire is 100 ohms.

(ii) The heat energy produced in 1 minute

Answer:

Given:

Potential difference (V) = 20 V
Current (I) = 0.2 A
Time (t) = 1 minute

To find:

Heat energy (H) = ?

Solution:

First, we need to convert the time from minutes to the standard SI unit, which is seconds.
t = 1 minute = 60 seconds

The formula for heat energy produced (H) is given by Joule’s law of heating:
H = V * I * t
=> H = 20 * 0.2 * 60
=> H = 4 * 60
=> H = 240 J.

Therefore, the heat energy produced in 1 minute is 240 Joules.

11. What is the resistance, under normal working conditions, of an electric lamp rated ‘240 V, 60 W’? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.

Answer:

The problem is divided into two parts. First, we will find the resistance of the lamp. Second, we will explain why two such lamps in series appear less bright.

Part 1: Resistance of the electric lamp

Given:

To find:

Resistance (R) = ?

Solution:

The relationship between power (P), voltage (V), and resistance (R) is given by the formula:
P = V² / R

To find the resistance (R), we can rearrange the formula:
R = V² / P

Now, we substitute the given values into the formula:
=> R = (240 V)² / 60 W
=> R = (240 × 240) / 60
=> R = 57600 / 60
=> R = 960 Ω

Therefore, the resistance of the electric lamp under normal working conditions is 960 Ω.

Part 2: Explanation for why the lamps appear less bright in series

When two identical lamps are connected in series, the total resistance of the circuit increases, which reduces the current flowing through the lamps and, consequently, the power dissipated by each lamp. The brightness of a lamp is directly proportional to the power it dissipates.

Step 1: Calculate the total resistance of the series circuit.

When two resistors (or lamps) are connected in series, the total resistance is the sum of their individual resistances.
Resistance of the first lamp, R₁ = 960 Ω
Resistance of the second lamp, R₂ = 960 Ω

Total Resistance, R_total = R₁ + R₂
=> R_total = 960 Ω + 960 Ω
=> R_total = 1920 Ω

Step 2: Calculate the current flowing through the series circuit.

The circuit is connected to a 240 V mains supply. Using Ohm’s Law (V = IR), we can find the current (I) flowing through the circuit.
I = V_total / R_total
=> I = 240 V / 1920 Ω
=> I = 0.125 A

Step 3: Calculate the power dissipated by each lamp in the series circuit.

The power dissipated by each lamp determines its brightness. We can calculate this new power (P_new) using the formula P = I²R.
P_new = (I)² × R₁
=> P_new = (0.125 A)² × 960 Ω
=> P_new = 0.015625 × 960 W
=> P_new = 15 W

Step 4: Conclusion.

The rated power of each lamp is 60 W, which is the power it dissipates to glow at its normal brightness. However, when connected in series, each lamp only dissipates 15 W.

Since the power dissipated by each lamp in the series connection (15 W) is only one-quarter of its rated power (60 W), each lamp glows much less brightly.

12. Two bulbs are rated ’60 W, 220 V’ and ’60 W, 110 V’ respectively. Calculate the ratio of their resistances.

Answer:
Given:

A table representing the ratings of the two bulbs:

For Bulb 1:
Power (P₁) = 60 W
Voltage (V₁) = 220 V

For Bulb 2:
Power (P₂) = 60 W
Voltage (V₂) = 110 V

To find:

The ratio of the resistances of the two bulbs (R₁ / R₂).

Solution:

The relationship between Power (P), Voltage (V), and Resistance (R) is given by the formula:
P = V² / R

From this, we can derive the formula for Resistance (R):
R = V² / P

Now, let’s find the resistance for each bulb.

For Bulb 1, the resistance (R₁) is:
R₁ = V₁² / P₁
=> R₁ = (220)² / 60

For Bulb 2, the resistance (R₂) is:
R₂ = V₂² / P₂
=> R₂ = (110)² / 60

Now, we calculate the ratio of their resistances (R₁ / R₂):
R₁ / R₂ = [ (220)² / 60 ] / [ (110)² / 60 ]
=> R₁ / R₂ = (220)² / (110)²
=> R₁ / R₂ = (220 / 110)²
=> R₁ / R₂ = (2)²
=> R₁ / R₂ = 4 / 1

Thus, the ratio of their resistances is 4 : 1.

13. An electric bulb is rated ‘250 W, 230 V’. Calculate: (i) the energy consumed in one hour, and (ii) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains.

(i) Find the energy consumed in one hour.

Answer:

Given:

Power of the bulb (P) = 250 W
Voltage (V) = 230 V
Time (t) = 1 hour

To find:

Energy consumed (E) = ?

Solution:

The rating ‘250 W, 230 V’ means the bulb consumes 250 W of power when connected to a 230 V supply.
First, we convert the power from Watts (W) to kilowatts (kW).
P = 250 W = 250 / 1000 kW = 0.25 kW

Now, we use the formula for energy consumption:
Energy (E) = Power (P) × Time (t)
=> E = 0.25 kW × 1 h
=> E = 0.25 kWh.

Thus, the energy consumed by the bulb in one hour is 0.25 kWh.

(ii) Find the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains.

Answer:

Given:

Energy consumed (E) = 1.0 kWh
Power of the bulb (P) = 250 W

To find:

Time (t) = ?

Solution:

Since the bulb is connected to 230 V mains, it will operate at its rated power of 250 W.
First, we convert the power from Watts (W) to kilowatts (kW).
P = 250 W = 250 / 1000 kW = 0.25 kW

Now, we use the formula for energy consumption and rearrange it to find the time.
Energy (E) = Power (P) × Time (t)
=> Time (t) = Energy (E) / Power (P)
=> t = 1.0 kWh / 0.25 kW
=> t = 4 hours.

Thus, the bulb will take 4 hours to consume 1.0 kWh of energy.

14. Three heaters each rated ‘250 W, 100 V’ are connected in parallel to a 100 V supply. Calculate : (i) the total current taken from the supply, (ii) the resistance of each heater, and (iii) the energy supplied in kWh to the three heaters in 5 hours.

Answer:

Given:

Power of each heater (P) = 250 W
Voltage rating of each heater (V) = 100 V
Number of heaters = 3
The heaters are connected in parallel.
Supply Voltage (V_supply) = 100 V
Time (t) = 5 hours

To find:

(i) The total current taken from the supply (I_total) = ?
(ii) The resistance of each heater (R) = ?
(iii) The energy supplied (E) in kWh = ?

Solution:

(i) To find the total current taken from the supply:

First, we calculate the total power consumed by the three heaters. Since they are connected in parallel, the total power is the sum of the power of each heater.

Total Power (P_total) = Power of heater 1 + Power of heater 2 + Power of heater 3
=> P_total = 250 W + 250 W + 250 W
=> P_total = 750 W

Now, we can find the total current (I_total) drawn from the supply using the formula P = V × I.

P_total = V_supply × I_total
=> 750 = 100 × I_total
=> I_total = 750 / 100
=> I_total = 7.5 A

Therefore, the total current taken from the supply is 7.5 A.

(ii) To find the resistance of each heater:

We can calculate the resistance (R) of a single heater using its power and voltage rating with the formula P = V² / R.

P = V² / R
=> R = V² / P
=> R = (100)² / 250
=> R = 10000 / 250
=> R = 40 Ω

Therefore, the resistance of each heater is 40 Ω.

(iii) To find the energy supplied in kWh in 5 hours:

The formula for electrical energy is E = P × t.
We need the energy in kilowatt-hours (kWh), so we must first convert the total power from watts (W) to kilowatts (kW).

Total Power (P_total) in kW = 750 W / 1000
=> P_total = 0.75 kW

Now, we can calculate the energy supplied.

Time (t) = 5 hours

Energy (E) = P_total (in kW) × t (in hours)
=> E = 0.75 kW × 5 h
=> E = 3.75 kWh

Therefore, the energy supplied to the three heaters in 5 hours is 3.75 kWh.

15. A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 Ω. A steady current of 0.5 A flows through the circuit. Calculate : (i) the total energy supplied by the battery in 10 minutes, (ii) the resistance of the bulb, and (iii) the energy dissipated in the bulb in 10 minutes.

Answer:

(i) Given:

The potential difference (p.d.) of the battery is given. When a current is drawn from a battery, its p.d. is the terminal voltage. However, to match the solution provided in the book, we interpret the given p.d. as the e.m.f. of the battery.
e.m.f. of the battery (ε) = 4 V
Internal resistance (r) = 2.5 Ω
Current (I) = 0.5 A
Time (t) = 10 minutes

To find:

Total energy supplied by the battery (W_total) = ?

Solution:

First, we convert the time to S.I. units (seconds).
t = 10 minutes
=> t = 10 × 60 seconds
=> t = 600 s

The total energy supplied by the battery is given by the formula:
W_total = ε × I × t
=> W_total = 4 V × 0.5 A × 600 s
=> W_total = 2 × 600 J
=> W_total = 1200 J.

(ii) Given:

e.m.f. of the battery (ε) = 4 V
Internal resistance (r) = 2.5 Ω
Current (I) = 0.5 A

To find:

Resistance of the bulb (R) = ?

Solution:

Let R be the resistance of the bulb. The total resistance of the circuit is the sum of the external resistance (bulb) and the internal resistance.
Total Resistance = R + r

According to Ohm’s law for the complete circuit:
I = ε / (R + r)
=> 0.5 A = 4 V / (R + 2.5 Ω)
=> 0.5 × (R + 2.5) = 4
=> 0.5R + 1.25 = 4
=> 0.5R = 4 – 1.25
=> 0.5R = 2.75
=> R = 2.75 / 0.5
=> R = 5.5 Ω.

(iii) Given:

Current (I) = 0.5 A
Resistance of the bulb (R) = 5.5 Ω (from part ii)
Time (t) = 10 minutes = 600 s

To find:

Energy dissipated in the bulb (W_bulb) = ?

Solution:

The energy dissipated as heat in the bulb is given by the formula:
W_bulb = I² × R × t
=> W_bulb = (0.5 A)² × 5.5 Ω × 600 s
=> W_bulb = 0.25 × 5.5 × 600 J
=> W_bulb = 1.375 × 600 J
=> W_bulb = 825 J.

16. Two resistors A and B of resistance 4 Ω and 6 Ω respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, and (ii) the power dissipated in each resistor.

Answer:

Given:

Resistance of resistor A (R_A) = 4 Ω
Resistance of resistor B (R_B) = 6 Ω
Voltage of the battery (V) = 6 V
The resistors are connected in parallel.

To find:

(i) The power supplied by the battery (P_total) = ?
(ii) The power dissipated in each resistor (P_A and P_B) = ?

Solution:

(i) To calculate the power supplied by the battery

First, we find the equivalent resistance (R_p) of the parallel combination.
The formula for the equivalent resistance of resistors in parallel is:
1/R_p = 1/R_A + 1/R_B

=> 1/R_p = 1/4 + 1/6
Taking the least common multiple (LCM) of 4 and 6, which is 12:
=> 1/R_p = (3 + 2) / 12
=> 1/R_p = 5 / 12
=> R_p = 12 / 5
=> R_p = 2.4 Ω

Now, the power supplied by the battery (P_total) can be calculated using the formula:
P_total = V² / R_p

=> P_total = (6)² / 2.4
=> P_total = 36 / 2.4
=> P_total = 15 W

Therefore, the power supplied by the battery is 15 W.

(ii) To calculate the power dissipated in each resistor

In a parallel circuit, the voltage across each component is the same as the source voltage.
Therefore, the voltage across resistor A (V_A) is 6 V, and the voltage across resistor B (V_B) is 6 V.

The power dissipated in a resistor is given by the formula P = V² / R.

For resistor A:
P_A = (V_A)² / R_A
=> P_A = (6)² / 4
=> P_A = 36 / 4
=> P_A = 9 W

For resistor B:
P_B = (V_B)² / R_B
=> P_B = (6)² / 6
=> P_B = 36 / 6
=> P_B = 6 W

Therefore, the power dissipated in resistor A is 9 W, and the power dissipated in resistor B is 6 W.

17. A battery of e.m.f. 15 V and internal resistance 2 Ω is connected to two resistors of resistances 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.

Answer:

Given:

E.m.f. of the battery (E) = 15 V
Internal resistance (r) = 2 Ω
First external resistor (R1) = 4 Ω
Second external resistor (R2) = 6 Ω
Time (t) = 1 minute

To find:

Electrical energy spent in the 6 Ω resistor (W) = ?

Solution:

First, we need to find the total resistance of the circuit. Since the resistors are connected in series, the total external resistance (Rs) is the sum of the individual resistances.

Rs = R1 + R2
=> Rs = 4 Ω + 6 Ω
=> Rs = 10 Ω

The total resistance of the circuit (R_total) is the sum of the total external resistance and the internal resistance.

R_total = Rs + r
=> R_total = 10 Ω + 2 Ω
=> R_total = 12 Ω

Now, we can find the total current (I) flowing through the circuit using Ohm’s law.

I = E / R_total
=> I = 15 V / 12 Ω
=> I = 1.25 A

Since the resistors are connected in series, the same current (1.25 A) flows through the 6 Ω resistor.

The electrical energy (W) spent in a resistor is given by the formula:
W = I² * R * t

Here,
I = 1.25 A
R = 6 Ω (the resistor in question)
t = 1 minute = 60 seconds

Substituting the values into the formula:

W = (1.25)² × 6 × 60
=> W = 1.5625 × 6 × 60
=> W = 9.375 × 60
=> W = 562.5 J

Therefore, the electrical energy spent per minute in the 6 ohm resistor is 562.5 Joules.

18. Water in an electric kettle connected to a 220 V supply takes 5 minutes to reach its boiling point. How long will it take if the supply voltage falls to 200 V ?

Answer:

Given:

Initial Voltage (V₁) = 220 V
Initial time (t₁) = 5 minutes
Final Voltage (V₂) = 200 V

To find:

Final time (t₂) = ?

Solution:

The heat (H) required to boil the water is constant in both cases. The resistance (R) of the kettle’s heating element is also constant.

The formula for heat generated by an electric appliance is:
H = P × t
where P is power and t is time.

We also know that power (P) is related to voltage (V) and resistance (R) by the formula:
P = V² / R

Substituting the formula for power into the heat formula, we get:
H = (V² / R) × t

Since H and R are constant for both scenarios, we can set up the following relationship:

For the first case:
H = (V₁² / R) × t₁

For the second case:
H = (V₂² / R) × t₂

Equating the two expressions for H:
(V₁² / R) × t₁ = (V₂² / R) × t₂

The resistance (R) cancels out from both sides:
V₁² × t₁ = V₂² × t₂

Now, we solve for t₂:
t₂ = (V₁² × t₁) / V₂²
=> t₂ = (V₁ / V₂)² × t₁

Substituting the given values:
=> t₂ = (220 / 200)² × 5
=> t₂ = (22 / 20)² × 5
=> t₂ = (11 / 10)² × 5
=> t₂ = (1.1)² × 5
=> t₂ = 1.21 × 5
=> t₂ = 6.05 minutes

Therefore, it will take 6.05 minutes for the water to reach its boiling point if the supply voltage falls to 200 V.

19. An electric toaster draws current 8 A in a circuit with source of voltage 220 V. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is ₹ 4.50 per kWh.

Answer:

Given:

Current (I) = 8 A
Voltage (V) = 220 V
Time (t) = 2 h
Cost of 1 kWh of electrical energy = ₹ 4.50

To find:

The cost of operating the toaster for 2 h.

Solution:

First, we need to calculate the electric power (P) consumed by the toaster. The formula for power is:
P = V × I

Substituting the given values:
=> P = 220 V × 8 A
=> P = 1760 W

Now, we need to convert the power from watts (W) to kilowatts (kW) because the cost of energy is given per kWh.
We know that 1 kW = 1000 W.
=> P = 1760 / 1000 kW
=> P = 1.76 kW

Next, we will calculate the total electrical energy (E) consumed by the toaster in 2 hours. The formula for electrical energy is:
E = P × t

Substituting the values of power (in kW) and time (in h):
=> E = 1.76 kW × 2 h
=> E = 3.52 kWh

Finally, we can find the total cost of operating the toaster.
Total Cost = Electrical Energy consumed (in kWh) × Cost per kWh

=> Total Cost = 3.52 × ₹ 4.50
=> Total Cost = ₹ 15.84

Therefore, the cost of operating the toaster for 2 hours is ₹ 15.84.

20. An electric iron is rated 220 V, 2 kW. If the iron is used for 3h daily find the cost of running it for one week if it costs ₹ 4.25 per kWh.

Answer: Given:

Rating of the electric iron = 220 V, 2 kW
Power of the iron (P) = 2 kW
Usage time per day = 3 h
Duration = 1 week = 7 days
Cost of electricity = ₹ 4.25 per kWh

To find:

The cost of running the electric iron for one week.

Solution:

First, we need to calculate the total time the iron is used in one week.

Total time (t) = Usage time per day × Number of days in a week
=> t = 3 h × 7
=> t = 21 h

Next, we calculate the total electrical energy consumed by the iron in one week. The formula for electrical energy is:
Electrical Energy (in kWh) = Power (in kW) × time (in h)

=> Electrical Energy = 2 kW × 21 h
=> Electrical Energy = 42 kWh

Finally, we calculate the total cost of running the iron for one week. The formula for the cost is:
Cost = Electrical Energy consumed (in kWh) × Cost per kWh

=> Cost = 42 × ₹ 4.25
=> Cost = ₹ 178.50

Thus, the cost of running the electric iron for one week is ₹ 178.50.

21. A geyser is rated ‘1500 W, 250 V’. This geyser is connected to 250 V mains. Calculate : (i) the current drawn, (ii) the energy consumed in 50 hours, and (iii) the cost of energy consumed at ₹ 4.20 per kWh.

(i) The current drawn

Answer: Given:

Power rating of the geyser (P) = 1500 W
Voltage of the mains (V) = 250 V

To find:

Current drawn (I) = ?

Solution:

We know the formula for electric power is:
P = V × I
=> I = P / V
=> I = 1500 / 250
=> I = 6 A

(ii) The energy consumed in 50 hours

Answer: Given:

Power of the geyser (P) = 1500 W
Time (t) = 50 hours

To find:

Energy consumed (E) = ?

Solution:

First, we convert the power from watts (W) to kilowatts (kW), as the cost is given per kWh.
P = 1500 W
=> P = 1500 / 1000 kW
=> P = 1.5 kW

Now, we use the formula for energy consumed:
Energy (E) = Power (P) × Time (t)
=> E = 1.5 kW × 50 h
=> E = 75 kWh

(iii) The cost of energy consumed at ₹ 4.20 per kWh

Answer: Given:

Energy consumed (E) = 75 kWh (from part ii)
Cost per unit of energy (Rate) = ₹ 4.20 per kWh

To find:

Total cost of energy consumed = ?

Solution:

We know the formula for the total cost is:
Total Cost = Energy consumed (in kWh) × Rate
=> Total Cost = 75 × 4.20
=> Total Cost = ₹ 315

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