Get summaries, questions, answers, solutions, notes, extras, PDF and guides for Chapter 11 Electricity: Class 10 Science textbook, which is part of the syllabus for students studying under SEBA (Assam Board), NBSE (Nagaland Board), TBSE (Tripura Board), CBSE (Central Board), MBOSE (Meghalaya Board), BSEM (Manipur Board), WBBSE (West Bengal Board), and all other boards following the NCERT books. These solutions, however, should only be treated as references and can be modified/changed.
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Summary
Electricity is a very useful form of energy in our daily lives, powering our homes, schools, and hospitals. It involves the flow of tiny particles called electric charges. When these charges move through a conductor, like a metal wire, we say there is an electric current. Think of it like water flowing in a river; flowing water makes a water current, and flowing charges make an electric current. For current to flow continuously, it needs a complete, unbroken path called an electric circuit. A switch is a device that can open or close this path, turning devices on or off.
Electric current is measured by how much charge passes a point in a certain amount of time. The unit for electric charge is the coulomb, and the unit for electric current is the ampere. An instrument called an ammeter measures the current in a circuit and is always connected in a line with other components. Conventionally, the direction of electric current is considered to be from the positive end to the negative end of a power source, which is opposite to the direction in which tiny negative charges called electrons actually move.
For charges to flow, there needs to be a difference in electric pressure, similar to how water flows from a higher level to a lower one due to a pressure difference. This electric pressure difference is called potential difference, and it is the work done to move a unit charge from one point to another. Its unit is the volt. A voltmeter measures potential difference and is connected across the points where the difference is to be measured. Cells or batteries provide this potential difference in a circuit.
Ohm’s law describes the relationship between potential difference, current, and resistance. It states that the current flowing through a conductor is directly related to the potential difference across its ends, as long as the temperature stays the same. The constant in this relationship is called resistance, which is the property of a conductor to oppose the flow of charges. Its unit is the ohm. Resistance of a conductor depends on its length, its cross-sectional area, and the material it is made from. A property called resistivity is unique to each material.
Resistors can be connected in two main ways. In a series connection, resistors are joined end-to-end, and the same current flows through each. The total resistance is the sum of individual resistances. If one part breaks, the whole circuit stops working. In a parallel connection, resistors are connected across the same two points. The potential difference is the same across each, but the current divides among them. The total resistance in parallel is less than the smallest individual resistance. This is how most household appliances are wired, allowing them to operate independently.
When current flows through a resistor, some electrical energy is converted into heat. This is called the heating effect of electric current. The amount of heat produced depends on the current, the resistance, and the time for which the current flows. This effect is used in devices like electric heaters and irons. Electric bulbs also use this effect; their filaments, often made of tungsten, get very hot and glow. Fuses are safety devices that use this heating effect to melt and break a circuit if the current becomes too high, protecting appliances. Electric power is the rate at which electrical energy is used. Its unit is the watt. The energy we use is often measured in kilowatt-hours.
Textbook solutions
Intext Questions and Answers I
1. What does an electric circuit mean?
Answer: An electric circuit means a continuous and closed path of an electric current.
2. Define the unit of current.
Answer: The unit of current is the ampere (A). One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s.
3. Calculate the number of electrons constituting one coulomb of charge.
Answer: To calculate the number of electrons constituting one coulomb of charge, we use the information that an electron possesses a negative charge of 1.6 × 10⁻¹⁹ C.
Let Q be the total charge, which is 1 coulomb (1 C).
Let e be the magnitude of the charge on one electron, which is 1.6 × 10⁻¹⁹ C.
The number of electrons (n) constituting one coulomb of charge is calculated as:
n = Q / e
⇒ n = 1 C / (1.6 × 10⁻¹⁹ C)
⇒ n = 0.625 × 10¹⁹
⇒ n = 6.25 × 10¹⁸ electrons.
The SI unit of electric charge, the coulomb (C), is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons, and our calculation provides a more precise number based on the given charge of an electron.
Intext Questions and Answers II
1. Name a device that helps to maintain a potential difference across a conductor.
Answer: A battery, consisting of one or more electric cells, is a device that helps to maintain a potential difference across a conductor. The chemical action within a cell generates the potential difference across the terminals of the cell. In order to maintain the current in a given electric circuit, the cell has to expend its chemical energy stored in it.
2. What is meant by saying that the potential difference between two points is 1 V?
Answer: Saying that the potential difference between two points is 1 V means that one volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer: The electric potential difference (V) between two points in an electric circuit carrying some current is defined as the work done (W) to move a unit charge (Q) from one point to the other, so Potential difference (V) = Work done (W)/Charge (Q). From this, the work done W, which represents the energy given, is equal to VQ. For a 6 V battery, the potential difference V is 6 V. For each coulomb of charge, Q is 1 C. Thus, the amount of energy given to each coulomb of charge is 6 V multiplied by 1 C, which equals 6 Joules.
Intext Questions and Answers III
1. On what factors does the resistance of a conductor depend?
Answer: The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.
2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer: Current will flow more easily through a thick wire of the same material when connected to the same source. This is because the resistance of a uniform metallic conductor is inversely proportional to its area of cross-section. A thicker wire has a larger cross-sectional area, which means it offers less resistance to the flow of current.
3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer: If the resistance of an electrical component remains constant while the potential difference across its two ends decreases to half of its former value, the current through it will also decrease to half of its former value. This is because, according to Ohm’s law, current (I) is directly proportional to the potential difference (V) when resistance (R) is constant (I = V/R).
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because the resistivity of an alloy is generally higher than that of its constituent metals. Furthermore, alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices.
5. Use the data in Table 11.2 to answer the following –
| Category | Material | Resistivity (Ω m) |
|---|---|---|
| Conductors | Silver | 1.60 × 10⁻⁸ |
| Copper | 1.62 × 10⁻⁸ | |
| Aluminium | 2.63 × 10⁻⁸ | |
| Tungsten | 5.20 × 10⁻⁸ | |
| Nickel | 6.84 × 10⁻⁸ | |
| Iron | 10.0 × 10⁻⁸ | |
| Chromium | 12.9 × 10⁻⁸ | |
| Mercury | 94.0 × 10⁻⁸ | |
| Manganese | 1.84 × 10⁻⁶ | |
| Alloys | Constantan (alloy of Cu and Ni) | 49 × 10⁻⁶ |
| Manganin (alloy of Cu, Mn and Ni) | 44 × 10⁻⁶ | |
| Nichrome (alloy of Ni, Cr, Mn and Fe) | 100 × 10⁻⁶ | |
| Insulators | Glass | 10¹⁰ – 10¹⁴ |
| Hard rubber | 10¹³ – 10¹⁶ | |
| Ebonite | 10¹⁵ – 10¹⁷ | |
| Diamond | 10¹² – 10¹³ | |
| Paper (dry) | 10¹² |
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer: (a) Among iron and mercury, iron is a better conductor. This is because iron has a resistivity of 10.0 × 10⁻⁸ Ω m, while mercury has a resistivity of 94.0 × 10⁻⁸ Ω m. A material with lower resistivity is a better conductor.
(b) Based on the data in Table 11.2, silver is the best conductor, as it has the lowest resistivity of 1.60 × 10⁻⁸ Ω m.
Intext Questions and Answers IV
1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
Net potential = 6 V
Using Ohm’s law V = IR, we have,
6 = I × 25
⟹ I = 6/25
⟹ I = 0.24 Ampere
Now for the 12 Ω resistor, current = 0.24 A
So, using Ohm’s law V = 0.24 × 12 V
⟹ V = 2.88 V
Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.
Intext Questions and Answers V
1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.
Answer: The formula for the equivalent resistance (Rp) of resistors connected in parallel is: 1/Rp = 1/R1 + 1/R2 + 1/R3 + …
(a) For 1 Ω and 10⁶ Ω in parallel:
Given, R1 = 1 Ω
R2 = 10⁶ Ω
We know,
1/Rp = 1/R1 + 1/R2
⟹ 1/Rp = 1/1 + 1/10⁶
⟹ 1/Rp = (10⁶ + 1) / 10⁶
⟹ Rp = 10⁶ / (10⁶ + 1)
⟹ Rp = 1000000 / 1000001
⟹ Rp = 0.999999 Ω or 1 Ω
(b) For 1 Ω, 10³ Ω, and 10⁶ Ω in parallel:
Give,
R1 = 1 Ω
R2 = 10³ Ω
R3 = 10⁶ Ω
We know,
1/Rp = 1/R1 + 1/R2 + 1/R3
⟹ 1/Rp = 1/1 + 1/10³ + 1/10⁶
⟹ 1/Rp = 1/1 + 1/1000 + 1/1000000
⟹ 1/Rp = (1000000 + 1000 + 1) / 1000000
⟹ 1/Rp = 1001001 / 1000000
⟹ Rp = 1000000 / 1001001
⟹ Rp = 0.999 Ω or 1 Ω
2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer: Given:
Resistance of lamp, R_lamp = 100 Ω
Resistance of toaster, R_toaster = 50 Ω
Resistance of water filter, R_filter = 500 Ω
Voltage of the source, V = 220 V
First, we calculate the total current drawn by the three appliances connected in parallel. The total current is the sum of the currents through each appliance.
Current through lamp, I_lamp = V / R_lamp
⟹ I_lamp = 220 V / 100 Ω
⟹ I_lamp = 2.2 A
Current through toaster, I_toaster = V / R_toaster
⟹ I_toaster = 220 V / 50 Ω
⟹ I_toaster = 4.4 A
Current through water filter, I_filter = V / R_filter
⟹ I_filter = 220 V / 500 Ω
⟹ I_filter = 0.44 A
Total current,
I_total = I_lamp + I_toaster + I_filter
I_total = 2.2 A + 4.4 A + 0.44 A = 7.04 A
The electric iron takes the same amount of current.
Current through the iron, I_iron = 7.04 A.
Now, we find the resistance of the electric iron using Ohm’s law (R = V/I).
Resistance of the iron, R_iron = V / I_iron
⟹ R_iron = 220 V / 7.04 A
⟹ R_iron = 31.25 Ω
The resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.
3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: Connecting electrical devices in parallel with a battery instead of in series has several advantages:
One advantage is that a parallel circuit divides the current through the electrical gadgets. This is helpful particularly when each gadget has different resistance and requires different current to operate properly. In a series circuit, the current is constant throughout, which is impracticable if devices need widely different currents.
Another advantage is related to component failure. In a series circuit, when one component fails, the circuit is broken and none of the components works. In a parallel circuit, if one device fails (opens), the other devices connected in parallel can continue to operate independently as they still have a complete path for current.
Additionally, the total resistance in a parallel circuit is decreased. Each device in a parallel circuit gets the full battery voltage, allowing them to operate at their intended power.
4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer: Given resistors: R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω.
(a) To get a total resistance of 4 Ω:
This requires a mixed combination. Let’s connect the 3 Ω and 6 Ω resistors in parallel and then connect this combination in series with the 2 Ω resistor.
First, calculate the equivalent resistance of the parallel combination (Rp) of 3 Ω and 6 Ω:
1/Rp = 1/3 + 1/6
⟹ 1/Rp = (2 + 1) / 6
⟹ 1/Rp = 3/6
⟹ 1/Rp = 1/2
⟹ Rp = 2 Ω
Now, add the resistance of the 2 Ω resistor in series with this combination:
Total Resistance, R_total = Rp + 2 Ω
⟹ R_total = 2 Ω + 2 Ω
⟹ R_total = 4 Ω
So, to get 4 Ω, connect the 3 Ω and 6 Ω resistors in parallel, and connect this combination in series with the 2 Ω resistor.
(b) To get a total resistance of 1 Ω:
Let’s try connecting all three resistors in parallel.
⟹ 1/R_total = 1/2 + 1/3 + 1/6
⟹ 1/R_total = (3 + 2 + 1) / 6
⟹ 1/R_total = 6/6
⟹ 1/R_total = 1
⟹ R_total = 1 Ω
So, to get 1 Ω, connect all three resistors (2 Ω, 3 Ω, and 6 Ω) in parallel.
5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer: Given resistors: R1 = 4 Ω, R2 = 8 Ω, R3 = 12 Ω, R4 = 24 Ω.
(a) The highest total resistance:
To obtain the highest resistance, connect the resistors in series.
R_highest = R1 + R2 + R3 + R4
⟹ R_highest = 4 Ω + 8 Ω + 12 Ω + 24 Ω
⟹ R_highest = 48 Ω
(b) The lowest total resistance:
To obtain the lowest resistance, connect the resistors in parallel.
1/R_lowest = 1/R1 + 1/R2 + 1/R3 + 1/R4
⟹ 1/R_lowest = 1/4 + 1/8 + 1/12 + 1/24
⟹ 1/R_lowest = (6 + 3 + 2 + 1) / 24
⟹ 1/R_lowest = 12 / 24 = 1/2
⟹ R_lowest = 2 Ω
Intext Questions and Answers VI
1. Why does the cord of an electric heater not glow while the heating element does?
Answer: The cord of an electric heater does not glow while the heating element does because the heat produced in a resistor is directly proportional to its resistance for a given current, according to Joule’s law of heating. The heating element is made of a material with high resistance, so it becomes very hot and glows when current passes through it; this heating effect is utilised in devices such as electric heaters. The cord, however, is made of a material with very low resistance, such as copper (which has low resistivity and is generally used for electrical transmission lines), and therefore, it generates significantly less heat for the same current and does not glow.
2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer: The heat generated can be computed using the relationship where the work done in moving a charge Q through a potential difference V is VQ, and this energy is dissipated as heat.
Given:
Charge, Q = 96000 C
Potential difference, V = 50 V
Time, t = 1 hour = 3600 s (Note: time is not directly needed if using H = VQ)
The heat generated H is given by:
H = VQ
Substituting the given values:
H = 50 V × 96000 C
H = 4,800,000 J
Thus, the heat generated is 4,800,000 joules.
3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer: The heat developed can be calculated using Joule’s law of heating, which states that the amount of heat H produced in time t is H = I²Rt.
Given:
Resistance, R = 20 Ω
Current, I = 5 A
Time, t = 30 s
The heat developed H is given by:
H = I²Rt
Substituting the given values:
H = (5 A)² × 20 Ω × 30 s
H = 25 A² × 20 Ω × 30 s
H = 500 × 30 J
H = 15000 J
Thus, the heat developed in 30 s is 15000 joules.
Intext Questions and Answers VII
1. What determines the rate at which energy is delivered by a current?
Answer: The rate at which electric energy is dissipated or consumed in an electric circuit is termed electric power. The power P is given by P = VI, or P = I²R = V²/R.
2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer: The formula for electric power (P) is: P = V × I
Substitute the given values into the formula:
P = 220 V × 5 A
⟹ P = 1100 W
The formula for electrical energy (E) is: E = P × t
t = 2 hours
t = 2 × 60 minutes/hour × 60 seconds/minute
⟹ t = 7200 s
Now,
E = 1100 W × 7200 s
E = 1100 J/s × 7200 s
⟹ E = 7,920,000 J
This can also be expressed in scientific notation: E = 7.92 × 10⁶ J
Exercise Questions and Answers
1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is –
(a) 1/25
(b) 1/5
(c) 5 (d) 25
Answer: Let the initial resistance of the wire be R.
When the wire is cut into five equal parts, the resistance of each part will be R/5, because resistance is directly proportional to length.
Let R₁ = R₂ = R₃ = R₄ = R₅ = R/5.
These five parts are connected in parallel. The equivalent resistance R’ is given by the formula:
1/R’ = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅
⟹ 1/R’ = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
⟹ 1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R
⟹ 1/R’ = 5 × (5/R)
⟹ 1/R’ = 25/R
⟹ R’ = R/25
We need to find the ratio R/R’.
R/R’ = R / (R/25)
⟹ R/R’ = R × (25/R)
⟹ R/R’ = 25
Therefore, the correct option is (d) 25.
2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer: The electrical power P is given by P = VI. Alternatively, using Ohm’s Law (V=IR), we can write P = (IR)I = I²R, and P = V(V/R) = V²/R. The term IR² is not included among these expressions for electrical power.
Therefore, the correct option is (b) IR²
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer: First, we calculate the resistance of the bulb, which remains constant.
Rated Voltage (V_rated) = 220 V
Rated Power (P_rated) = 100 W
Using the formula P = V²/R, we find the resistance R:
R = V_rated² / P_rated
⟹ R = (220 V)² / 100 W
⟹ R = 48400 / 100 Ω
⟹ R = 484 Ω
Now, the bulb is operated at a new voltage (V_new) = 110 V.
The new power consumed (P_new) is:
P_new = V_new² / R
⟹ P_new = (110 V)² / 484 Ω
⟹ P_new = 12100 / 484 W
⟹ P_new = 25 W
Therefore, the correct option is (d) 25 W.
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: Let the resistance of each of the two identical wires be R.
The potential difference (V) is the same in both cases.
The formula for heat produced is H = (V²/R_total) × t.
Case 1: Series Combination
The total resistance in series is R_s = R + R = 2R.
The heat produced in series is H_s = (V² / R_s) × t = (V² / 2R) × t.
Case 2: Parallel Combination
The total resistance in parallel is given by 1/R_p = 1/R + 1/R = 2/R ⟹ R_p = R/2.
The heat produced in parallel is H_p = (V² / R_p) × t = (V² / (R/2)) × t = (2V² / R) × t.
Ratio of Heat Produced
We need to find the ratio H_s / H_p.
H_s / H_p = [ (V² / 2R) × t ] / [ (2V² / R) × t ]
H_s / H_p = (V² / 2R) × (R / 2V²)
H_s / H_p = 1 / 4
⟹ H_s : H_p = 1:4
Therefore, the correct option is (c) 1:4.
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.
6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer: Part 1: Finding the length of the wire.
Given:
Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m
Radius (r) = d/2 = 0.25 × 10⁻³ m
Resistivity (ρ) = 1.6 × 10⁻⁸ Ω m
Resistance (R) = 10 Ω
First, calculate the cross-sectional area (A):
A = πr² = 3.14 × (0.25 × 10⁻³ m)²
⟹ A = 3.14 × 0.0625 × 10⁻⁶ m²
⟹ A = 0.19625 × 10⁻⁶ m²
Now, use the resistance formula R = ρ(l/A) to find the length (l):
l = (R × A) / ρ
⟹ l = (10 Ω × 0.19625 × 10⁻⁶ m²) / (1.6 × 10⁻⁸ Ω m)
⟹ l = (1.9625 × 10⁻⁶) / (1.6 × 10⁻⁸) m
⟹ l = 122.7 m
Part 2: Change in resistance when the diameter is doubled.
Resistance is inversely proportional to the area of cross-section (R ∝ 1/A), and area is proportional to the square of the diameter (A ∝ d²).
Therefore, R ∝ 1/d².
If the diameter is doubled (d’ = 2d), the new resistance (R’) will be:
R’ ∝ 1/(2d)² ∝ 1/(4d²)
This means R’ = R/4.
New resistance R’ = 10 Ω / 4 = 2.5 Ω.
The resistance changes from 10 Ω to 2.5 Ω, becoming one-fourth of its original value.
7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –
| I (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
| V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
Calculation of Resistance:
The resistance (R) is the slope of the V-I graph.
Slope = ΔV / ΔI
Let’s take two points from the data, for instance, the first and the last:
R = (13.2 V – 1.6 V) / (4.0 A – 0.5 A)
⟹ R = 11.6 V / 3.5 A
⟹ R = 3.31 Ω
8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer: Given:
Voltage (V) = 12 V
Current (I) = 2.5 mA = 0.0025 A
Using Ohm’s Law, R = V/I:
R = 12 V / 0.0025 A
⟹ R = 4800 Ω or 4.8 kΩ
9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer: In a series circuit, the current is the same through all components.
First, calculate the total equivalent resistance (R_s):
R_s = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
⟹ R_s = 13.4 Ω
Now, calculate the total current (I) using Ohm’s Law, I = V/R_s:
I = 9 V / 13.4 Ω
⟹ I = 0.67 A
Since the current is the same everywhere in a series circuit, the current flowing through the 12 Ω resistor is 0.67 A.
10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer: Let ‘n’ be the number of resistors required.
Given:
Voltage (V) = 220 V
Total Current (I) = 5 A
Resistance of one resistor (R₁) = 176 Ω
First, find the total equivalent resistance (R_eq) required for the circuit:
R_eq = V / I
⟹ R_eq = 220 V / 5 A
⟹ R_eq = 44 Ω
For ‘n’ identical resistors connected in parallel, the equivalent resistance is R_eq = R₁ / n.
n = R₁ / R_eq
⟹ n = 176 Ω / 44 Ω
⟹ n = 4
Four resistors are required.
11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer: Let the three resistors be R₁, R₂, and R₃, each being 6 Ω.
(i) To get a total resistance of 9 Ω:
Connect two resistors (R₁ and R₂) in parallel, and then connect this combination in series with the third resistor (R₃).
Resistance of the parallel part (R_p):
1/R_p = 1/R₁ + 1/R₂
⟹ R_p = 1/6 + 1/6
⟹ R_p = 2/6
⟹ R_p = 1/3
⟹ R_p = 3 Ω.
Total resistance (R_total) = R_p + R₃ = 3 Ω + 6 Ω = 9 Ω.
(ii) To get a total resistance of 4 Ω:
Connect two resistors (R₁ and R₂) in series, and then connect this combination in parallel with the third resistor (R₃).
Resistance of the series part (R_s):
R_s = R₁ + R₂
= 6 Ω + 6 Ω
= 12 Ω.
Total resistance (R_total):
1/R_total = 1/R_s + 1/R₃
⟹ 1/R_total = 1/12 + 1/6
⟹ 1/R_total = (1 + 2)/12
⟹ 1/R_total = 3/12
⟹ 1/R_total = 1/4
⟹ R_total = 4 Ω.
12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer: Let ‘n’ be the number of lamps.
Given:
Power of one bulb (P) = 10 W
Voltage (V) = 220 V
Maximum allowable current (I_max) = 5 A
First, find the current drawn by a single bulb (I_bulb):
I_bulb = P / V
⟹ I_bulb = 10 W / 220 V
⟹ I_bulb = 0.04545 A
Now,
n × I_bulb = I_max
⟹ n = I_max / I_bulb
⟹ n = 5 A / 0.04545 A
⟹ n = 110
A maximum of 110 lamps can be connected.
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer: Given:
V = 220 V
R_A = 24 Ω
R_B = 24 Ω.
Case 1: Used separately
The current will be the same for each coil.
I_separate = V / R_A
⟹ I_separate = 220 V / 24 Ω
⟹ I_separate = 9.17 A
Case 2: Used in series
Total resistance R_s = R_A + R_B
⟹ R_s = 24 Ω + 24 Ω
⟹ R_s = 48 Ω.
Current I_series = V / R_s
⟹ I_series = 220 V / 48 Ω
⟹ I_series = 4.58 A
Case 3: Used in parallel
Total resistance 1/R_p = 1/R_A + 1/R_B
⟹ R_p = 1/24 + 1/24
⟹ R_p = 2/24
⟹ R_p = 1/12
⟹ R_p = 12 Ω.
Current I_parallel = V / R_p
⟹ I_parallel = 220 V / 12 Ω
⟹ I_parallel = 18.33 A
14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer: Circuit (i):
V = 6 V.
Resistors 1 Ω and 2 Ω are in series.
Total resistance R_s = 1 Ω + 2 Ω = 3 Ω.
Current in the circuit I₁ = V / R_s = 6 V / 3 Ω = 2 A.
The current through the 2 Ω resistor is 2 A.
Power used in the 2 Ω resistor is P₁ = I₁² × R
⟹ P₁ = (2 A)² × 2 Ω
⟹ P₁ = 8 W.
Circuit (ii):
V = 4 V.
Resistors 12 Ω and 2 Ω are in parallel.
The voltage across the 2 Ω resistor is the same as the battery voltage, which is 4 V.
Power used in the 2 Ω resistor is P₂ = V² / R
⟹ P₂ = (4 V)² / 2 Ω
⟹ P₂ = 16 / 2 W
⟹ P₂ = 8 W.
Comparison: The power used in the 2 Ω resistor is the same in both circuits (P₁ = P₂ = 8 W). The ratio of power used is 1:1.
Extras
Additional MCQs (Knowledge Based)
1. What is the SI unit of electric charge?
A. Coulomb
B. Ampere
C. Volt
D. Ohm
Answer: A. Coulomb
40. In an electric circuit, the direction of conventional current is taken as the direction of flow of:
A. Electrons
B. Negative charges
C. Neutral particles
D. Positive charges
Answer: D. Positive charges
Additional MCQs (Competency Based)
1. Assertion (A): In an electric circuit, the conventional direction of current is considered from the positive terminal to the negative terminal of the source.
Reason (R): When electricity was first discovered, electrons had not yet been identified, and current was assumed to be the flow of positive charges.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.
24. A student observes that the cord of an electric heater does not glow while its heating element does. This is because:
(a) The cord has a much higher resistance than the heating element.
(b) The heating element has a much higher resistance than the cord.
(c) The current in the cord is less than the current in the heating element.
(d) The cord is better insulated than the heating element.
Answer: (b) The heating element has a much higher resistance than the cord.
Additional Questions and Answers
1. What is electric current?
Answer: If an electric charge flows through a conductor, for example, through a metallic wire, we say that there is an electric current in the conductor. Electric current is expressed by the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges.
28. Explain why domestic electrical appliances are connected in parallel rather than in series.
Answer: Domestic electrical appliances are connected in parallel rather than in series for several important reasons:
- Independent Operation and Correct Current: In a parallel circuit, each appliance can be operated independently with its own switch. A parallel circuit divides the current through the electrical gadgets. This is helpful, particularly when each gadget has different resistance and requires different current to operate properly. In a series circuit, the current is constant throughout, which is impracticable for appliances needing widely different current values.
- Continued Operation if One Fails: If one appliance in a parallel circuit fails or is switched off, the other appliances continue to operate. In a series circuit, if one component fails, the circuit is broken, and none of the components work.
- Constant Voltage: Each appliance in a parallel circuit receives the same voltage from the source (e.g., 220 V in domestic supply lines). This ensures that all appliances operate at their rated voltage.
- Lower Overall Resistance: The total resistance in a parallel circuit is decreased, which allows for a larger total current to be drawn from the supply if needed, without excessive voltage drop across the connecting wires.
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