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NCERT Class 10 Science Electricity: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 10 Science book, chapter “Electricity”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   A wire of resistance R is cut into five equal parts joined in parallel, giving R′. The ratio R/R′ is (a) 1/25 (b) 1/5 (c) 5 (d) 25.

   1. The original wire has resistance R.
   2. Cutting it into five equal parts makes each part have resistance R/5.
   3. When five equal resistors are joined in parallel, the equivalent resistance R' is given by 1/R' = 5/(R/5).
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   4. Simplifying the equation gives 1/R' = 25/R, so R' = R/25.
   5. The ratio R/R' is therefore R/(R/25) = 25.
   6. This matches option (d).

   Answer: (d) 25

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2. Exercises Q2

   Which does not represent electrical power: (a) I²R (b) IR² (c) VI (d) V²/R?

   1. Electrical power is defined as the product of voltage and current, so P = VI.
      why?

      Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

   2. Using Ohm's law V = IR, we can substitute V in the power formula to get P = (IR)I = I²R.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   3. Alternatively, substituting I = V/R into P = VI gives P = V(V/R) = V²/R.
      why?

      Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

   4. Option (a) I²R and option (c) VI are valid expressions for power.
   5. Option (d) V²/R is also a valid expression for power.
   6. Option (b) IR² does not match any standard formula for electrical power.
   7. Therefore, IR² is the expression that does not represent electrical power.

   Answer: (b) IR² does not represent electrical power

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3. Exercises Q3

   A bulb rated 220 V, 100 W runs on 110 V. The power used is (a) 100 W (b) 75 W (c) 50 W (d) 25 W.

   1. The resistance of the bulb is constant and can be found using the rated values with the formula R = V²/P.
   2. Substituting the rated voltage and power gives R = 220²/100 = 48400/100 = 484 Ω.
   3. When the bulb is connected to 110 V, the actual power consumed is calculated using P = V²/R.
      why?

      Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

   4. Substituting the new voltage and resistance gives P = 110²/484 = 12100/484 = 25 W.

   Answer: (d) 25 W

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4. Exercises Q4

   Two identical wires are connected first in series, then in parallel, across the same voltage. The ratio of heat produced (series : parallel) is (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1.

   1. Let the resistance of each identical wire be R and the applied voltage be V.
   2. When connected in series, the total resistance is R_s = R + R = 2R.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

   3. The heat produced in series is H_s = V²t / R_s = V²t / 2R.
   4. When connected in parallel, the total resistance is R_p = (R × R) / (R + R) = R/2.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   5. The heat produced in parallel is H_p = V²t / R_p = V²t / (R/2) = 2V²t / R.
      why?

      Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

   6. The ratio of heat produced in series to parallel is H_s : H_p = (V²t / 2R) : (2V²t / R).
   7. Simplifying the ratio gives 1/2 : 2, which is 1 : 4.
   8. This matches option (c).

   Answer: (c) 1 : 4

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5. Exercises Q5

   How is a voltmeter connected to measure the potential difference between two points?

   1. A voltmeter is always connected in parallel with the component or between the two points where the potential difference is to be measured.
   2. This connection ensures that the voltmeter does not alter the current flowing through the main circuit.
   3. The positive terminal of the voltmeter is connected to the point at higher potential and the negative terminal to the point at lower potential.

   Answer: A voltmeter is connected in parallel across the two points to measure the potential difference.

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6. Exercises Q6

   A copper wire of diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ω m must have resistance 10 Ω. Find its length. How does the resistance change if the diameter is doubled?

   1. The formula for resistance is R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area.
      why?

      R = ρl/A: a wire's resistance grows with its length and falls as its cross-section grows; ρ (resistivity) depends only on the material and temperature.

   2. The diameter d is 0.5 mm, so the radius r is 0.25 mm, which equals 0.25 × 10⁻³ m.
   3. The cross-sectional area A = πr² = π × (0.25 × 10⁻³)² ≈ 1.9635 × 10⁻⁷ m² (using π ≈ 3.1416).
   4. Rearranging the formula for length gives L = RA/ρ.
   5. Substituting the values, L = (10 × 1.9625 × 10⁻⁷) / (1.6 × 10⁻⁸).
   6. So L = (10 × 1.9635 × 10⁻⁷) ÷ (1.6 × 10⁻⁸) ≈ 122.72 m (about 122.7 m).
   7. If the diameter is doubled, the new radius is 0.5 mm, so the new area A' is π(0.5 × 10⁻³)² = 7.85 × 10⁻⁷ m².
   8. Since area is proportional to the square of the diameter, doubling the diameter makes the area 4 times larger.
   9. Resistance is inversely proportional to area, so the new resistance R' is R/4.
   10. Calculating the new resistance, R' = 10/4 = 2.5 Ω.

   Answer: Length = 122.72 m; New resistance = 2.5 Ω

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7. Exercises Q7

   For a resistor, I = 0.5, 1.0, 2.0, 3.0, 4.0 A gives V = 1.6, 3.4, 6.7, 10.2, 13.2 V. Plot V against I and find the resistance.

   1. List the given pairs of current (I) and voltage (V): (0.5, 1.6), (1.0, 3.4), (2.0, 6.7), (3.0, 10.2), (4.0, 13.2).
   2. Draw a graph with current I on the x-axis and voltage V on the y-axis.
   3. Plot the five points on the graph and draw a straight line of best fit passing through the origin.
   4. The graph is a straight line passing through the origin, which confirms that the resistor obeys Ohm's law.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   5. Calculate the resistance using the formula R = V ÷ I for each data point.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   6. For I = 0.5 A and V = 1.6 V, R = 1.6 ÷ 0.5 = 3.2 Ω.
   7. For I = 1.0 A and V = 3.4 V, R = 3.4 ÷ 1.0 = 3.4 Ω.
   8. For I = 2.0 A and V = 6.7 V, R = 6.7 ÷ 2.0 = 3.35 Ω.
   9. For I = 3.0 A and V = 10.2 V, R = 10.2 ÷ 3.0 = 3.4 Ω.
   10. For I = 4.0 A and V = 13.2 V, R = 13.2 ÷ 4.0 = 3.3 Ω.
   11. The calculated values of resistance are approximately 3.2 Ω, 3.4 Ω, 3.35 Ω, 3.4 Ω, and 3.3 Ω.
   12. Taking the average of these values gives a resistance of about 3.3 Ω.

   Answer: about 3.3 Ω

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8. Exercises Q8

   A 12 V battery across an unknown resistor gives 2.5 mA. Find the resistance.

   1. The voltage of the battery is given as V = 12 V.
   2. The current flowing through the resistor is I = 2.5 mA.
   3. Convert the current from milliamperes to amperes: I = 2.5 × 10⁻³ A = 0.0025 A.
   4. According to Ohm's law, resistance R is equal to voltage divided by current: R = V / I.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   5. Substitute the values into the formula: R = 12 / 0.0025.
   6. Calculate the result: R = 4800 Ω.

   Answer: The resistance of the unknown resistor is 4800 Ω.

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9. Exercises Q9

   A 9 V battery is in series with 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω resistors. What current flows through the 12 Ω resistor?

   1. The total resistance of the series circuit is the sum of all individual resistances.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

   2. R_total = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω.
   3. Using Ohm's law, the current in the circuit is I = V ÷ R_total.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   4. I = 9 ÷ 13.4 ≈ 0.67 A.
   5. In a series circuit, the current is the same through every component.
   6. Therefore, the current through the 12 Ω resistor is also 0.67 A.

   Answer: 0.67 A

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10. Exercises Q10

    How many 176 Ω resistors in parallel carry 5 A on a 220 V line?

    1. The total resistance of the parallel combination is found using Ohm's law, R = V ÷ I.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    2. Substituting the given values, R = 220 ÷ 5 = 44 Ω.
    3. For n identical resistors of resistance R₀ in parallel, the equivalent resistance is R = R₀ ÷ n.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    4. Rearranging the formula to solve for n gives n = R₀ ÷ R.
    5. Substituting the values, n = 176 ÷ 44 = 4.

    Answer: 4

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11. Exercises Q11

    How do you connect three 6 Ω resistors to get (i) 9 Ω (ii) 4 Ω?

    1. (i) Connect two 6 Ω resistors in parallel to get an equivalent resistance of 3 Ω.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    2. (i) Connect the third 6 Ω resistor in series with this 3 Ω combination.
       why?

       In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    3. (i) The total resistance is 6 + 3 = 9 Ω.
    4. (ii) Connect two 6 Ω resistors in series to get an equivalent resistance of 12 Ω.
       why?

       In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    5. (ii) Connect the third 6 Ω resistor in parallel with this 12 Ω combination.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    6. (ii) The total resistance is (12 × 6) ÷ (12 + 6) = 72 ÷ 18 = 4 Ω.

    Answer: (i) 9 Ω; (ii) 4 Ω

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12. Exercises Q12

    How many 10 W, 220 V lamps can be connected in parallel across a 220 V line if the maximum current is 5 A?

    1. The power of each lamp is 10 W and the voltage is 220 V.
    2. The current drawn by one lamp is I = P/V = 10/220 A.
    3. The maximum total current allowed on the line is 5 A.
    4. Let n be the number of lamps connected in parallel.
    5. The total current is the sum of currents through all lamps, so n × (10/220) = 5.
    6. Solving for n gives n = 5 × 220 / 10.
    7. n = 1100 / 10 = 110.
    8. Therefore, 110 lamps can be connected in parallel.

    Answer: 110 lamps

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13. Exercises Q13

    A hot plate on 220 V has two 24 Ω coils used separately, in series or in parallel. Find the current in each case.

    1. When coil A is used alone, the resistance is 24 Ω.
    2. Current I = V ÷ R = 220 ÷ 24 = 9.17 A.
       why?

       Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    3. When coil B is used alone, the resistance is 24 Ω.
    4. Current I = V ÷ R = 220 ÷ 24 = 9.17 A.
       why?

       Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    5. When coils are in series, total resistance R = 24 + 24 = 48 Ω.
       why?

       In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    6. Current I = V ÷ R = 220 ÷ 48 = 4.58 A.
       why?

       Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    7. When coils are in parallel, 1/R = 1/24 + 1/24 = 2/24.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    8. Total resistance R = 24 ÷ 2 = 12 Ω.
    9. Current I = V ÷ R = 220 ÷ 12 = 18.33 A.
       why?

       Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    Answer: (i) 9.17 A; (ii) 9.17 A; (iii) 4.58 A; (iv) 18.33 A

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14. Exercises Q14

    Compare the power in the 2 Ω resistor in (i) a 6 V battery in series with 1 Ω and 2 Ω (ii) a 4 V battery in parallel with 12 Ω and 2 Ω.

    1. (i) Total resistance in series is 1 Ω + 2 Ω = 3 Ω.
       why?

       In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    2. (i) Current through the circuit is I = V/R = 6 V / 3 Ω = 2 A.
       why?

       Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    3. (i) Power in the 2 Ω resistor is P = I²R = (2 A)² × 2 Ω = 8 W.
       why?

       Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

    4. (ii) In parallel, the voltage across the 2 Ω resistor is 4 V.
       why?

       In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

    5. (ii) Power in the 2 Ω resistor is P = V²/R = (4 V)² / 2 Ω = 16/2 = 8 W.
       why?

       Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

    6. The power in both cases is equal.

    Answer: (i) 8 W; (ii) 8 W; equal

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15. Exercises Q15

    Lamps of 100 W and 60 W (both 220 V) are in parallel on 220 V mains. What current is drawn?

    1. The total power consumed by the lamps in parallel is the sum of their individual powers.
    2. Total Power = 100 W + 60 W = 160 W.
    3. The voltage across the parallel combination is the mains voltage, which is 220 V.
    4. Using the formula Power = Voltage × Current, we can find the total current.
    5. Current = Total Power ÷ Voltage.
    6. Current = 160 W ÷ 220 V.
    7. Current ≈ 0.727 A.
    8. Rounding to two decimal places, the current is 0.73 A.

    Answer: 0.73 A

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16. Exercises Q16

    Which uses more energy: a 250 W TV in 1 hour or a 1200 W toaster in 10 minutes?

    1. Energy used by the TV is calculated as Power × Time.
    2. The TV runs for 1 hour, which is 3600 seconds.
    3. Energy for TV = 250 W × 3600 s = 900000 J.
    4. Convert 900000 J to kJ by dividing by 1000, giving 900 kJ.
    5. Energy used by the toaster is calculated as Power × Time.
    6. The toaster runs for 10 minutes, which is 600 seconds.
    7. Energy for toaster = 1200 W × 600 s = 720000 J.
    8. Convert 720000 J to kJ by dividing by 1000, giving 720 kJ.
    9. Compare the two energies: 900 kJ is greater than 720 kJ.

    Answer: TV; 900 kJ; 720 kJ

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17. Exercises Q17

    A 44 Ω heater draws 5 A for 2 hours. Find the rate at which heat is produced.

    1. Given resistance R = 44 Ω and current I = 5 A.
    2. The rate of heat production is the power P, calculated using the formula P = I²R.
       why?

       Electric power P = VI = I²R = V²/R; energy = power × time, and 1 kWh = 3.6 × 10⁶ J.

    3. Substitute the values: P = (5)² × 44.
    4. Calculate the square of the current: 5² = 25.
    5. Multiply by the resistance: 25 × 44 = 1100.
    6. The unit of power is Watts (W).

    Answer: 1100 W

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18. Exercises Q18

    Explain: (a) why tungsten is used for lamp filaments (b) why heating devices use alloys, not pure metals (c) why series wiring is not used at home (d) how a wire's resistance varies with its area of cross-section (e) why copper and aluminium are used for transmission wires.

    1. (a) Tungsten has a very high melting point, so it does not melt when the filament gets extremely hot.
    2. (b) Alloys have higher resistivity than pure metals, so they produce more heat, and they do not oxidise (burn) easily at high temperatures.
       why?

       Oxidation is gain of oxygen (or loss of hydrogen); reduction is loss of oxygen (or gain of hydrogen). In a redox reaction one substance is oxidised while another is reduced.

    3. (c) In series wiring, the voltage is shared among appliances, so they do not get the full supply voltage needed to work properly.
    4. (d) Resistance is inversely proportional to the area of cross-section, so a thicker wire has lower resistance.
    5. (e) Copper and aluminium have low resistivity, which minimizes energy loss as heat during long-distance transmission.
       why?

       R = ρl/A: a wire's resistance grows with its length and falls as its cross-section grows; ρ (resistivity) depends only on the material and temperature.

    Answer: (a) High melting point; (b) Higher resistivity; (c) Voltage sharing; (d) Inversely proportional; (e) Low resistivity

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