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Electrical energy and power; series/parallel resistor combinations: NEET MCQs with answers

88 questions on this topic of Current Electricity (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Consider the following two statements regarding electrical power and heating: Statement I: For a given potential difference, the power dissipated in a resistor is inversely proportional to its resistance. Statement II: When electric current passes through a conductor, the rate of heat production depends only on the current and not on the resistance of the conductor. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because P = V²/R, so for a fixed V, power P is inversely proportional to R. Statement II is incorrect because Joule's heating rate is P = I²R, which directly depends on the resistance of the conductor as well as the current.

medium

2. A bulb is rated 100 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 110 V is:

  1. 100 W
  2. 50 W
  3. 25 W
  4. 400 W
Show answer and explanation

Answer: (C) 25 W

R = V²/P is fixed, so P' = P(V'/V)² = 100 × (110/220)² = 25 W.

hard

3. A variable resistor R is connected across a source of constant emf E and internal resistance r. The power dissipated in the variable resistor R is 16 W when R = 4 Ω and also 16 W when R = 9 Ω. What is the value of the internal resistance r and the emf E?

  1. r = 5 Ω, E = 18 V
  2. r = 6 Ω, E = 20 V
  3. r = 6 Ω, E = 15 V
  4. r = 4.5 Ω, E = 24 V
Show answer and explanation

Answer: (B) r = 6 Ω, E = 20 V

The power dissipated in an external resistor R connected to a cell (E, r) is P = E²R / (R + r)². For two different external resistances R₁ and R₂ giving the same power dissipation, the internal resistance is the geometric mean: r = √(R₁R₂) = √(4 × 9) = 6 Ω. Using P = E²R₁ / (R₁ + r)², we get 16 = E²(4) / (4 + 6)² = 4E² / 100, which gives E² = 400, so E = 20 V.

medium

4. Resistors of 4 Ω and 11 Ω are connected in parallel, and this combination is connected in series with a 7 Ω resistor. The equivalent resistance is:

  1. 9.93 Ω
  2. 2.93 Ω
  3. 22 Ω
  4. 4.77 Ω
Show answer and explanation

Answer: (A) 9.93 Ω

Parallel part = 4×11/15 = 2.93 Ω. Adding 7 Ω in series gives 9.93 Ω.

easy

5. Three identical resistors are connected to a constant voltage source. In combination X, all three are connected in parallel. In combination Y, two are in parallel and the third is connected in series with this pair. What is the ratio of total power consumed in combination X to that in combination Y?

  1. 2 : 9
  2. 9 : 2
  3. 3 : 2
  4. 4 : 9
Show answer and explanation

Answer: (B) 9 : 2

Let each resistor have resistance R and source voltage be V. For combination X (three parallel), R_eq = R/3, so P_X = V² / (R/3) = 3V²/R. For combination Y, R_eq = R/2 + R = 3R/2, so P_Y = V² / (3R/2) = 2V²/(3R). The ratio P_X / P_Y = (3) / (2/3) = 9/2.

hard

6. Two resistors R and 2R are connected in parallel across an ideal battery. If the power dissipated in R is P, the total power supplied by the battery is:

  1. 3P/2
  2. 2P
  3. 3P
  4. P/2
Show answer and explanation

Answer: (A) 3P/2

Both parallel resistors have the same voltage V. P_R=V²/R=P. For 2R, power=V²/(2R)=P/2. Total=P+P/2=3P/2.

Quick tip: Parallel branches share voltage, not current.

NCERT Class 12, Chapter 3, page 84

medium

7. An electric tea maker operates at 220 V and draws 4 A of current. It takes 5 minutes to heat a certain amount of water to boiling point. If the operational cost of electricity is ₹ 6.00 per kilowatt-hour (kWh), what is the cost of boiling water in this appliance 10 times?

  1. ₹ 2.20
  2. ₹ 4.40
  3. ₹ 0.44
  4. ₹ 8.80
Show answer and explanation

Answer: (B) ₹ 4.40

Power of the tea maker P = V × I = 220 V × 4 A = 880 W = 0.88 kW. Time for one boiling cycle = 5 minutes = 5/60 h = 1/12 h. Energy consumed per cycle = 0.88 kW × (1/12) h = 0.88 / 12 kWh. For 10 cycles, total energy = 10 × (0.88 / 12) kWh = 8.8 / 12 kWh = 11/15 kWh ≈ 0.7333 kWh. Total cost = (11/15) × 6.00 = 66 / 15 = ₹ 4.40.

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8. A bulb is rated 60 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 110 V is:

  1. 240 W
  2. 60 W
  3. 30 W
  4. 15 W
Show answer and explanation

Answer: (D) 15 W

R = V²/P is fixed, so P' = P(V'/V)² = 60 × (110/220)² = 15 W.

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9. A bulb is rated 100 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 120 V is:

  1. 54.5 W
  2. 100 W
  3. 29.8 W
  4. 336 W
Show answer and explanation

Answer: (C) 29.8 W

R = V²/P is fixed, so P' = P(V'/V)² = 100 × (120/220)² = 29.8 W.

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10. A bulb is rated 40 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 165 V is:

  1. 30 W
  2. 22.5 W
  3. 40 W
  4. 71.1 W
Show answer and explanation

Answer: (B) 22.5 W

R = V²/P is fixed, so P' = P(V'/V)² = 40 × (165/220)² = 22.5 W.

medium

11. A bulb is rated 40 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 176 V is:

  1. 25.6 W
  2. 62.5 W
  3. 40 W
  4. 32 W
Show answer and explanation

Answer: (A) 25.6 W

R = V²/P is fixed, so P' = P(V'/V)² = 40 × (176/220)² = 25.6 W.

medium

12. A bulb is rated 200 W at 220 V. Assuming its resistance stays constant, the power it consumes when connected to 120 V is:

  1. 59.5 W
  2. 200 W
  3. 672 W
  4. 109 W
Show answer and explanation

Answer: (A) 59.5 W

R = V²/P is fixed, so P' = P(V'/V)² = 200 × (120/220)² = 59.5 W.

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