Formula
Power dissipation
P = V I = I² R = V²/R. Energy dissipated as heat (Joule heating). Hourly: 1 kWh = 3.6 × 10⁶ J.
-- NCERT Class 12 Physics, Ch. 3, p. 93Electrical Energy Power
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2022, 2024Official key: NTA-verifiedLast updated: 21 Sep 2026
Electrical Energy Power, explained for NEET
The trap: picking the wrong form of the power formula. P = I²R says power grows with R; P = V²/R says power falls with R. Both are correct — they answer different questions. Use I²R when the current is the same for the resistors being compared (series). Use V²/R when the voltage is the same (parallel, or a load across a fixed supply). Grabbing I²R for a fixed-voltage comparison inverts the answer.
The concept. When a steady current I flows through a conductor with potential difference V across it, the energy dissipated in time Δt is ΔW = IVΔt, so the power is P = IV (NCERT Class 12 Physics Part I, Chapter 3, Section 3.9, page 92). For a resistor, substituting V = IR gives P = I²R = V²/R (page 93). This energy appears as heat in the resistor (Joule heating).
Why transmission lines use high voltage. A station sending power P through cables of resistance R_c at voltage V carries current I = P/V, so the cable loss is P_c = I²R_c = P²R_c/V² (NCERT, page 93). Loss falls as 1/V², which is why power is transmitted at high voltage.
Bridge to NEET. The in-scope question pattern compares heating in two resistors connected in series or in parallel. Series and parallel combinations are listed in the NTA NEET (UG) 2026 Physics syllabus, Unit 12; the standard results are: series — same current, R_eq = R₁ + R₂; parallel — same voltage, 1/R_eq = 1/R₁ + 1/R₂. Recent papers include a 2022 item on heat in 100 Ω and 200 Ω in parallel (ratio 2 : 1) and a 2024 item on two heaters in series versus parallel (ratio 2 : 9).
Watch-out: before you write a power formula, ask "what is common — current or voltage?" A lower-rated bulb has higher resistance (R = V²/P at its rated voltage), so in series it dissipates more power.
Can you answer these Electrical Energy Power MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A resistor of resistance R carries a steady current I with a potential difference V across it. Which expression gives the power dissipated in it?
Show answer and why every option is right or wrong
Answer: C. C is correct: P = IV, and with V = IR this becomes P = I²R (equivalently V²/R) (NCERT Class 12 Physics Part I, Chapter 3, page 93).
Why A is wrong: A is wrong because V²R multiplies by R instead of dividing; the voltage form is V²/R.
Why B is wrong: B is wrong because IR² squares the resistance instead of the current; the current form is I²R.
Why D is wrong: D is wrong because V/I is the resistance R (Ohm's relation), not a power; it has units of ohm, not watt.
MCQ 2Easy RecallPractice
According to NCERT, why is electrical power transmitted over long distances at high voltage?
Show answer and why every option is right or wrong
Answer: A. A is correct: I = P/V, so the cable loss P_c = I²R_c = P²R_c/V² falls as 1/V² (NCERT Class 12 Physics Part I, Chapter 3, page 93).
Why B is wrong: B is wrong because the power P to be delivered is fixed by demand; raising V lowers the current for that same P rather than increasing P.
Why C is wrong: C is wrong because R_c depends on the cable's material, dimensions and temperature, not on the applied voltage.
Why D is wrong: D is wrong because the loss P²R_c/V² is directly proportional to R_c; high voltage reduces the loss but does not remove its dependence on R_c.
MCQ 3Easy RecallPractice
A steady current I flows through a conductor with a potential difference V across it for a time t. The electrical energy dissipated is:
Show answer and why every option is right or wrong
Answer: D. D is correct: energy ΔW = IVΔt (NCERT Class 12 Physics Part I, Chapter 3, Section 3.9, page 92). Power is the rate, IV; energy is power × time.
Why A is wrong: A is wrong because it divides by time instead of multiplying; IV/t has units of W/s, not joule.
Why B is wrong: B is wrong because it uses V/I (the resistance) in place of the product IV; Vt/I has units of Ω·s, not joule.
Why C is wrong: C is wrong because IV is the power (rate of energy dissipation), not the energy; the time factor is missing.
MCQ 4Direct ApplicationPractice
A 6 Ω resistor is connected across an ideal 12 V battery. The power dissipated in the resistor is:
Show answer and why every option is right or wrong
Answer: B. B is correct: the voltage is given, so use P = V²/R = (12)²/6 = 144/6 = 24 W (NCERT Class 12 Physics Part I, Chapter 3, page 93).
Why A is wrong: A is wrong because 2 A is the current V/R = 12/6; stopping at the current and writing it in watts skips the power step.
Why C is wrong: C is wrong because 864 comes from V²R = 144 × 6; the voltage form divides by R, it does not multiply.
Why D is wrong: D is wrong because 72 comes from V × R = 12 × 6; power is V × I or V²/R, never V × R.
MCQ 5Direct ApplicationPractice
A steady current of 3 A flows through a 4 Ω resistor for 7 s. The heat produced is:
Show answer and why every option is right or wrong
Answer: D. D is correct: the current is given, so use energy = I²Rt = (3)² × 4 × 7 = 9 × 4 × 7 = 252 J (NCERT Class 12 Physics Part I, Chapter 3, pages 92–93).
Why A is wrong: A is wrong because 84 J = I × R × t = 3 × 4 × 7; the current was not squared.
Why B is wrong: B is wrong because 336 J = I × R² × t = 3 × 16 × 7; the resistance was squared instead of the current.
Why C is wrong: C is wrong because 36 = I²R = 9 × 4 is the power in watts; multiplying by the 7 s time was skipped.
MCQ 6Direct ApplicationPractice
A 3 Ω resistor and a 6 Ω resistor are connected in parallel across an ideal battery. The ratio of heat produced in the 3 Ω resistor to that in the 6 Ω resistor in the same time is:
Show answer and why every option is right or wrong
Answer: B. B is correct: in parallel the voltage is common, so use P = V²/R; P ∝ 1/R gives P₃ : P₆ = (1/3) : (1/6) = 2 : 1. Parallel combination is in the NTA NEET (UG) 2026 Physics syllabus, Unit 12; the power form is from NCERT Class 12 Physics Part I, Chapter 3, page 93.
Why A is wrong: A is wrong because 1 : 2 comes from using P = I²R with a common current (P ∝ R); in parallel the voltage, not the current, is common.
Why C is wrong: C is wrong because 4 : 1 treats power as proportional to 1/R²; with a common voltage P = V²/R is proportional to 1/R, not 1/R².
Why D is wrong: D is wrong because 1 : 4 both uses the wrong (common-current) form and squares the resistance ratio.
MCQ 7CalculationPractice
Bulb X is rated 25 W and bulb Y is rated 75 W, both at the same voltage V. They are connected in series across a supply of voltage V. Assuming the resistances stay constant, which bulb dissipates more power, and what is P_X : P_Y?
Show answer and why every option is right or wrong
Answer: C. C is correct. Step 1: at rated voltage R = V²/P, so R_X = V²/25 and R_Y = V²/75, giving R_X = 3R_Y. Step 2: in series the current is common, so use P = I²R; P ∝ R gives P_X : P_Y = 3 : 1. The lower-rated bulb X dissipates more (NCERT Class 12 Physics Part I, Chapter 3, page 93; series combination per the NTA NEET (UG) 2026 syllabus, Unit 12).
Why A is wrong: A is wrong because it reuses the rated powers 25 : 75 directly, which applies only when each bulb has voltage V across it; in series the current is common, so P ∝ R.
Why B is wrong: B is wrong because it squares the resistance ratio (3² = 9); with a common current P = I²R is proportional to R, not R².
Why D is wrong: D is wrong because it assumes equal current means equal power; equal current gives P ∝ R, and the resistances differ by a factor of 3.
MCQ 8CalculationPractice
A power station delivers a fixed power P through transmission cables of resistance R_c. If the transmission voltage is increased by a factor of 10, the power lost in the cables becomes:
Show answer and why every option is right or wrong
Answer: A. A is correct. Step 1: current I = P/V falls to 1/10. Step 2: loss P_c = I²R_c = P²R_c/V² falls to (1/10)² = 1/100 (NCERT Class 12 Physics Part I, Chapter 3, page 93).
Why B is wrong: B is wrong because it takes the loss as proportional to I (or 1/V); the loss is I²R_c, so the factor 1/10 must be squared.
Why C is wrong: C is wrong because it uses V²/R_c with the transmission voltage, but that voltage is not the voltage across the cables; the cable loss is I²R_c with I = P/V.
Why D is wrong: D is wrong because a fixed delivered power does not fix the cable loss; the loss depends on the current P/V, which changes with V.
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Electrical Energy Power: quick recall before you leave
How do you solve a Electrical Energy Power question? A worked example
- 1
Given
Two heaters rated 1 kW and 2 kW at the same supply voltage V. They are connected first in series and then in parallel across that same fixed supply V. Assume the heater resistances stay constant.
- 2
Required
Ratio of total power drawn in series to total power drawn in parallel, P_series : P_parallel.
- 3
Concept
The supply voltage is fixed, so compare total powers with P = V²/R_eq. Each heater's resistance comes from its rating: R = V²/P_rated. Series: R_eq = R₁ + R₂. Parallel: each heater has the full V across it, so it draws its rated power.
- 4
Formula
P = V²/R (NCERT Class 12 Physics Part I, Chapter 3, page 93); R_series = R₁ + R₂ (series combination, NTA NEET (UG) 2026 syllabus, Unit 12).
- 5
Substitution
R₁ = V²/1000 (1 kW = 1000 W exactly), R₂ = V²/2000.
Series: R_eq = V²/1000 + V²/2000 = 3V²/2000.
P_series = V² / (3V²/2000) = 2000/3 W.
Parallel: P_parallel = 1000 W + 2000 W = 3000 W. - 6
Calculation
P_series : P_parallel = (2000/3) : 3000 = 2000 : 9000 = 2 : 9.
The factors 1000 and 2000 are exact unit conversions from the kW ratings, and the ratio is a pure number, so they do not contribute to any significant-figure count. - 7
Final answer
P_series : P_parallel = 2 : 9
- 8
Common trap
Using P = I²R for the series case with an assumed "same current" in both arrangements gives P ∝ R_eq and inverts the result to 9 : 2. The current is not the same in the two arrangements; only the supply voltage is. Another slip is adding the ratings in series (1 + 2 = 3 kW) — ratings add only in parallel, where each heater gets its rated voltage.
(This series-versus-parallel heater comparison appeared in NEET 2024, with answer 2 : 9.) - 9
Similar NEET-style question
"A 3 Ω and a 6 Ω resistor are connected in series across a fixed ideal battery, and then in parallel across the same battery. Find the ratio of total power in series to total power in parallel."
Strategy: R_series = 9 Ω, R_parallel = 3 × 6/(3 + 6) = 2 Ω. Fixed V, so P ∝ 1/R_eq: P_series : P_parallel = 2 : 9.
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What to remember before solving Electrical Energy Power questions
Which Electrical Energy Power formulas do you need for NEET?
1 formula — click to collapse
Electrical power
Power dissipated as heat in resistor (Joule heating).
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | power | W |
| V | voltage | V |
| I | current | A |
| R | resistance | Ω |
Valid when
- Resistor (ohmic)
- DC or instantaneous
More in Current Electricity: 3 exam traps and mistakes · 6 formulas · 5 question patterns from its other lessons.
Electrical Energy Power questions from past NEET papers
2 questions from NEET 2022, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
11 : 1
22 : 9
31 : 2
42 : 3
NTA Answer: Option 2(revised_final)
NEET 2022
14 : 1
21 : 2
32 : 1
41 : 4
NTA Answer: Option 3(final)
How does NEET ask about Electrical Energy Power?
1 recurring pattern from past papers — click to collapse
Heating power: P = V^2/R = I^2*R. Compare two resistors in series/parallel.
Direct ApplicationEasy
Common distractors
uses wrong form
Picks I^2 R when V is constant or vice versa
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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