Formula
Power in AC circuit
P_avg = V_rms I_rms cos φ. cos φ is the power factor. Pure R: cos φ = 1; pure L or C: cos φ = 0 (wattless current — no average power dissipated).
-- NCERT Class 12 Physics, Ch. 7, p. 191Power in AC
8 MCQs1 revision card9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Power in AC, explained for NEET
The costly habit in AC power is treating it like DC: multiplying the two meter readings and stopping. A circuit drawing 5.0 A at 2.0 × 10² V is not consuming 1.0 × 10³ W unless the phase angle happens to be zero. That product is apparent power, measured in volt-amperes. Real power — the heat the resistor actually delivers, the number on the electricity bill — carries a third factor.
NCERT Class 12 Physics Chapter 7, page 191, gives it as
P_avg = V_rms · I_rms · cos φ
with φ the phase angle between source voltage and circuit current, and cos φ called the power factor. Three readings are involved, not two. Note what each symbol must be: RMS values, not peak. Substituting peak values inflates the answer by a factor of 2.
Why the cosine appears: instantaneous power p = vi oscillates, and over a full cycle the part of the current in quadrature with the voltage contributes nothing net — it returns as much energy to the source as it draws. Only the in-phase component does sustained work. The average of that product over one cycle is exactly V_rms I_rms cos φ.
Three consequences worth holding:
- A pure resistor has φ = 0, so cos φ = 1 and P = V_rms I_rms. This is the only case where the DC-style product is right.
- An ideal inductor or capacitor has φ = ±90°, so cos φ = 0 and average power is zero, whatever the current. (The wattless-current lesson develops this.)
- cos φ is never negative for a passive circuit, and never exceeds 1. An answer outside [0, 1] means an arithmetic slip, not a strange circuit.
The formula assumes steady-state single-frequency sinusoidal drive. Watch the sign convention too: φ is the voltage-minus-current phase, but since cosine is even, a leading and a lagging circuit with the same |φ| dissipate identical power. Sign confusion here changes nothing; forgetting the cosine changes everything.
Can you answer these Power in AC 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
In the relation P_avg = V_rms I_rms cos φ for a steady-state AC circuit, the quantity cos φ is called the
Show answer and why every option is right or wrong
Answer: C. cos φ is named the power factor; NCERT Class 12 Physics Chapter 7, page 191, introduces the term alongside the average-power expression.
Why A is wrong: A is wrong because the form factor is the ratio of RMS to average-rectified value of a waveform, a property of the waveform shape rather than of the voltage-current phase relationship.
Why B is wrong: B is wrong because the quality factor describes the sharpness of a resonance curve, a separate quantity that does not appear in the average-power expression.
Why D is wrong: D is wrong because no standard quantity is called the impedance ratio, and cos φ relates phase to power rather than comparing two impedances.
MCQ 2Easy RecallPractice
The SI unit in which average power in an AC circuit is expressed is the
Show answer and why every option is right or wrong
Answer: B. Average power is real power and is measured in watts; NCERT Class 12 Physics Chapter 7, page 191, expresses P_avg in watts.
Why A is wrong: A is wrong because the volt-ampere is reserved for apparent power, the bare product V_rms I_rms before the power factor is applied.
Why C is wrong: C is wrong because the volt-ampere reactive labels reactive power, the component associated with energy shuttled back and forth rather than dissipated.
Why D is wrong: D is wrong because the joule is a unit of energy, not of power; power is energy per unit time.
MCQ 3Easy RecallPractice
For a passive AC circuit the power factor cos φ can take values only within the range
Show answer and why every option is right or wrong
Answer: A. A passive circuit cannot generate energy, so its average power is non-negative and cos φ lies between 0 and 1 inclusive, as follows from P_avg = V_rms I_rms cos φ in NCERT Class 12 Physics Chapter 7, page 191.
Why B is wrong: B is wrong because a negative power factor would mean average power flowing back to the source, which a passive circuit containing only R, L and C cannot sustain.
Why C is wrong: C is wrong because it quotes the range of the angle φ itself rather than of its cosine; cos φ is a dimensionless number, not an angle in radians.
Why D is wrong: D is wrong because the cosine of any real angle is bounded by 1 in magnitude, so no choice of circuit elements can push cos φ outside that bound.
MCQ 4Direct ApplicationPractice
An AC circuit operates at an RMS voltage of 2.0 × 10² V and draws an RMS current of 5.0 A with a power factor of 0.60. The average power consumed is
Show answer and why every option is right or wrong
Answer: B. P_avg = V_rms I_rms cos φ = (2.0 × 10²)(5.0)(0.60) = 6.0 × 10² W, applying the average-power relation from NCERT Class 12 Physics Chapter 7, page 191.
Why A is wrong: A is wrong because it is the apparent power V_rms I_rms with the cos φ factor omitted, which is the central error this topic guards against.
Why C is wrong: C is wrong because it divides by the power factor instead of multiplying, inverting the correction rather than applying it.
Why D is wrong: D is wrong because 3.0 × 10² W is half the correct power, as if the power factor were 0.30. P = V I cos φ = 2.0 × 10² × 5.0 × 0.60 = 6.0 × 10² W.
MCQ 5Direct ApplicationPractice
A device draws an RMS current of 4.0 A from a 2.5 × 10² V RMS supply and dissipates an average power of 5.0 × 10² W. Its power factor is
Show answer and why every option is right or wrong
Answer: D. Rearranging the relation of NCERT Class 12 Physics Chapter 7, page 191, cos φ = P_avg/(V_rms I_rms) = 5.0 × 10²/[(2.5 × 10²)(4.0)] = 5.0 × 10²/1.0 × 10³ = 0.50.
Why A is wrong: A is wrong because it divides the average power by 2.0 × 10³, doubling the apparent power in the denominator.
Why B is wrong: B is wrong because it takes the ratio of current to voltage-derived figures incorrectly; the correct denominator is the full product V_rms I_rms = 1.0 × 10³ V·A.
Why C is wrong: C is wrong because a unity power factor would require the average power to equal the apparent power of 1.0 × 10³ W, twice the stated dissipation.
MCQ 6Concept TrapPractice
Two AC circuits are driven by identical sources and draw equal RMS currents. In circuit P the current lags the voltage by 30°; in circuit Q the current leads the voltage by 30°. Their average power consumptions compare as
Show answer and why every option is right or wrong
Answer: C. The power factor depends on φ only through cos φ, which is an even function, so a lead and a lag of equal magnitude give identical average power under the relation of NCERT Class 12 Physics Chapter 7, page 191.
Why A is wrong: A is wrong because the direction of the phase shift does not determine whether energy is dissipated; the magnitude of the phase angle alone sets the power factor.
Why B is wrong: B is wrong because average power is taken over a complete cycle, so the point within the cycle at which energy transfer peaks is irrelevant to the average.
Why D is wrong: D is wrong because cos(−30°) = cos(30°) = 0.87, a positive value; a leading phase does not produce a negative power factor.
MCQ 7CalculationPractice
An AC source of RMS voltage 1.2 × 10² V drives a circuit in which the current is 3.0 A RMS, lagging the voltage by 60°. If the phase lag is then reduced to 0° while the RMS voltage and RMS current are held unchanged, the average power increases by
Show answer and why every option is right or wrong
Answer: A. Initially P = (1.2 × 10²)(3.0)cos 60° = (3.6 × 10²)(0.50) = 1.8 × 10² W; finally P = (3.6 × 10²)(1) = 3.6 × 10² W, so the increase is 1.8 × 10² W, using NCERT Class 12 Physics Chapter 7, page 191.
Why B is wrong: B is wrong because 3.6 × 10² W is the final average power itself, not the increase from the initial value of 1.8 × 10² W.
Why C is wrong: C is wrong because it doubles the apparent power, which no combination of the given quantities produces.
Why D is wrong: D is wrong because the power factor rises from 0.50 to 1.00, so the average power cannot stay unchanged.
MCQ 8Direct ApplicationPractice
A load draws an RMS current of 2.0 A from a supply and dissipates 1.9 × 10² W with a power factor of 0.95. The supply RMS voltage is then doubled and the RMS current rises to 3.0 A, the power factor staying at 0.95. The new average power is closest to
Show answer and why every option is right or wrong
Answer: D. Since P = V_rms I_rms cos φ and cos φ is unchanged, the new power is the old power scaled once by the voltage ratio (2×) and once by the current ratio (3.0/2.0 = 1.5×), applied together as a single use of the proportionality: P_new = 1.9 × 10² × 2 × 1.5 = 5.7 × 10² W, per NCERT Class 12 Physics Chapter 7, page 191.
Why A is wrong: A is wrong because it keeps the original voltage of 1.0 × 10² V while updating only the current, ignoring the stated doubling of the supply voltage.
Why B is wrong: B is wrong because it doubles the original power, updating the voltage but leaving the current at 2.0 A.
Why C is wrong: C is wrong because 4.3 × 10² W is 1.9 × 10² × (3.0/2.0)²: it scales the power by I² as if the load's resistance were fixed, ignoring that the voltage doubled. The first reading gives V = 1.9 × 10²/(2.0 × 0.95) = 1.0 × 10² V, so the new power is 2.0 × 10² × 3.0 × 0.95 ≈ 5.7 × 10² W.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Power in AC: quick recall before you leave
How do you solve a Power in AC question? A worked example
- 1
Given
V_rms = 2.20 × 10² V
I_rms = 8.0 A
φ = 60° (current lags voltage) — angle treated as exact
cos 60° = 0.500 (exact mathematical value) - 2
Required
The average power dissipated in the circuit, and the apparent power, so the two can be compared.
- 3
Concept
Instantaneous power in an AC circuit oscillates. Averaged over a complete cycle, only the component of current in phase with the voltage contributes; the quadrature component returns as much energy to the source as it draws. The average of the product over one cycle is the RMS voltage times the RMS current times the cosine of the phase angle between them. The bare product V_rms I_rms is the apparent power and is an upper bound, attained only when φ = 0.
- 4
Formula
P_avg = V_rms · I_rms · cos φ
Apparent power S = V_rms · I_rms - 5
Substitution
P_avg = (2.20 × 10² V)(8.0 A)(0.500)
S = (2.20 × 10² V)(8.0 A) - 6
Calculation
S = 2.20 × 10² × 8.0 = 1.76 × 10³ V·A
P_avg = 1.76 × 10³ × 0.500 = 8.8 × 10² W
The angle 60° is an exact problem-defined value and cos 60° = 0.500 is an exact mathematical constant; neither contributes to the significant-figure count. The measured quantities are 2.20 × 10² V (3 s.f.) and 8.0 A (2 s.f.), so the result carries 2 significant figures. - 7
Final answer
P_avg = 8.8 × 10² W, against an apparent power of 1.8 × 10³ V·A. The circuit is billed for roughly half of what the two meter readings alone suggest.
- 8
Common trap
Reporting 1.8 × 10³ W — the apparent power — as the average power. The units are the giveaway: apparent power is quoted in volt-amperes, real power in watts, and they coincide only at unity power factor. A second variant is substituting peak values for RMS: with V₀ = √2 V_rms and I₀ = √2 I_rms, that inflates the answer by exactly 2. Both errors produce options that sit ready in NEET distractor sets.
- 9
Similar NEET-style question
An AC circuit operating at 2.4 × 10² V RMS draws 5.0 A RMS at a power factor of 0.80. Find (a) the apparent power and (b) the average power, and state the unit of each. *(Answers: 1.2 × 10³ V·A; 9.6 × 10² W.)*
What to remember before solving Power in AC questions
Which Power in AC formulas do you need for NEET?
1 formula — click to collapse
Average AC power
Average power dissipated in AC circuit. cos(phi) = power factor.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P_avg | average power | W |
| V_rms, I_rms | RMS | V, A |
| phi | phase angle | rad |
Valid when
- Steady-state AC
- Pure sinusoid
More in Electromagnetic Induction and Alternating Currents: 4 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Power in AC questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 20 past-paper questions from Electromagnetic Induction and Alternating Currents →
Sources
NCERT refs: Class 12 Physics Chapter 7, p.191
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →