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. 191Wattless Current
8 MCQs1 revision card9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsOfficial key: NTA-verifiedLast updated: 23 Sep 2026
Wattless Current, explained for NEET
Several amperes flow through the coil, yet the wattmeter reads nothing. That is not a fault in the meter. The common slip on this topic is arithmetic rather than conceptual: asked for the wattless current, students compute I_rms cos φ — the component that does carry power — and report that.
Split the RMS current into two components measured against the applied voltage. The component in phase with the voltage, I_rms cos φ, is the active or "watt-full" current; multiplied by V_rms it gives the average power, P_avg = V_rms I_rms cos φ (NCERT Class 12 Physics, Chapter 7, printed page 192). The component at 90° to the voltage, I_rms sin φ, is the wattless current. Over a complete cycle it drives energy into the reactive element for one quarter cycle and takes all of it back during the next, so its net contribution to average power is exactly zero.
Two limiting cases are worth fixing. In an ideal inductor or capacitor — no resistance — φ = ±90°, cos φ = 0, and the entire current is wattless: the source delivers zero average power however large that current is. At resonance in a series circuit the opposite holds, φ = 0, and the wattless component vanishes.
In NEET questions you are usually handed two of {V_rms, I_rms, P_avg, power factor} and asked for the wattless component. The routine is short: get cos φ, then sin φ = √(1 − cos²φ), then multiply by I_rms.
"Wattless" describes the ideal reactance only. A real coil has resistance, and that resistance dissipates power through its own in-phase component.
Watch-out: sin and cos are not interchangeable here. cos φ → power. sin φ → wattless current. If your answer equals the power factor times the current, you have computed the other component.
Can you answer these Wattless Current 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
Wattless current is the name given to
Show answer and why every option is right or wrong
Answer: C. The wattless component is the part of the current 90° out of phase with the applied voltage; since average power carries a cos φ factor, this component contributes nothing to it — see the average-power relation in NCERT Class 12 Physics, Chapter 7, printed page 192.
Why A is wrong: A is wrong because a purely resistive circuit has φ = 0, so its current is entirely in phase with the voltage and dissipates the full V_rms I_rms — the opposite of wattless.
Why B is wrong: B is wrong because unity power factor means cos φ = 1 and sin φ = 0, so the wattless component is zero there, not maximum.
Why D is wrong: D is wrong because a wattless current is a steady-state sinusoidal current that persists indefinitely; the name refers to zero average power, not to short duration.
MCQ 2Easy RecallPractice
In a circuit where the RMS current I_rms lags the applied voltage by a phase angle φ, the wattless component of the current is
Show answer and why every option is right or wrong
Answer: D. Resolving the current about the voltage direction gives an in-phase part I_rms cos φ and a quadrature part I_rms sin φ; the quadrature part is the wattless component, consistent with P_avg = V_rms I_rms cos φ in NCERT Class 12 Physics, Chapter 7, printed page 192.
Why A is wrong: A is wrong because I_rms cos φ is the active (in-phase) component — the one that does deliver power.
Why B is wrong: B is wrong because tan φ gives the net-reactance-to-resistance ratio, not a current component; it is unbounded and can exceed I_rms.
Why C is wrong: C is wrong because dividing by cos φ produces a value larger than the RMS current itself, which no component of that current can be.
MCQ 3Easy RecallPractice
Over one complete cycle, the average power drawn by an ideal resistance-free inductor connected to an AC source is
Show answer and why every option is right or wrong
Answer: A. For an ideal inductor φ = 90°, so cos φ = 0 and P_avg = V_rms I_rms cos φ = 0; energy taken from the source in one quarter cycle is returned in the next. See NCERT Class 12 Physics, Chapter 7, printed page 192.
Why B is wrong: B is wrong because V_rms I_rms is the apparent power; the average power is that quantity multiplied by cos φ, which is zero at 90°.
Why C is wrong: C is wrong because the factor of one-half belongs to peak-value forms of AC power, not to the RMS form, and in any case cos 90° = 0.
Why D is wrong: D is wrong because dividing by √2 is a peak-to-RMS conversion, not a power factor; the average power here is exactly zero.
MCQ 4Direct ApplicationPractice
An AC circuit draws an RMS current of 5.0 A, the current lagging the applied voltage by 60°. The wattless component of this current is closest to
Show answer and why every option is right or wrong
Answer: B. The wattless component is I_rms sin φ = 5.0 × sin 60° = 5.0 × 0.866 ≈ 4.3 A, using the quadrature decomposition behind P_avg = V_rms I_rms cos φ (NCERT Class 12 Physics, Chapter 7, printed page 192).
Why A is wrong: A is wrong because 5.0 × cos 60° = 2.5 A is the active component; the wattless component takes sin φ, not cos φ.
Why C is wrong: C is wrong because the whole current is wattless only when φ = 90°, and here φ is 60°.
Why D is wrong: D is wrong because the wattless component vanishes only at φ = 0, when the current is entirely in phase with the voltage.
MCQ 5Direct ApplicationPractice
A load draws an RMS current of 1.0 × 10¹ A at a power factor of 0.80, the current lagging. The wattless component of the current is
Show answer and why every option is right or wrong
Answer: D. cos φ = 0.80 gives sin φ = √(1 − 0.64) = 0.60, so the wattless component is 1.0 × 10¹ × 0.60 = 6.0 A — the part of the current that drops out of P_avg = V_rms I_rms cos φ (NCERT Class 12 Physics, Chapter 7, printed page 192).
Why A is wrong: A is wrong because I_rms/cos φ is not a component of the current; no component can exceed the RMS current itself.
Why B is wrong: B is wrong because the entire current would be wattless only at zero power factor, not at 0.80.
Why C is wrong: C is wrong because 1.0 × 10¹ × 0.80 = 8.0 A is the active component, obtained by using cos φ where sin φ is required.
MCQ 6Direct ApplicationPractice
A capacitor of negligible resistance is connected across an AC source of RMS voltage 2.0 × 10² V and draws an RMS current of 4.0 A. The average power drawn from the source over a complete cycle is
Show answer and why every option is right or wrong
Answer: A. For a pure capacitor the current leads the voltage by 90°, so cos φ = 0 and P_avg = V_rms I_rms cos φ = 0; the full 4.0 A is wattless current. See NCERT Class 12 Physics, Chapter 7, printed page 192.
Why B is wrong: B is wrong because V_rms I_rms = 8.0 × 10² W is the apparent power; average power needs the cos φ factor, which is zero here.
Why C is wrong: C is wrong because halving the apparent power corresponds to a power factor of 0.5, not to the 90° phase difference of a pure capacitor.
Why D is wrong: D is wrong because dividing by √2 is a peak-to-RMS conversion applied where none is needed — both quantities given are already RMS.
MCQ 7CalculationPractice
A circuit draws an RMS current of 5.0 A from a source of RMS voltage 2.0 × 10² V. The wattless component of the current is measured as 3.0 A. The average power consumed by the circuit is closest to
Show answer and why every option is right or wrong
Answer: C. The two components are perpendicular, so the active component is √(5.0² − 3.0²) = 4.0 A, and P_avg = V_rms × (active component) = 2.0 × 10² × 4.0 = 8.0 × 10² W, applying P_avg = V_rms I_rms cos φ from NCERT Class 12 Physics, Chapter 7, printed page 192.
Why A is wrong: A is wrong because it multiplies the source voltage by the wattless component 3.0 A; only the in-phase component carries power.
Why B is wrong: B is wrong because it treats the whole 5.0 A as in phase, which would require a power factor of 1.
Why D is wrong: D is wrong because it halves the correct result, importing a factor of one-half that belongs to peak-value power expressions.
MCQ 8CalculationPractice
In a series AC circuit the RMS current is 4.0 A and lags the source voltage by 30°. The source frequency is then tuned to bring the circuit to resonance, the RMS current still being 4.0 A. The wattless component of the current changes from
Show answer and why every option is right or wrong
Answer: B. Initially the wattless component is 4.0 × sin 30° = 2.0 A; at resonance the current is in phase with the voltage, so φ = 0, sin φ = 0 and the wattless component is zero — all the current then becomes the active component in P_avg = V_rms I_rms cos φ (NCERT Class 12 Physics, Chapter 7, printed page 192).
Why A is wrong: A is wrong because the wattless component falls to zero at resonance, where φ = 0; it cannot rise to the full current.
Why C is wrong: C is wrong because 4.0 × cos 30° = 3.5 A is the active component at 30°, not the wattless one; the final value of zero is right but the initial value is not.
Why D is wrong: D is wrong because at resonance the current is entirely in phase with the voltage, so the wattless component is zero, not the full 4.0 A.
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Wattless Current: quick recall before you leave
How do you solve a Wattless Current question? A worked example
- 1
Given
V_rms = 2.20 × 10² V; I_rms = 2.50 A; P_avg = 3.30 × 10² W. All three are three-significant-figure measurements.
- 2
Required
The power factor cos φ, and the wattless component I_rms sin φ, in A.
- 3
Concept
The RMS current splits into a component in phase with the voltage and a component at 90° to it. Only the in-phase component appears in the average power; the quadrature component is the wattless current.
- 4
Formula
P_avg = V_rms I_rms cos φ, so cos φ = P_avg / (V_rms I_rms).
Then sin φ = √(1 − cos²φ), and the wattless component is I_rms sin φ. - 5
Substitution
cos φ = (3.30 × 10²) / [(2.20 × 10²) × 2.50]
- 6
Calculation
Denominator: (2.20 × 10²) × 2.50 = 5.50 × 10² V·A
cos φ = (3.30 × 10²) / (5.50 × 10²) = 0.600
sin φ = √(1 − 0.600²) = √(1 − 0.360) = √0.640 = 0.800
Wattless component = I_rms sin φ = 2.50 × 0.800 = 2.00 A
The 1 inside √(1 − cos²φ) is an exact mathematical constant from the Pythagorean identity and does not limit the significant figures; the three-figure input data sets the precision. - 7
Final answer
Power factor cos φ = 0.600; wattless component of the current = 2.00 A.
Check: the active component is 2.50 × 0.600 = 1.50 A, and (2.20 × 10²) × 1.50 = 3.30 × 10² W, which matches the wattmeter reading. - 8
Common trap
Reporting 1.50 A as the wattless current. That is I_rms cos φ, the component that carries the whole 3.30 × 10² W. The wattless component uses sin φ. A quick self-check: the wattless component and the active component must combine as √(1.50² + 2.00²) = 2.50 A, the stated RMS current.
- 9
Similar NEET-style question
An AC source of RMS voltage 1.00 × 10² V supplies a circuit drawing 3.00 A RMS at a power factor of 0.600. Find the wattless component of the current and the average power. *(Answer: sin φ = 0.800, wattless component = 2.40 A; P_avg = 1.80 × 10² W.)*
What to remember before solving Wattless Current questions
Which Wattless Current 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.
Wattless Current 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
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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