PhysicsISC Class 12 · Alternating Current

Power in AC Circuits: Average Power, Power Factor and Wattless Current

Q: Why does a pure inductor take current but no power?

Its current is 90° out of phase with the voltage. In one quarter cycle it takes energy into its magnetic field and in the next it returns it, so the average is zero.

Q: What is the difference between apparent power and true power?

Apparent power V_{rms}I_{rms} (VA) is what the supply must carry; true power V_{rms}I_{rms}\cos\phi (W) is what is actually converted to heat or work.

Q: Why is a choke preferred to a resistor for dropping voltage?

A choke drops voltage through its reactance with a power factor near zero, so it wastes almost no energy, while a resistor turns the extra energy into heat.

Q: What does the ISC 2027 syllabus need on power?

P = \frac{1}{2}V_0I_0\cos\phi = V_{rms}I_{rms}\cos\phi = I_{rms}^2R, power factor \frac{R}{Z}, special cases for pure R, L and C, the choke coil (analytical only), wattless current and maximum power at resonance.

Q: How has power been tested recently?

ISC 2025 Question 18(ii) (5 marks) asked for the power factor of a circuit whose current lagged the emf by \frac{\pi}{3}, along with I_{rms}, frequency and impedance.

Updated 15 October 2026 · Fact-checked

The average power in an a.c. circuit is $P = V_{rms}I_{rms}\cos\phi = \frac{1}{2}V_0I_0\cos\phi$, where φ is the phase difference between voltage and current. The factor $\cos\phi = \frac{R}{Z}$ is the power factor. Power is used only in the resistance, so $P = I_{rms}^2R$; a pure inductor or capacitor uses none.

Only the part of the current along V (I cosφ) delivers power; the 90° part, I sinφ, is wattless, and cosφ = R/Z is the power factor.

Illustration drawn for QuestionBuddy.

Understand Power in AC Circuits and Power Factor

Instantaneous power. Let $v = V_{0} sin ω t$ and $i = I_{0} sin (ω t - ϕ)$ . The power at any instant is
$p = v i = V_{0} I_{0} sin ω t sin (ω t - ϕ) .$
Using $2 sin A sin B = cos (A - B) - cos (A + B)$ :
$p = \frac{V _{0} I _{0}}{2} [cos ϕ - cos (2 ω t - ϕ)] .$

Average over one cycle. The term $cos (2 ω t - ϕ)$ goes through two full cycles in one period, so its average is zero. Hence
$P = \frac{V _{0} I _{0}}{2} cos ϕ = \frac{V _{0}}{2} \cdot \frac{I _{0}}{2} cos ϕ = V_{r m s} I_{r m s} cos ϕ .$
The product $V_{r m s} I_{r m s}$ (in volt-ampere, VA) is the apparent power; P (in watt) is the true power.

Power factor. From the impedance triangle $cos ϕ = \frac{R}{Z}$ . Substituting $V_{r m s} = I_{r m s} Z$ :
$P = I_{r m s} Z \cdot I_{r m s} \cdot \frac{R}{Z} = I_{r m s}^{2} R .$
So only the resistance dissipates energy. The inductor and capacitor store energy for part of a cycle and give it back for the rest.

Special cases.

Pure R: φ = 0, $cos ϕ = 1$ , $P = V_{r m s} I_{r m s}$ .
Pure L or pure C: φ = ±90°, $cos ϕ = 0$ , P = 0 even though a current flows.
Series LCR at resonance: Z = R, $cos ϕ = 1$ , and P is maximum, $\frac{V _{r m s}^{2}}{R}$ .
General LCR: $0 < cos ϕ < 1$ .

Wattless current. Resolve $I_{r m s}$ (which lags V by φ) into two parts: $I_{r m s} cos ϕ$ along V and $I_{r m s} sin ϕ$ at 90° to V. Only the first carries power. The component $I_{r m s} sin ϕ$ is the wattless current: it flows in the wires but delivers no average power. In a pure inductor or capacitor the whole current is wattless.

Choke coil. To run a low-voltage device (say a lamp or a fluorescent tube) on the mains, the extra voltage must be dropped somewhere. A series resistor drops it but wastes $I^{2} R$ as heat. A choke, an inductor of large L and very small resistance, drops the voltage through its reactance $X_{L}$ , so its power factor is nearly 0 and it wastes almost no power. Only the small resistance of its wire dissipates energy.

Why power factor matters. For a fixed true power P at voltage V, the current is $I = \frac{P}{V c o s ϕ}$ . A low power factor needs a larger current, and the loss in the supply cables, $I^{2} R_{l in e}$ , rises. Supply companies therefore want loads with power factor close to 1. Adding a suitable capacitor across an inductive load supplies a leading wattless current that cancels the lagging one and raises the power factor.

Power in AC Circuits and Power Factor: formulas and key facts

Average power: $P = V_{r m s} I_{r m s} cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ$; Watts.
Power factor: $cos ϕ = \frac{R}{Z} = \frac{P}{V _{r m s} I _{r m s}}$; Between 0 and 1.
Power in R only: $P = I_{r m s}^{2} R$; L and C dissipate nothing.
Apparent power: $S = V_{r m s} I_{r m s}$; Volt-ampere (VA).
Wattless current: $I_{r m s} sin ϕ$; Carries no average power.
Power-carrying current: $I_{r m s} cos ϕ$; In phase with V.

How to solve Power in AC Circuits and Power Factor questions

Find the phase angle (or R and Z) first; the power always follows.

1From equations: read $V_{0}$ , $I_{0}$ and φ; then $P = \frac{1}{2} V_{0} I_{0} cos ϕ$ .
2From components: find Z, then $I_{r m s} = \frac{V _{r m s}}{Z}$ and $cos ϕ = \frac{R}{Z}$ .
3Compute $P = V_{r m s} I_{r m s} cos ϕ$ and check it equals $I_{r m s}^{2} R$ .
4From meter readings (V, I, P): $cos ϕ = \frac{P}{V I}$ , $R = \frac{P}{I ^{2}}$ , $Z = \frac{V}{I}$ , $X = Z^{2} - R^{2}$ .
5Wattless current = $I_{r m s} sin ϕ$ ; power-carrying part = $I_{r m s} cos ϕ$ .
6For a choke: find the total Z needed, subtract R by Pythagoras (not by plain subtraction) to get $X_{L}$ , then $L = \frac{X _{L}}{2 π f}$ .

Power in AC Circuits and Power Factor in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

What the electricity meter counts

Suppose a ceiling fan on 230 V (r.m.s.) mains draws 0.4 A with a power factor of 0.8 and runs for 10 hours.

Find the apparent power, the true power and the energy billed.

Try it first, then show the solutionSolution

Apparent power $= 230 \times 0.4 = 92$ VA.
True power $= 92 \times 0.8 = 73.6$ W.
Energy $= 73.6 \times 10 = 736$ Wh $= 0.736$ kWh (units).

Answer: 92 VA; 73.6 W; about 0.74 unit. The meter bills the true power.

Memory hook: Volt-amps flow in the wire; watts show up on the bill.

Why the supply company wants power factor near 1

Imagine a workshop that needs 4.6 kW from a 230 V supply through cables of total resistance 0.5 Ω, first with power factor 1 and then with power factor 0.5.

Compare the current and the heat lost in the cables.

Try it first, then show the solutionSolution

At cos φ = 1: $I = \frac{4600}{230} = 20$ A; loss $= 2 0^{2} \times 0.5 = 200$ W.
At cos φ = 0.5: $I = \frac{4600}{230 \times 0.5} = 40$ A; loss $= 4 0^{2} \times 0.5 = 800$ W.

Answer: Halving the power factor doubles the current and quadruples the cable loss (200 W to 800 W).

Memory hook: Low power factor: same work, more current, more waste.

Tricks to remember Power in AC Circuits and Power Factor

Short memory aids. Say each one aloud once, then use it on a question the same day.

Half, peak, peak, cos
$P = \frac{1}{2} V_{0} I_{0} cos ϕ$ ; with r.m.s. values the half disappears.
Use it when: Choosing the right form.
Only R pays the bill
$P = I^{2} R$ : L and C borrow energy and return it.
Use it when: Checking any power answer.
Cos for watts, sin for wattless
$I cos ϕ$ does the work; $I sin ϕ$ just circulates.
Use it when: Wattless current questions.
R over Z
Power factor is the base over the hypotenuse of the impedance triangle.
Use it when: Quick power factor.

Power in AC Circuits and Power Factor: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Power in a pure capacitor · 1 mark · MCQBoard level

The average power consumed by an ideal capacitor connected to an a.c. source over one cycle is:
(a) $V_{r m s} I_{r m s}$
(b) $\frac{1}{2} V_{0} I_{0}$
(c) zero
(d) $I_{r m s}^{2} X_{C}$

Try it first, then show the solutionSolution

φ = 90°, so $cos ϕ = 0$ and $P = V_{r m s} I_{r m s} cos 90° = 0$ .

Answer: (c).

Example 2: Power from readings and phase · 2 marks · Short answerBoard level

In a circuit, the r.m.s. voltage is 200 V, the r.m.s. current is 5 A and the current lags the voltage by 60°. Find the power factor and the average power.

Try it first, then show the solutionSolution

Power factor $= cos 60° = 0.5$ . (1)
$P = 200 \times 5 \times 0.5 = 500$ W. (1)

Answer: 0.5; 500 W.

Example 3: Derivation of average power · 3 marks · Long answerHarder board level

For $v = V_{0} sin ω t$ and $i = I_{0} sin (ω t - ϕ)$ , show that the average power over a cycle is $V_{r m s} I_{r m s} cos ϕ$ . Hence show that a pure inductor consumes no power.

Try it first, then show the solutionSolution

$p = v i = V_{0} I_{0} sin ω t sin (ω t - ϕ) = \frac{V _{0} I _{0}}{2} [cos ϕ - cos (2 ω t - ϕ)]$ . (1)
The average of $cos (2 ω t - ϕ)$ over a full cycle is zero, so $P = \frac{V _{0} I _{0}}{2} cos ϕ = V_{r m s} I_{r m s} cos ϕ$ . (1)
For a pure inductor φ = 90°: $cos ϕ = 0$ , so P = 0. (1)

Answer: $P = V_{r m s} I_{r m s} cos ϕ$ ; zero for a pure inductor.

Example 4: Power and wattless current in a coil · 3 marks · Short answerHarder board level

A coil of resistance 12 Ω and inductive reactance 16 Ω is connected to a 100 V (r.m.s.) a.c. supply. Find the current, the power factor, the average power and the wattless component of the current.

Try it first, then show the solutionSolution

$Z = 1 2^{2} + 1 6^{2} = 20$ Ω; $I = \frac{100}{20} = 5$ A. (1)
$cos ϕ = \frac{12}{20} = 0.6$ ; $P = 100 \times 5 \times 0.6 = 300$ W (check: $5^{2} \times 12 = 300$ W). (1)
$sin ϕ = \frac{16}{20} = 0.8$ ; wattless current $= 5 \times 0.8 = 4$ A. (1)

Answer: 5 A; 0.6; 300 W; 4 A.

Example 5: Coil constants from three meters · 4 marks · Long answerHOTS / competency

A coil is connected to a 100 V (r.m.s.), 50 Hz supply. An ammeter reads 2 A and a wattmeter reads 120 W. Find the power factor, the resistance, the reactance and the inductance of the coil.

Try it first, then show the solutionSolution

$cos ϕ = \frac{P}{V I} = \frac{120}{200} = 0.6$ . (1)
$R = \frac{P}{I ^{2}} = \frac{120}{4} = 30$ Ω; $Z = \frac{100}{2} = 50$ Ω (check: $\frac{R}{Z} = 0.6$ ). (1)
$X_{L} = 5 0^{2} - 3 0^{2} = 40$ Ω. (1)
$L = \frac{X _{L}}{2 π f} = \frac{40}{100 π} \approx 0.127$ H. (1)

Answer: 0.6; 30 Ω; 40 Ω; ≈ 0.127 H.

Example 6: Choke coil or resistor? · 4 marks · Case-basedHOTS / competency

Suppose a lamp rated 100 V, 50 W must be run from a 230 V (r.m.s.), 50 Hz supply.
(i) Find the lamp current and resistance. (1 mark)
(ii) Find the inductance of a pure choke in series that lets it work normally. (1 mark)
(iii) If a series resistor were used instead, find its value and the power it wastes. (1 mark)
(iv) Find the power factor with the choke and show the supply still delivers 50 W. (1 mark)

Try it first, then show the solutionSolution

(i) $I = \frac{50}{100} = 0.5$ A; $R = \frac{100}{0.5} = 200$ Ω.
(ii) Total $Z = \frac{230}{0.5} = 460$ Ω; $X_{L} = 46 0^{2} - 20 0^{2} \approx 414.2$ Ω; $L = \frac{414.2}{100 π} \approx 1.32$ H.
(iii) Total resistance 460 Ω, so extra R = 260 Ω, wasting $0. 5^{2} \times 260 = 65$ W.
(iv) $cos ϕ = \frac{200}{460} \approx 0.435$ ; $P = 230 \times 0.5 \times 0.435 \approx 50$ W, all in the lamp.

Answer: (i) 0.5 A, 200 Ω; (ii) ≈ 1.32 H; (iii) 260 Ω, 65 W wasted; (iv) ≈ 0.435, 50 W.

Where students lose marks in Power in AC Circuits and Power Factor

The slips examiners see most often, and what to write instead.

Mistake: Giving the phase angle instead of the power factor
The ISC 2025 examiners noted that candidates wrote the phase-angle formula when asked for the power factor.
Wrong working
Power factor $= \frac{π}{3}$ .
Right working
Power factor $= cos \frac{π}{3} = 0.5$ .
Do this: Power factor is $cos ϕ$ , a pure number between 0 and 1.
Usually costs: 1 mark.
Mistake: Mixing peak and r.m.s. values
Also noted in 2025: $I_{0}$ used where $I_{r m s}$ was needed.
Wrong working
$P = V_{0} I_{0} cos ϕ$ .
Right working
$P = \frac{1}{2} V_{0} I_{0} cos ϕ = V_{r m s} I_{r m s} cos ϕ$ .
Do this: The half appears only with peak values.
Usually costs: 1 mark.
Mistake: Using $P = I^{2} Z$
Z includes reactance, which dissipates nothing.
Do this: $P = I_{r m s}^{2} R$ .
Usually costs: 1 mark.
Mistake: Subtracting R from Z to get X
Z, R and X form a right triangle, not a line.
Do this: $X = Z^{2} - R^{2}$ .
Usually costs: 1 mark.

Exam tips

In the derivation, state clearly that the average of $cos (2 ω t - ϕ)$ over a cycle is zero.
Always cross-check $V_{r m s} I_{r m s} cos ϕ$ against $I_{r m s}^{2} R$ .
Explain a choke in two points: large $X_{L}$ limits the current; near-zero R means almost no power loss.
Give power in W, apparent power in VA, and power factor without a unit.

Practice questions on Power in AC Circuits and Power Factor

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The power factor of a purely resistive a.c. circuit is:
(a) 0
(b) 0.5
(c) 1
(d) infinite
Hint
φ = 0.
Answer
(c).
Q2 · 2 marks · Short answerBoard level
For $v = 100 sin ω t$ V and $i = 4 sin (ω t - \frac{π}{3})$ A, find the average power.
Hint
Use peak values with the factor one half.
Answer
$P = \frac{1}{2} \times 100 \times 4 \times cos 60° = 100$ W.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The power factor of a series LCR circuit is greatest at resonance.
Reason (R): At resonance the impedance equals the resistance.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$cos ϕ = \frac{R}{Z}$ .
Answer
(a).
Q4 · 3 marks · Short answerHarder board level
A series circuit has R = 40 Ω, $X_{L}$ = 50 Ω and $X_{C}$ = 20 Ω on 250 V (r.m.s.). Find the current, the power factor and the average power.
Hint
Z = 50 Ω.
Answer
5 A; 0.8; 1000 W (= $5^{2} \times 40$ ).
Q5 · 2 marks · Short answerHarder board level
An r.m.s. current of 10 A flows in a circuit of power factor 0.6. Find the power-carrying and wattless components of the current.
Hint
$sin ϕ = 1 - cos^{2} ϕ$ .
Answer
6 A and 8 A.
Q6 · 3 marks · Short answerHOTS / competency
A load draws 10 A and 1.2 kW from a 240 V (r.m.s.) supply. Find the power factor, the phase angle, the resistance and the reactance of the load.
Hint
$R = \frac{P}{I ^{2}}$ .
Answer
0.5; 60°; R = 12 Ω; Z = 24 Ω, X ≈ 20.8 Ω.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an inductive heater coil on a 220 V (r.m.s.), 50 Hz supply takes 2 kW with the current lagging by 45°.
(i) Find the current. (1 mark)
(ii) Find Z and R. (1 mark)
(iii) Find the inductance. (1 mark)
(iv) Find the wattless current and suggest how the power factor could be raised. (1 mark)
Hint
$I = \frac{P}{V c o s ϕ}$ .
Answer
(i) ≈ 12.9 A. (ii) Z ≈ 17.1 Ω, R ≈ 12.1 Ω. (iii) $X_{L} = R \approx 12.1$ Ω, L ≈ 38.5 mH. (iv) ≈ 9.1 A; connect a suitable capacitor to supply a leading wattless current that cancels it.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Average power in an a.c. circuit?Show answer
$P = V_{r m s} I_{r m s} cos ϕ$ .
2.Power factor in terms of R and Z?Show answer
$cos ϕ = \frac{R}{Z}$ .
3.Power in a pure inductor?Show answer
Zero.
4.What is wattless current?Show answer
The component $I_{r m s} sin ϕ$ , at 90° to the voltage, which carries no average power.
5.220 V, 2 A, power factor 0.5: power?Show answer
220 W.

Power in AC Circuits and Power Factor: frequently asked questions

Why does a pure inductor take current but no power?

Its current is 90° out of phase with the voltage. In one quarter cycle it takes energy into its magnetic field and in the next it returns it, so the average is zero.

What is the difference between apparent power and true power?

Apparent power V_{rms}I_{rms} (VA) is what the supply must carry; true power V_{rms}I_{rms}\cos\phi (W) is what is actually converted to heat or work.

Why is a choke preferred to a resistor for dropping voltage?

A choke drops voltage through its reactance with a power factor near zero, so it wastes almost no energy, while a resistor turns the extra energy into heat.

What does the ISC 2027 syllabus need on power?

P = \frac{1}{2}V_0I_0\cos\phi = V_{rms}I_{rms}\cos\phi = I_{rms}^2R, power factor \frac{R}{Z}, special cases for pure R, L and C, the choke coil (analytical only), wattless current and maximum power at resonance.

How has power been tested recently?

ISC 2025 Question 18(ii) (5 marks) asked for the power factor of a circuit whose current lagged the emf by \frac{\pi}{3}, along with I_{rms}, frequency and impedance.