MHT-CET Physics · AC Circuits

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

Only the resistor takes power in an a.c. circuit: the average power is V_rms I_rms cos φ, where the power factor cos φ = R/Z — zero for a pure inductor or capacitor, one at resonance.

All AC Circuits notes MHT-CET Physics strategy

Why this matters

25 PYQs, one HARD. Three shapes: the average power from given equations or components, the power factor and how it changes when the frequency changes or a component is added, and the wattless current of a purely reactive circuit.

Concept 1 of 3: Average Power

Inductors and capacitors store energy for part of a cycle and hand it all back, so over a cycle only the resistor takes energy. Hence P = I²R, which is the same as V_rms I_rms cos φ.

Definition

$P = V_{\text{rms}}I_{\text{rms}}\cos\phi = \dfrac{V_0I_0}{2}\cos\phi = I_{\text{rms}}^2R$ .
From equations: $e = 160\sin(100t)$ , $i = 0.25\sin\left(100t + \dfrac{\pi}{3}\right)$ ⇒ $P = \dfrac{160 \times 0.25}{2} \times \dfrac{1}{2} = 10$ W.
LR circuit with $E = E_0\cos\omega t$ : $P = \dfrac{E_0^2R}{2Z^2}$ . $X_L = R$ ⇒ $\dfrac{E_0^2}{4R}$ ; $X_L = 2R$ ⇒ $\dfrac{E_0^2}{10R}$ ; $X_L = \sqrt{3}R$ ⇒ $\dfrac{E_0^2}{8R}$ .
Given $R$ and $Z$ : $P = \dfrac{V_{\text{rms}}^2R}{Z^2}$ .

Average power

P = V_{\text{rms}}I_{\text{rms}}\cos\phi = I_{\text{rms}}^2R

Worked example

v = 200\sin\omega t

volt and

i = 4\sin\left(\omega t - \dfrac{\pi}{3}\right)

A. Average power?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q31Moderate

Example 1 · AC Circuits · Power in AC Circuit — Average, Factor, Wattless

In LCR series circuit, an alternating e.m.f. 'e' and current 'i' are given by equations

e = 160\sin(100t)

V and

i = 250\sin\!\left(100t + \frac{\pi}{3}\right)

mA. The average power dissipated in the circuit is

Using peak values without the ½

P = \frac{V_0I_0}{2}\cos\phi

: the ½ turns two peaks into r.m.s. values. Dropping it doubles the power and lands on a printed option.

Drill 11 more on average power

Concept 2 of 3: Power Factor

The power factor cos φ = R/Z says what fraction of V_rms I_rms is actually used. Anything that moves the circuit toward resonance raises it; anything that adds net reactance lowers it.

Definition

$\cos\phi = \dfrac{R}{Z} = \dfrac{\text{true power}}{\text{apparent power}}$ ; apparent power $= V_{\text{rms}}I_{\text{rms}}$ , and their ratio the other way is $\dfrac{Z}{R}$ .
LR with power factor $\dfrac{1}{\sqrt{2}}$ ( $X_L = R$ ): frequency doubled ⇒ $X_L = 2R$ , factor $\dfrac{1}{\sqrt{5}}$ . CR with $\dfrac{1}{\sqrt{2}}$ : frequency HALVED ⇒ $X_C = 2R$ , also $\dfrac{1}{\sqrt{5}}$ .
Adding $X_C = X_L$ to an LR circuit makes the factor 1.
Removing L leaves the phase at $\dfrac{\pi}{3}$ , removing C also $\dfrac{\pi}{3}$ ⇒ $X_L = X_C$ , so with both the factor is 1.

Power factor

\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}

Worked example

R = 60\,\Omega

X_L = 100\,\Omega

X_C = 20\,\Omega

. Power factor?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q9Moderate

Example 2 · AC Circuits · Power in AC Circuit — Average, Factor, Wattless

The power factor of an R-L circuit is

\frac{1}{\sqrt{2}}

. If the frequency of AC is doubled the power factor will now be

Doubling the frequency of an RC circuit

Frequency UP lowers

X_C

and RAISES an RC circuit's power factor. To make

X_C = 2R

the frequency must be halved — the RC and LR versions of the same question move in opposite directions.

Drill 9 more on power factor

Concept 3 of 3: Wattless Current

In a pure inductor or capacitor the current is 90° out of step with the voltage, so the power it takes in one quarter-cycle it returns in the next. The current flows but no net power is used — a wattless current.

Definition

Pure $L$ or pure $C$ : $\phi = 90^\circ$ , power factor 0, average power 0.
$i = 5\sin\left(100t - \dfrac{\pi}{2}\right)$ with $e = 200\sin 100t$ : 0 W.
Wattless current: the component $I_{\text{rms}}\sin\phi$ , at $90^\circ$ to the voltage.

Purely reactive circuit

\phi = 90^\circ \;\Rightarrow\; P = V_{\text{rms}}I_{\text{rms}}\cos 90^\circ = 0

Worked example

An ideal inductor carries 3 A r.m.s. on 230 V. Average power?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q4Easy

Example 3 · AC Circuits · Power in AC Circuit — Average, Factor, Wattless

Average power associated with an ideal inductor and ideal capacitor over a complete cycle of a.c. is respectively

Current means power

A pure coil can carry amperes and consume nothing. Multiplying

V_{\text{rms}}I_{\text{rms}}

without

\cos\phi

gives the apparent power, not the power used.

Drill 2 more on wattless current

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Average Power
Average power
$P = V_{\text{rms}}I_{\text{rms}}\cos\phi = I_{\text{rms}}^2R$
Power Factor
Power factor
$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$
Wattless Current
Purely reactive circuit
$\phi = 90^\circ \;\Rightarrow\; P = V_{\text{rms}}I_{\text{rms}}\cos 90^\circ = 0$

Watch out for (3)

Using peak values without the ½
→ Average Power
Doubling the frequency of an RC circuit
→ Power Factor
Current means power
→ Wattless Current

Test yourself on AC Circuits

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the AC Circuits chapter test Drill all 25 questions

Related notes

Series LCR — Impedance and Phase →