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Physics · Alternating Current

Power in an AC Circuit and Power Factor

When an alternating voltage drives a current that is out of phase with it, the energy actually consumed is not simply the product of voltage and current. NCERT §7.7 introduces the average power $P = V_{\text{rms}} I_{\text{rms}} \cos\varphi$ and names the factor $\cos\varphi$ the power factor. This subtopic builds that result from the instantaneous power, develops the impedance-triangle reading of $\cos\varphi = R/Z$, and works through the four standard cases that NEET tests every year — including the recurring power-loss numericals.

From Instantaneous to Average Power

A sinusoidal voltage $v = v_m \sin\omega t$ applied to a series LCR circuit drives a current $i = i_m \sin(\omega t + \varphi)$, where $i_m = v_m / Z$ and $\varphi = \tan^{-1}\!\big[(X_C - X_L)/R\big]$ is the phase difference between the source voltage and the current. The instantaneous power delivered by the source is the product of the two:

$$p = v\,i = v_m i_m \,\sin\omega t\,\sin(\omega t + \varphi).$$

Expanding the product of sines with a standard identity rewrites this as the sum of a constant term and a term that oscillates at twice the source frequency:

$$p = \tfrac{1}{2}\,v_m i_m\big[\cos\varphi - \cos(2\omega t + \varphi)\big].$$

Term in $p$Time behaviourAverage over one cycle
$\tfrac{1}{2}v_m i_m \cos\varphi$Constant (no $t$)$\tfrac{1}{2}v_m i_m \cos\varphi$
$-\tfrac{1}{2}v_m i_m \cos(2\omega t + \varphi)$Oscillates at $2\omega$Zero — positive half cancels negative half

Only the first term survives the cycle average. Writing $v_m / \sqrt2 = V$ and $i_m / \sqrt2 = I$ for the rms values, the average power becomes the central result of this subtopic:

$$P = \frac{v_m i_m}{2}\cos\varphi = \frac{v_m}{\sqrt2}\cdot\frac{i_m}{\sqrt2}\cos\varphi = V\,I\,\cos\varphi.$$

Because $V = IZ$, this can also be written $P = I^2 Z \cos\varphi$. The quantity $\cos\varphi$ is the power factor: the average power dissipated depends not only on the rms voltage and current but also on the cosine of the phase angle between them.

t P̂ = VI cosφ (average) p(t): oscillates at 2ω v(t)
Instantaneous power $p(t)$ oscillates at twice the source frequency about a positive mean. The mean of the $2\omega$ term is zero, leaving the average $\bar P = VI\cos\varphi$.

The Power Factor and the Impedance Triangle

In a series LCR circuit the impedance forms a right triangle with the resistance $R$ along the base, the net reactance $(X_L - X_C)$ vertical, and the impedance $Z$ as the hypotenuse. The phase angle $\varphi$ sits at the base, so the power factor is read directly off the triangle:

$$\cos\varphi = \frac{R}{Z}, \qquad Z = \sqrt{R^2 + (X_L - X_C)^2}.$$

This is why $P = VI\cos\varphi$ collapses to a familiar dc-like form. Substituting $\cos\varphi = R/Z$ and $V = IZ$ gives $P = I(IZ)(R/Z) = I^2 R$. The reactive elements drop out entirely — confirming that only the resistor dissipates energy.

φ R X  L − X C Z cosφ = R / Z   |   tanφ = (X L − X C) / R
Impedance triangle. The power factor $\cos\varphi = R/Z$ is the cosine of the base angle; a circuit is "more resistive" when $R$ dominates the hypotenuse.
NEET Trap

Power factor $\cos\varphi$ is not the phase angle $\varphi$

Questions often state a phase angle and ask for the power factor, or vice versa. The power factor is the cosine of $\varphi$, a pure number between 0 and 1 — never the angle itself and never the current. A phase angle of $60^\circ$ gives a power factor of $0.5$, not $60$. Also note $\cos\varphi$ is even: a lagging $\varphi = +53^\circ$ and a leading $\varphi = -53^\circ$ give the same power factor $0.6$, because power is always positive.

Rule: $\cos\varphi = R/Z \in [0,1]$. Larger reactance → smaller $\cos\varphi$ → less power dissipated.

The Power Triangle: True, Reactive, Apparent

Multiplying every side of the impedance triangle by $I^2$ scales it into the power triangle. The base becomes the true (average) power $P = I^2 R = VI\cos\varphi$, the vertical side becomes the reactive power $Q = I^2(X_L - X_C) = VI\sin\varphi$, and the hypotenuse becomes the apparent power $S = I^2 Z = VI$. The power factor is then the ratio of true to apparent power.

φ P = VI cosφ Q = VI sinφ S = VI
The power triangle (impedance triangle $\times\,I^2$). True power $P$ is dissipated in $R$; reactive power $Q$ shuttles back and forth in $L$ and $C$; apparent power $S = VI$ is what the source supplies.
QuantitySymbolExpressionMeaning
True (average) power$P$$VI\cos\varphi = I^2R$Energy actually dissipated, in $R$ only
Reactive power$Q$$VI\sin\varphi$Energy exchanged with $L$ and $C$; net zero
Apparent power$S$$VI = I^2Z$Product of rms voltage and rms current
Power factor$\cos\varphi$$P/S = R/Z$Fraction of supplied power that is dissipated
Build the foundation

The impedance $Z$ and phase angle $\varphi$ used here come straight from the phasor analysis of the LCR loop. Revisit Series LCR Circuit & Resonance if the $Z$ triangle is unfamiliar.

The Four Standard Cases

NCERT §7.7 organises the power factor into four cases. Memorising the value of $\cos\varphi$ for each is the single most reliable way to clear power-factor MCQs.

CasePhase $\varphi$$\cos\varphi$Average power $P$
(i) Purely resistive ($R$ only)$0$$1$$VI$ — maximum dissipation
(ii) Pure $L$ or pure $C$$\pm\,\pi/2$$0$$0$ — wattless current
(iii) Series LCR (general)$\tan^{-1}\!\frac{X_C - X_L}{R}$$R/Z$$VI\cos\varphi = I^2R$
(iv) LCR at resonance$0$  ($X_L = X_C$)$1$$I^2Z = I^2R$ — maximum

Cases (i) and (iv) both give $\cos\varphi = 1$ for the same underlying reason: the net reactance is zero, so the circuit behaves as a pure resistance. Case (iii) covers every RL, RC, or RLC circuit away from resonance, where $0 < \cos\varphi < 1$ and power is dissipated only in the resistor even though $L$ and $C$ are present.

Wattless Current

In a purely inductive or purely capacitive circuit the current is exactly $\pi/2$ out of phase with the voltage, so $\cos\varphi = 0$ and $P = VI\cos\varphi = 0$. A current flows, yet over a complete cycle no net energy is dissipated. Such a current is called the wattless current. During one quarter cycle the reactive element stores energy drawn from the source; in the next quarter cycle it returns exactly that energy, so the bookkeeping closes at zero.

More generally, the current in any circuit can be resolved into two components relative to the applied voltage. The component $I\cos\varphi$ in phase with the voltage is the power component; it accounts for all the dissipation. The component $I\sin\varphi$ perpendicular to the voltage is the wattless component; it contributes zero average power.

NEET Trap

Wattless current still flows — zero power is not zero current

A common mistake is to assume that because a pure inductor or capacitor consumes no power, no current flows through it. The current $I = V/X_L$ (or $V/X_C$) is perfectly real and can be large. What is zero is the average power, because the energy taken in is handed straight back. The reactive element opposes current through reactance, not by dissipating energy.

Rule: Pure $L$ or pure $C$ → current present, but $\cos\varphi = 0$ ⇒ $P = 0$. Dissipation requires a resistive component.

Worked Examples

NCERT Example 7.8

A sinusoidal voltage of peak value $283\ \text{V}$ and frequency $50\ \text{Hz}$ is applied to a series LCR circuit with $R = 3\ \Omega$, $L = 25.48\ \text{mH}$, and $C = 796\ \mu\text{F}$. Find the impedance, the phase angle, the power dissipated, and the power factor.

The reactances are $X_L = 2\pi\nu L = 8\ \Omega$ and $X_C = 1/(2\pi\nu C) = 4\ \Omega$. The impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + (8-4)^2} = 5\ \Omega$.

The phase angle is $\varphi = \tan^{-1}\!\big[(X_C - X_L)/R\big] = \tan^{-1}(-4/3) = -53.1^\circ$; the current lags the voltage. The rms current is $I = i_m/\sqrt2 = (v_m/Z)/\sqrt2 = (283/5)/\sqrt2 = 40\ \text{A}$. Power dissipated $P = I^2 R = (40)^2 \times 3 = 4800\ \text{W}$.

Power factor $= \cos\varphi = \cos(-53.1^\circ) = 0.6$.

NCERT Example 7.7

For power-transmission circuits, why does a low power factor imply large power loss, and how does adding a capacitor help?

From $P = VI\cos\varphi$, delivering a fixed power $P$ at a fixed voltage $V$ requires $I = P/(V\cos\varphi)$. A small $\cos\varphi$ forces a large current, and the line loss $I^2R$ grows as the square of the current — so a low power factor wastes energy in transmission.

If the current lags the voltage (inductive load), the lagging wattless component $I_q$ can be cancelled by an equal leading wattless current supplied by a capacitor of suitable value connected in parallel. With $I_q$ neutralised, the impedance tends to $R$, $\cos\varphi$ tends to $1$, and the power becomes effectively $I_p V$.

Quick Recap

Power and power factor at a glance

  • Average power: $P = V_{\text{rms}} I_{\text{rms}}\cos\varphi = I^2 R$. The $2\omega$ term in instantaneous power averages to zero.
  • Power factor: $\cos\varphi = R/Z$, a number in $[0,1]$ — the cosine of the phase angle, not the angle.
  • Pure $R$: $\cos\varphi = 1$, maximum power. Pure $L$ or $C$: $\cos\varphi = 0$, wattless current, zero power.
  • At resonance $X_L = X_C$, so $\cos\varphi = 1$ and dissipation $P = I^2R$ is maximum.
  • Only the resistor dissipates energy; $L$ and $C$ only store and return it.
  • Low $\cos\varphi$ → large current for fixed power → large $I^2R$ transmission loss; corrected with a capacitor.

NEET PYQ Snapshot — Power in an AC Circuit and Power Factor

Real NEET questions on average power, power factor, and wattless current, with worked answers.

NEET 2020

A series LCR circuit is connected to a voltage source. When $L$ is removed from the circuit, the phase difference between current and voltage is $\pi/3$. If instead $C$ is removed, the phase difference is again $\pi/3$. The power factor of the circuit is:

  1. 0.5
  2. 1.0
  3. −1.0
  4. zero
Answer: (2) 1.0

Removing $L$ gives $\tan(\pi/3) = X_C/R$; removing $C$ gives $\tan(\pi/3) = X_L/R$. Equal angles force $X_L = X_C$, the resonance condition. Then $Z = R$ and $\cos\varphi = R/Z = 1$.

NEET 2018

An inductor of $20\ \text{mH}$, a capacitor of $100\ \mu\text{F}$ and a resistor of $50\ \Omega$ are connected in series across a source $V = 10\sin 314t$. The power loss in the circuit is:

  1. 0.79 W
  2. 0.43 W
  3. 2.74 W
  4. 1.13 W
Answer: (1) 0.79 W

$X_L = \omega L = 314 \times 0.02 = 6.28\ \Omega$; $X_C = 1/(\omega C) = 31.85\ \Omega$. Using $P = \dfrac{v_0^2 R}{2[R^2 + (X_L - X_C)^2]} = \dfrac{100 \times 50}{2[2500 + 653.8]} \approx 0.79\ \text{W}$. The $\tfrac12$ converts peak voltage to rms.

NEET 2016

An inductor $20\ \text{mH}$, a capacitor $50\ \mu\text{F}$ and a resistor $40\ \Omega$ are connected in series across a source $V = 10\sin 340t$. The power loss in the AC circuit is:

  1. 0.67 W
  2. 0.76 W
  3. 0.89 W
  4. 0.51 W
Answer: (4) 0.51 W

$X_L = \omega L = 6.8\ \Omega$; $X_C = 1/(\omega C) = 58.8\ \Omega$; $Z = \sqrt{40^2 + 52^2} \approx 65.6\ \Omega$. With $P = V_{\text{rms}} I_{\text{rms}}\cos\varphi$ and $\cos\varphi = R/Z$, $P = \left(\dfrac{10}{\sqrt2}\right)\!\left(\dfrac{10/\sqrt2}{65.6}\right)\!\left(\dfrac{40}{65.6}\right) \approx 0.51\ \text{W}$.

NEET 2016

A small signal voltage $V(t) = V_0\sin\omega t$ is applied across an ideal capacitor $C$. Which statement is correct?

  1. Over a full cycle the capacitor $C$ does not consume any energy from the voltage source
  2. Current $I(t)$ is in phase with voltage $V(t)$
  3. Current $I(t)$ leads voltage $V(t)$ by 180°
  4. Current $I(t)$ lags voltage $V(t)$ by 90°
Answer: (1)

In an ideal capacitor the current leads the voltage by $90^\circ$, so $\varphi = \pi/2$ and $\cos\varphi = 0$. The average power $P = VI\cos\varphi = 0$ — the wattless-current case. The capacitor stores and returns energy without net consumption.

FAQs — Power in an AC Circuit and Power Factor

The questions students most often confuse on power and the power factor.

What is the formula for average power in an AC circuit?
The average power dissipated over one complete cycle is P = Vrms Irms cos φ, where Vrms and Irms are the rms voltage and current and φ is the phase angle between them. The factor cos φ is called the power factor. It can also be written as P = I²Z cos φ or, since Z cos φ = R, as P = I²R.
What is the power factor and how is it related to impedance?
The power factor is cos φ, the cosine of the phase angle between voltage and current. From the impedance triangle, cos φ = R/Z, where R is the resistance and Z is the impedance. It ranges from 0 (purely reactive circuit) to 1 (purely resistive circuit or at resonance) and measures how close the circuit is to dissipating maximum power.
What is wattless current?
Wattless current is the current that flows in a purely inductive or purely capacitive circuit, where the phase difference between voltage and current is π/2, so cos φ = 0. Although a current flows, the average power dissipated over a complete cycle is zero. It is the component of current perpendicular to the applied voltage.
Why is the average power zero in a pure inductor or pure capacitor?
In a pure inductor or pure capacitor the current is exactly π/2 out of phase with the voltage, so cos φ = 0 and P = Vrms Irms cos φ = 0. Physically, the element absorbs energy during one quarter cycle and returns the same energy to the source in the next quarter cycle, so over a full cycle no net energy is dissipated. Only a resistor dissipates energy.
What is the power factor at resonance in a series LCR circuit?
At resonance the inductive and capacitive reactances cancel (XL = XC), so the phase angle φ = 0 and cos φ = 1. The impedance equals R, and the power dissipated is maximum: P = I²Z = I²R. Resonance is the condition of maximum power transfer to the circuit.
Why does a low power factor cause large power loss in transmission lines?
To deliver a fixed power P at a fixed voltage V, the current required is I = P / (V cos φ). A small cos φ forces a large current, and the heat loss in the line resistance is I²R, which grows with the square of the current. A low power factor therefore means a large transmission loss, which is why the power factor is improved by adding a capacitor of suitable value.
Related Topics
Alternating Current Back to the full chapter overview and all subtopics. Series LCR Circuit & Resonance Impedance Z, phase angle φ and the resonance condition that fixes cos φ = 1. AC Applied to a Resistor The pure-R case: voltage and current in phase, cos φ = 1. AC Applied to an Inductor Current lags by π/2, zero average power — the wattless inductor. AC Applied to a Capacitor Current leads by π/2, the capacitive wattless-current case. Phasors The rotating-vector picture behind the phase angle φ. Transformers Why power is transmitted at high voltage to cut I²R loss.