Power in AC Circuit: The Power Factor

Alternating Current · CBSE

Theory

Derivation of the average power

A source $v = v_m\sin\omega t$ applied to a series LCR circuit drives $i = i_m\sin(\omega t+\phi)$ , with $i_m = v_m/Z$ .
The instantaneous power supplied is

$p = vi = v_m i_m \sin\omega t\,\sin(\omega t+\phi) = \dfrac{v_m i_m}{2}\left[\cos\phi - \cos(2\omega t+\phi)\right]$

Averaging over one complete cycle: the first term is constant, and the average of the second is zero (its positive half cancels its negative half). Hence

$P_{av} = \dfrac{v_m i_m}{2}\cos\phi = \dfrac{v_m}{\sqrt2}\cdot\dfrac{i_m}{\sqrt2}\cos\phi = V I\cos\phi$

Equivalently $P_{av} = I^2 Z\cos\phi = I^2R$ , since $Z\cos\phi = R$ .
The same result written with $i = i_0\sin(\omega t-\phi)$ and the identity $\sin\omega t\,\sin(\omega t-\phi) = \dfrac{1}{2}\left[(1-\cos2\omega t)\cos\phi - \sin2\omega t\,\sin\phi\right]$ gives $P_{av} = \dfrac{V_0I_0}{2}\cos\phi$ after averaging, since $\langle\cos2\omega t\rangle = \langle\sin2\omega t\rangle = 0$ .

The power factor

$\cos\phi$ is called the power factor of the circuit:

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

It is dimensionless, lies between 0 and 1, and measures how close the circuit is to expending the maximum possible power for the given $V$ and $I$ .
Given a power factor, the phase difference follows at once: $\cos\phi = 0.5 \Rightarrow \phi = 60^\circ$ .

The four cases NCERT lists

Circuit	Phase angle	Power factor	Average power
Purely resistive	$\phi = 0$	$\cos\phi = 1$	maximum, $P = VI = I^2R$
Purely inductive or purely capacitive	$\phi = \pm\dfrac{\pi}{2}$	$\cos\phi = 0$	zero (wattless current)
General series LCR	$\phi = \tan^{-1}\dfrac{X_L-X_C}{R}$	$0 < \cos\phi < 1$	$VI\cos\phi = I^2R$ , dissipated only in $R$
Series LCR at resonance	$\phi = 0$ since $X_L = X_C$	$\cos\phi = 1$	maximum, $P = I^2R = V^2/R$

Only the resistor dissipates energy in an ac circuit; pure inductance and pure capacitance dissipate none.

Wattless current

When $\cos\phi = 0$ a current flows but no power is consumed over a cycle; such a current is called a wattless current. It occurs in a purely inductive or purely capacitive circuit (or an LC circuit of negligible resistance).
In general, resolve the current phasor $\mathbf I$ into two components: $I\cos\phi$ along $\mathbf V$ , the power component, which accounts for the whole of the dissipation; and $I\sin\phi$ perpendicular to $\mathbf V$ , the wattless component, which carries no power.
Consequence for an LC circuit with negligible $R$ : the current is $I = V/\lvert X_L - X_C\rvert$ , but the average power consumed is zero.

Improving the power factor

For transporting electric power, a low power factor means a large power loss. To deliver a given power $P$ at a given voltage $V$ , the current needed is $I = P/(V\cos\phi)$ ; a small $\cos\phi$ forces a large current and hence a large $I^2R$ loss in the transmission line.
Remedy: connect a capacitor of appropriate capacitance in the circuit (in parallel with the load) so that the leading wattless current it draws cancels the lagging wattless current of the inductive load. Then $Z$ tends to $R$ , $\cos\phi$ tends to 1, and the power delivered is effectively $I_pV$ .
In a series circuit the same idea appears as: unity power factor requires resonance, $X_L = X_C$ . If a capacitor $C_1$ is added in parallel with $C$ , the capacitance becomes $C + C_1$ and the condition is $\dfrac{1}{\omega(C+C_1)} = \omega L$ . For instance, with $\omega = 1000$ rad s $^{-1}$ , $L = 100$ mH and $C = 2\ \mu$ F: $\omega L = 100\ \Omega$ , so $C + C_1 = 10^{-5}$ F, giving $C_1 = 8\ \mu$ F.

Standard results and worked patterns

$P = VI\cos\phi = I^2R$ ; $\cos\phi = R/Z$ ; $P_{max} = VI$ at $\phi = 0$ ; $P = 0$ at $\phi = \pi/2$ .
Purely resistive numerical: $i = 10\sin(100\pi t)$ A through $R = 20\ \Omega$ gives $I = 10/\sqrt2$ and $P = I^2R = 50\times20 = 1000$ W.
Ideal inductor, zero power, in two lines: with $i = i_0\sin(\omega t-\pi/2)$ and $v = v_0\sin\omega t$ , $p = -\dfrac{v_0i_0}{2}\sin2\omega t$ , whose cycle average is zero. Physically the inductor absorbs energy while the field builds up in one quarter cycle and returns exactly the same energy as the field collapses in the next.
RL numerical pattern: $i = 14\sin(100\pi t)$ A through $R = 30\ \Omega$ in series with $L = 2/5\pi$ H gives $X_L = 40\ \Omega$ , $Z = 50\ \Omega$ , $I = 10$ A, $V_R = 300$ V, $V_L = 400$ V, and $\cos\phi = R/Z = 0.6$ .
If an RL circuit has $X_L = R$ its power factor is $1/\sqrt2$ ; adding a capacitor with $X_C = X_L$ brings it to resonance and the power factor to 1, so the ratio of the two power factors is $1/\sqrt2$ .
If an RL circuit shows $\phi = \pi/4$ and the same $R$ with a capacitor also shows $\phi = \pi/4$ , then $X_L = R = X_C$ ; putting $L$ , $C$ and $R$ together gives resonance, so $P = V^2/R = V_0^2/2R$ and $i = (V_0/R)\sin\omega t$ , in phase with the voltage.
$V_L = V_C \neq V_R$ in a series LCR circuit means $X_L = X_C$ , i.e. resonance, so the power factor is 1.
Series bulbs: with the same current through each, $P = I^2R$ and $R = V_{rated}^2/P_{rated}$ , so the lowest-wattage bulb has the largest resistance and dissipates the most power.
Elements identified from phase: current ahead of voltage by $\pi/4$ with $R = 100\ \Omega$ means the other element is a capacitor with $X_C = R = 100\ \Omega$ , so $Z = 100\sqrt2\ \Omega$ and $I = 141/(100\sqrt2) \approx 1$ A. Replacing the ac source by a dc source stops the current once the capacitor is charged.
RL circuit on dc: in the steady state an ideal inductor offers no opposition, so essentially the whole potential drop appears across $R$ (a transient drop appears across $L$ only while the current is building up).

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
An alternating current $I=(10\text{ A})\sin(100\pi t)$ is passed through a resistor of $20\ \Omega$ . What is the average power consumed by the resistor over a complete cycle?
1 markCBSE Class 12 Physics — Very Short Answer · AI 2021C · Section A, 1 markAnswer 1 ↓
2
The power factor of an ac circuit is $0.5$ . What is the phase difference between voltage and current in the circuit?
1 markCBSE Class 12 Physics — Very Short Answer · Foreign 2016, R · Section A, 1 markAnswer 2 ↓
3
Why is the use of ac voltage preferred over dc voltage? Give two reasons.
1 markCBSE Class 12 Physics — Very Short Answer · AI 2014 · Section A, 1 markAnswer 3 ↓
4
A resistor $R$ and an inductor $L$ are connected in series to a source of voltage $V=V_0\sin\omega t$ . The voltage is found to lead the current in phase by $\pi/4$ . If the inductor is replaced by a capacitor $C$ , the voltage lags behind the current in phase by $\pi/4$ . When $L$ , $C$ and $R$ are connected in series with the same source, find the (i) average power dissipated and (ii) instantaneous current in the circuit.
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 4 ↓
5
In a series LCR circuit, $V_L=V_C\neq V_R$ . What is the value of the power factor for this circuit?
2 marksCBSE Class 12 Physics — SA · 1/2, 2020 · Section B, 2 marksAnswer 5 ↓
6
In series LCR circuit obtain the conditions under which (i) the impedance of the circuit is minimum and (ii) wattless current flows in the circuit.
2 marksCBSE Class 12 Physics — SA · Foreign 2014, An · Section B, 2 marksAnswer 6 ↓
7
An alternating current $I=14\sin(100\pi t)$ A passes through a series combination of a resistor of $30\ \Omega$ and an inductor of $\left(\dfrac{2}{5\pi}\right)$ H. Taking $\sqrt2=1.4$ , calculate the (i) rms value of the voltage drops across the resistor and the inductor, and (ii) power factor of the circuit.
3 marksCBSE Class 12 Physics — SA · 2023 · Section C, 3 marksAnswer 7 ↓
8
(a) An ac circuit, as shown in the figure, has an inductor of inductance $L$ and a resistor of resistance $R$ connected in series. Using the phasor diagram, explain why the voltage in the circuit will lead the current in phase. (b) The potential difference across the resistor is $160$ V and that across the inductor is $120$ V. Find the effective value of the applied voltage. If the effective current in the circuit is $1.0$ A, calculate the total impedance of the circuit. (c) What will be the potential difference in the circuit when direct current is passed through the circuit?
3 marksCBSE Class 12 Physics — SA · 2019 · Section C, 3 marksAnswer 8 ↓
9
An ac circuit consists of a series combination of circuit elements $X$ and $Y$ . The current is ahead of the voltage in phase by $\pi/4$ . If element $X$ is a pure resistor of $100\ \Omega$ , (a) name the circuit element $Y$ . (b) calculate the rms value of current, if the rms value of voltage is $141$ V. (c) what will happen if the ac source is replaced by a dc source?
3 marksCBSE Class 12 Physics — SA · 2019 · Section C, 3 marksAnswer 9 ↓
10
Without making any other change, find the value of the additional capacitor $C_1$ , to be connected in parallel with the capacitor $C$ , in order to make the power factor of the circuit unity. Given: $L=100$ mH, $C=2\ \mu\text{F}$ , $R=400\ \Omega$ , source $V=V_0\sin(1000t+\phi)$ . (1/3, Delhi 2017) OR Calculate the value of the additional capacitor which may be joined suitably to the capacitor $C$ that would make the power factor of the circuit unity. ( $R=400\ \Omega$ in series with $C=2\ \mu\text{F}$ and $L=100$ mH, source $V=V_0\sin(1000t+\phi)$ .) (1/3, AI 2015)
3 marksCBSE Class 12 Physics — SA · Section C, 3 marksAnswer 10 ↓
11
(i) When an ac source is connected to an ideal inductor, show that the average power supplied by the source over a complete cycle is zero. (ii) A lamp is connected in series with an inductor and an ac source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor? Explain.
3 marksCBSE Class 12 Physics — SA · AI 2016 · Section C, 3 marksAnswer 11 ↓
12
A circuit containing an $80$ mH inductor and a $250$ mF capacitor in series is connected to a $240$ V, $100$ rad/s supply. The resistance of the circuit is negligible. (i) Obtain the rms value of the current. (ii) What is the total average power consumed by the circuit?
3 marksCBSE Class 12 Physics — SA · Delhi 2015C, Ap · Section C, 3 marksAnswer 12 ↓
13
A voltage $V=V_0\sin\omega t$ is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit?
3 marksCBSE Class 12 Physics — SA · AI 2014 · Section C, 3 marksAnswer 13 ↓
14
Show that an ideal inductor does not dissipate power in an ac circuit.
5 marksCBSE Class 12 Physics — Long Answer · 2/5, 2020, Ap · Section E, 5 marksAnswer 14 ↓
15
In a series LR circuit, $X_L=R$ and the power factor of the circuit is $P_1$ . When a capacitor with capacitance $C$ such that $X_L=X_C$ is put in series, the power factor becomes $P_2$ . Calculate $P_1/P_2$ .
5 marksCBSE Class 12 Physics — Long Answer · 2/5, Delhi 2016, An · Section E, 5 marksAnswer 15 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

16
Assertion (A): When three electric bulbs of power $200$ W, $100$ W and $50$ W are connected in series to a source, the power consumed by the $50$ W bulb is maximum. Reason (R): In a series circuit, the current is the same through each bulb, but the potential difference across each bulb is different.
- (a)
  Both A and R are true and R is the correct explanation of A.
- (b)
  Both A and R are true, but R is not the correct explanation of A.
- (c)
  A is true, but R is false.
- (d)
  A is false, and R is also false.
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 16 ↓
17
When an alternating voltage $E=E_0\sin\omega t$ is applied to a circuit, a current $I=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right)$ flows through it. The average power dissipated in the circuit is:
- (a)
  $E_{rms}I_{rms}$
- (b)
  $E_0I_0$
- (c)
  $\dfrac{E_0I_0}{\sqrt2}$
- (d)
  Zero
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 17 ↓
18
The power factor of a series LCR circuit at resonance will be:
- (a)
  1
- (b)
  0
- (c)
  $1/2$
- (d)
  $1/\sqrt2$
1 markCBSE Class 12 Physics — Multiple Choice · 2020, R · Section A, 1 markAnswer 18 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
$I_0=10$ A, $R=20\ \Omega$ , $\phi=0°$ (purely resistive circuit). $P=I_{rms}^2R=\left(\dfrac{I_0}{\sqrt2}\right)^2R=\left(\dfrac{10}{\sqrt2}\right)^2\times20=50\times20=1000\text{ W}$
2.↑ back to the question
$\cos\phi=0.5=\cos60° \Rightarrow \phi=60°$ So the phase difference between voltage and current is $60°$ .
3.↑ back to the question
(i) AC can be transmitted over long distances with much lower energy (transmission) losses than dc, since it can be stepped up to high voltage (low current) for transmission using a transformer. (ii) AC voltage can be conveniently stepped up or stepped down as per requirement using a transformer, which is not possible with dc.
4.↑ back to the question
For the RL circuit: $\tan\dfrac\pi4=\dfrac{X_L}{R}=1 \Rightarrow X_L=R$ For the RC circuit: $\tan\dfrac\pi4=\dfrac{X_C}{R}=1 \Rightarrow X_C=R$ So $X_L=X_C$ — when $L$ , $C$ , $R$ are together in series, this is exactly the resonance condition, so $Z=R$ and $\phi=0$ . (i) $P=V_{rms}I_{rms}\cos0°=\dfrac{V_{rms}^2}{R}=\dfrac{(V_0/\sqrt2)^2}{R}=\dfrac{V_0^2}{2R}$ (ii) Since $Z=R$ and $\phi=0$ : $I=I_0\sin\omega t=\dfrac{V_0}{R}\sin\omega t$ (current in phase with voltage).
5.↑ back to the question
$V_L=V_C \Rightarrow X_L=X_C$ — the resonance condition, so $Z=R$ . Power factor $\cos\phi=\dfrac RZ=\dfrac RR=1$
6.↑ back to the question
(i) $Z=\sqrt{R^2+\left(\omega L-\dfrac1{\omega C}\right)^2}$ is minimum (equal to $R$ ) when $\omega L=\dfrac1{\omega C}$ , i.e. at resonance, $\omega=1/\sqrt{LC}$ . (ii) $P_{av}=EI\cos\phi$ (rms values). For a wattless current, $P_{av}=0$ though current flows, requiring $\cos\phi=0$ , i.e. $\phi=\pi/2$ — this occurs when the circuit is purely inductive or purely capacitive (no resistance).
7.↑ back to the question
$I_0=14$ A, $\omega=100\pi$ rad/s $X_L=\omega L=100\pi\times\dfrac2{5\pi}=40\ \Omega$ $Z=\sqrt{R^2+X_L^2}=\sqrt{30^2+40^2}=50\ \Omega$ $I_{rms}=\dfrac{I_0}{\sqrt2}=\dfrac{14}{1.4}=10\text{ A}$ (i) $V_R=I_{rms}R=10\times30=300$ V; $V_L=I_{rms}X_L=10\times40=400$ V (ii) Power factor $\cos\phi=\dfrac RZ=\dfrac{30}{50}=0.6$
8.↑ back to the question
(a) In an R–L series circuit, $V_R$ is in phase with $I$ , while $V_L$ leads $I$ by $\pi/2$ . Adding these perpendicular phasors, the resultant applied voltage $V$ makes a positive angle $\phi=\tan^{-1}(X_L/R)$ with the current phasor — so the applied voltage leads the current in phase. (b) $\varepsilon_{net}=\sqrt{V_R^2+V_L^2}=\sqrt{160^2+120^2}=\sqrt{40000}=200\text{ V}$ $Z=\sqrt{R^2+X_L^2}$ ; with $R=\dfrac{V_R}{I}=160\ \Omega$ and $X_L=\dfrac{V_L}{I}=120\ \Omega$ : $Z=\sqrt{160^2+120^2}=\sqrt{40000}=200\ \Omega$ (c) With dc in the steady state, an ideal inductor behaves as a plain (zero-reactance) conductor, so essentially the entire potential drop appears across the resistance $R$ (only a brief transient drop appears across the inductor while the current is building up).
9.↑ back to the question
(a) Since the current leads the voltage, element $Y$ must be a capacitor. (b) $\tan\dfrac\pi4=\dfrac{X_C}{R}=1 \Rightarrow X_C=R=100\ \Omega$ $Z=\sqrt{R^2+X_C^2}=\sqrt{100^2+100^2}=100\sqrt2\ \Omega$ $I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{141}{100\sqrt2}\approx1\text{ A}$ (c) With a dc source, the capacitor blocks the steady current once fully charged, so there is no continuous flow of electric current through the circuit.
10.↑ back to the question
Unity power factor requires resonance: $X_L=X_{C,\text{new}}$ , where the new capacitance is $C+C_1$ (parallel combination). $\dfrac{1}{\omega(C+C_1)}=\omega L=1000\times100\times10^{-3}=100\ \Omega$ $C+C_1=\dfrac{1}{1000\times100}=10^{-5}\text{ F}$ $C_1=10^{-5}-C=10^{-5}-2\times10^{-6}=0.8\times10^{-5}\text{ F}=8\ \mu\text{F}$ So the required additional capacitor is $8\ \mu\text{F}$ , connected in parallel with the given capacitor $C$ .
11.↑ back to the question
(i) In an ideal inductor, $I_{rms}$ lags $V_{rms}$ by $\phi=\pi/2$ . $P_{av}=V_{rms}I_{rms}\cos\phi=V_{rms}I_{rms}\cos\left(\dfrac\pi2\right)=0$ So the average power supplied over a complete cycle is zero — the inductor only stores and returns energy, dissipating none. (ii) When the key is plugged in, the lamp lights up at a brightness set by $Z=\sqrt{R^2+X_L^2}$ . Inserting an iron rod increases $L$ (high permeability), so $X_L=\omega L$ increases; $Z$ increases and the current decreases — the brightness of the lamp decreases.
12.↑ back to the question
$X_L=\omega L=100\times80\times10^{-3}=8\ \Omega$ $X_C=\dfrac{1}{\omega C}=\dfrac{1}{100\times250\times10^{-3}}=\dfrac{1}{25}=0.04\ \Omega$ Net reactance $=|X_L-X_C|=|8-0.04|=7.96\ \Omega$ (i) $I_{rms}=\dfrac{V_{rms}}{\text{reactance}}=\dfrac{240}{7.96}\approx30.15\text{ A}$ (ii) Since resistance is negligible ( $R\approx0$ ), $P=I_{rms}^2R\approx0$ — the total average power consumed by the circuit is zero (a wattless current flows).
13.↑ back to the question
Let $I=I_0\sin(\omega t-\phi)$ be the current. Instantaneous power, $p=VI=V_0I_0\sin\omega t\,\sin(\omega t-\phi)=V_0I_0\left[\dfrac{1-\cos2\omega t}2\cos\phi-\dfrac{\sin2\omega t}2\sin\phi\right]$ Averaging over a full cycle, $\langle\cos2\omega t\rangle=\langle\sin2\omega t\rangle=0$ , so $P_{av}=\dfrac{V_0I_0}2\cos\phi=V_{rms}I_{rms}\cos\phi$ (i) No power is dissipated (wattless current) when $\cos\phi=0$ , i.e. $\phi=\pi/2$ — a purely inductive or purely capacitive circuit. (ii) Maximum power is dissipated when $\cos\phi=1$ , i.e. $\phi=0$ — a purely resistive circuit, or an LCR circuit at resonance ( $X_L=X_C$ ), giving $P_{av,max}=V_{rms}I_{rms}$ .
14.↑ back to the question
Let $V=V_0\sin\omega t$ be applied across an ideal inductor; the current lags by $\pi/2$ : $I=I_0\sin\left(\omega t-\dfrac\pi2\right)$ . $p=VI=V_0I_0\sin\omega t\,\sin\left(\omega t-\dfrac\pi2\right)=-V_0I_0\sin\omega t\cos\omega t=-\dfrac{V_0I_0}2\sin2\omega t$ $P_{av}=-\dfrac{V_0I_0}2\langle\sin2\omega t\rangle=0$ , since the average of $\sin2\omega t$ over a full cycle is zero. [Equivalently, $P_{av}=V_{rms}I_{rms}\cos\left(\dfrac\pi2\right)=0$ .] The inductor absorbs energy while the magnetic field builds up in one quarter cycle and returns exactly the same energy as the field collapses in the next — so it dissipates no net power.
15.↑ back to the question
LR circuit: $Z_1=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=\sqrt2\,R$ (using $X_L=R$ ) $P_1=\cos\phi_1=\dfrac{R}{Z_1}=\dfrac{R}{\sqrt2\,R}=\dfrac1{\sqrt2}$ LCR circuit with $X_L=X_C$ (resonance): $Z_2=R$ $P_2=\cos\phi_2=\dfrac{R}{Z_2}=1$ $\dfrac{P_1}{P_2}=\dfrac{1/\sqrt2}{1}=\dfrac1{\sqrt2}$
16.Answer: (B)↑ back to the question
At rated voltage, $P=\dfrac{V^2}{R} \Rightarrow R\propto\dfrac1P$ , so the $50$ W bulb has the largest resistance among the three. In series, the same current $I$ flows through all bulbs, so actual power dissipated is $P=I^2R$ ; since $R$ is largest for the $50$ W bulb, it dissipates the maximum power in this series combination — Assertion (A) is true. Reason (R) is also a true statement, but it is not, by itself, the correct explanation of why the $50$ W bulb draws maximum power (the real reason is $P=I^2R$ combined with $R$ being largest for the $50$ W bulb). Hence option (b).
17.Answer: (D)↑ back to the question
Phase difference $\phi=\pi/2$ . $P=E_{rms}I_{rms}\cos\phi=E_{rms}I_{rms}\cos\left(\dfrac\pi2\right)=0$ Hence option (d).
18.Answer: (A)↑ back to the question
At resonance, $X_L=X_C \Rightarrow Z=R$ . Power factor $\cos\phi=\dfrac RZ=\dfrac RR=1$ Hence option (a).