Power in AC circuit

AC Circuits · Maharashtra HSC

Theory

13.6 Power in an AC Circuit

Power is the rate of doing work. In a DC circuit, power is measured as the product of voltage and current. In an AC circuit the values of current and voltage change at every instant, so the power at a given instant is the product of the instantaneous voltage and the instantaneous current, $P = ei$ , and what is physically useful is its average over one complete cycle.

(a) Average Power Associated with a Resistance

In a pure resistor the alternating current is in phase with the applied alternating voltage: if $e = e_0\sin\omega t$ then $i = i_0\sin\omega t$ .

Instantaneous power:

$P = ei = (e_0\sin\omega t)(i_0\sin\omega t) = e_0i_0\sin^2\omega t \qquad \ldots(1)$

This varies with time, so the average power over a complete cycle is obtained by integrating (1):

$P_{av} = \dfrac{\text{work done by the emf on the charges in one cycle}}{\text{time for one cycle}} = \dfrac{\displaystyle\int_0^T P\,dt}{T} = \dfrac{\displaystyle\int_0^T e_0i_0\sin^2\omega t\,dt}{T}$

$\therefore\ P_{av} = \dfrac{e_0i_0}{T}\left(\int_0^T\sin^2\omega t\,dt\right) = \dfrac{e_0i_0}{T}\left(\dfrac{T}{2}\right) \qquad \left(\because \int_0^T\sin^2\omega t\,dt = \dfrac{T}{2}\right)$

$\therefore\ P_{av} = \dfrac{1}{2}e_0i_0 = \dfrac{e_0}{\sqrt2}\times\dfrac{i_0}{\sqrt2}$

$\therefore\ P_{av} = e_{rms}\times i_{rms}$

This product is also called the apparent power.

Power factor in a purely resistive circuit: $\cos\phi = \dfrac{R}{Z} = \dfrac{R}{\sqrt{R^2}} = \dfrac{R}{R} = 1$ , so $\phi = 0^\circ$ .

(b) Average Power Associated with an Inductor

In a purely inductive circuit the current lags the voltage by $\pi/2$ : if $e = e_0\sin\omega t$ then $i = i_0\sin(\omega t - \pi/2) = -i_0\cos\omega t$ .

$P = ei = (e_0\sin\omega t)\left(-i_0\cos\omega t\right) = -e_0i_0\sin\omega t\cos\omega t$

$P_{av} = \dfrac{\displaystyle\int_0^T P\,dt}{T} = -\dfrac{e_0i_0}{T}\int_0^T\sin\omega t\cos\omega t\,dt = -\dfrac{e_0i_0}{2T}\int_0^T\sin 2\omega t\,dt = 0$

$\therefore\ P_{av} = 0$

The average power over a complete cycle of AC through an ideal inductor is zero. This is the board's question "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".

(c) Average Power Associated with a Capacitor

In a purely capacitive circuit the current leads the emf by $\pi/2$ : if $e = e_0\sin\omega t$ then $i = i_0\sin(\omega t + \pi/2) = i_0\cos\omega t$ .

$P = ei = (e_0\sin\omega t)(i_0\cos\omega t) = e_0i_0\sin\omega t\cos\omega t$

$P_{av} = \dfrac{\displaystyle\int_0^T e_0i_0\sin\omega t\cos\omega t\,dt}{T} = 0$

The average power supplied to an ideal capacitor by the source over a complete cycle of AC is also zero — which proves that an ideal capacitor in an AC circuit does not dissipate power.

(d) Average Power in a Series LCR Circuit

Let $e = e_0\sin\omega t$ be applied across a series combination of a pure inductor, a capacitor and a resistor. There is a phase difference $\phi$ between the applied emf and the current, so

$i = i_0\sin(\omega t \pm \phi)$

Instantaneous power:

$P = ei = e_0i_0\sin\omega t\,\sin(\omega t \pm \phi)$

$= e_0i_0\sin\omega t\left(\sin\omega t\cos\phi \pm \cos\omega t\sin\phi\right)$

$= e_0i_0\left(\sin^2\omega t\cos\phi \pm \sin\omega t\cos\omega t\sin\phi\right)$

$\therefore\ P = e_0i_0\sin^2\omega t\cos\phi \pm \dfrac{e_0i_0}{2}\sin 2\omega t\sin\phi \qquad \ldots(1)$

This shows that $P$ varies with $t$ . Averaging over one cycle,

$P_{av} = \dfrac{\text{work done in one cycle}}{\text{time for one cycle}} = \dfrac{\displaystyle\int_0^T P\,dt}{T}$

$\therefore\ P_{av} = \dfrac{e_0i_0}{T}\left[\int_0^T\sin^2\omega t\cos\phi\,dt \pm \int_0^T\dfrac{\sin 2\omega t\,\sin\phi}{2}dt\right] \qquad \ldots(2)$

Now $\displaystyle\int_0^T\sin^2\omega t\,dt = \dfrac{T}{2}$ , so the first integral is $\dfrac{T}{2}\cos\phi$ , and $\displaystyle\int_0^T\sin 2\omega t\,dt = 0$ , so the second integral vanishes. Substituting in (2),

$P_{av} = \dfrac{e_0i_0}{T}\times\dfrac{T}{2}\cos\phi = \dfrac{e_0i_0}{2}\cos\phi = \dfrac{e_0}{\sqrt2}\cdot\dfrac{i_0}{\sqrt2}\cos\phi$

$\therefore\ P_{av} = e_{rms}\times i_{rms}\times\cos\phi$

This is the average power, or true power, dissipated in the AC circuit. It depends not only on the rms values of current and emf but also on the phase difference $\phi$ between them.

Power Factor

The factor $\cos\phi$ is called the power factor of the circuit.

$\cos\phi = \dfrac{P_{av}}{e_{rms}\times i_{rms}} = \dfrac{\text{True power}}{\text{Apparent power}}$

The average power $P_{av}$ is the true power, and the product $e_{rms}\times i_{rms}$ is the apparent power.

From the impedance triangle,

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

Non-inductive circuit ( $X_L = X_C$ ): $\cos\phi = R/\sqrt{R^2} = R/R = 1$ , so $\phi = 0$ .
Purely inductive or purely capacitive circuit: $\phi = 90^\circ$ , so the power factor is zero and $P_{av} = e_{rms}i_{rms}\cos 90^\circ = 0$ .

Wattless Current

The current through a pure inductor or an ideal capacitor, which consumes no power for its maintenance in the circuit, is called idle current or wattless current. Power is dissipated in a circuit only because of resistance.

Why a Low Power Factor Means Large Transmission Loss

For a given power delivered, $P_{av} = e_{rms}i_{rms}\cos\phi$ , a small $\cos\phi$ forces $i_{rms}$ to be large. The heat wasted in the transmission line is $i_{rms}^2R_{line}$ , which rises as the square of that current. Hence circuits used for transporting electric power are kept at a high power factor.

Standard Results

Pure resistor: $P_{av} = e_{rms}i_{rms} = \dfrac{e_0i_0}{2} = i_{rms}^2R$ ; power factor 1.
Pure inductor and ideal capacitor: $P_{av} = 0$ ; power factor 0.
Series LCR: $P_{av} = e_{rms}i_{rms}\cos\phi$ , $\cos\phi = R/Z$ .
Useful integrals: $\displaystyle\int_0^T\sin^2\omega t\,dt = \dfrac{T}{2}$ , $\displaystyle\int_0^T\sin\omega t\cos\omega t\,dt = 0$ , $\displaystyle\int_0^T\sin 2\omega t\,dt = 0$ .

Worked Pattern

A 100 $\Omega$ resistor across a 220 V, 50 Hz supply: $i_{rms} = 220/100 = 2.2$ A and $P_{av} = e_{rms}i_{rms} = 220\times2.2 = 484$ W (no $\cos\phi$ needed, since it is 1).

A series LCR circuit with $R = 3\ \Omega$ , $X_L = 8\ \Omega$ , $X_C = 4\ \Omega$ across $e_0 = 283$ V: $Z = \sqrt{3^2+(8-4)^2} = 5\ \Omega$ , $\tan\phi = 4/3$ so $\phi = 53.1^\circ$ , power factor $\cos\phi = 0.6$ , and $P_{av} = \dfrac{e_0}{\sqrt2}\times\dfrac{e_0}{\sqrt2\,Z}\times0.6 = 4805$ W.

Theory Questions Asked on This Subtopic

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

1
Obtain an expression for average power dissipated in a purely resistive A.C. circuit. Define power factor of the circuit and state its value for purely resistive A.C. circuit.
Maharashtra HSC board question — asked in October 2015Answer 1 ↓
2
Obtain an expression for average power dissipated in series LCR A.C. circuit. Hence obtain an expression for power factor of the circuit.
Maharashtra HSC board question — asked in July 2019Answer 2 ↓
3
Derive an expression for the impedance of an LCR circuit connected to an AC power supply. Draw phasor diagram.
Maharashtra HSC board question — asked in February 2024Answer 3 ↓

Numericals

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

4
A 100 $\Omega$ resistor is connected to a 220 V, 50 Hz supply. Calculate: i. r.m.s. value of current and ii. net power consumed over the full cycle
Maharashtra HSC board question — asked in February 2023Answer 4 ↓

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. Expression for average power in purely resistive circuit: a. Let, $e=e_0\sin\omega t$ be applied e.m.f across a resistor of resistance 'R' as shown in figure. At certain instant, current $i=i_0\sin\omega t$ is flowing through the resistor. In this case both 'e' and 'i' are in phase. b. Instantaneous power in circuit is given by, $P=ei=(e_0\sin\omega t)(i_0\sin\omega t)$ $\therefore P=e_0i_0\sin^2\omega t\qquad....(1)$ c. Average power for a complete cycle can be obtained by integrating equation (1). $\therefore P_{av}=\dfrac{\text{work done by emf on the charges in one cycle}}{\text{time for one cycle}}$ $=\dfrac{\displaystyle\int_0^T Pdt}{T}=\dfrac{\displaystyle\int_0^T e_0i_0\sin^2\omega t\,dt}{T}$ $\therefore P_{av}=\dfrac{e_0i_0}{T}\left(\int_0^T\sin^2\omega t\,dt\right)\qquad....(2)$ $\therefore P_{av}=\dfrac{e_0i_0}{T}\left(\dfrac{T}{2}\right)\qquad\left(\because\int_0^T\sin^2\omega t\,dt=\dfrac{T}{2}\right)$ $=\left(\dfrac{1}{2}\right)e_0i_0$ $\therefore P_{av}=\dfrac{e_0}{\sqrt{2}}\times\dfrac{i_0}{\sqrt{2}}$ $\therefore P_{av}=e_{rms}\times i_{rms}$ P is also called as apparent power. ii. Power factor for a circuit is given by, $\cos\phi=\dfrac{R}{Z}=\dfrac{\text{Resistance}}{\text{Impendance}}$ In purely resistive circuit, $\therefore\cos\phi=\dfrac{R}{\sqrt{R^2}}=\dfrac{R}{R}=1$ $\therefore\phi=0°$
2.↑ back to the question
i. Expression for average power dissipated in LCR circuit: a. Suppose that an alternating e.m.f $e=e_0\sin\omega t$ is applied across an AC circuit containing L, C and R as shown in figure. b. In such circuit, there is a phase difference $\phi$ between applied emf and current. Therefore, the instantaneous current is given by, $i=i_0\sin(\omega t\pm\phi)$ . c. Instantaneous power in the circuit is given by, $P=ei=e_0\sin\omega t\times i_0\sin(\omega t\pm\phi)$ $=e_0i_0\sin\omega t\times\sin(\omega t\pm\phi)$ $=e_0i_0\sin\omega t(\sin\omega t\cos\phi\pm\cos\omega t\sin\phi)$ $=e_0i_0(\sin^2\omega t\cos\phi\pm\sin\omega t\cos\omega t\sin\phi)$ $=e_0i_0\sin^2\omega t\cos\phi$ $\pm\dfrac{1}{2}e_0i_0\times2\sin\omega t\cos\omega t\sin\phi$ $\therefore P=e_0i_0\sin^2\omega t\cos\phi\pm\dfrac{e_0i_0}{2}\sin2\omega t\sin\phi\qquad....(1)$ This expression shows that P varies with t. d. Average power dissipated in the circuit can be obtained by integrating equation (1), $\therefore P_{av}=\dfrac{\text{Work done in one cycle}}{\text{time for one cycle}}=\dfrac{\displaystyle\int_0^T Pdt}{T}$ $\therefore P_{av}=\dfrac{e_0i_0}{T}\int_0^T\left[\sin^2(\omega t)\cos(\phi)\pm\dfrac{1}{2}\sin(2\omega t)\sin(\phi)\right]$ $\therefore P_{av}=\dfrac{e_0i_0}{T}\left[\int_0^T\sin^2\omega t\cos\phi\,dt\pm\int_0^T\dfrac{\sin2\omega t\sin\phi}{2}dt\right]\qquad....(2)$ $\because\int_0^T\sin^2\omega t\,dt=\dfrac{T}{2}$ $\therefore\int_0^T\sin^2\omega t\cos\phi\,dt=\dfrac{T}{2}\cos\phi\qquad....(3)$ e. Also, $\displaystyle\int_0^T\sin2\omega t\,dt=0$ $\therefore\int_0^T\dfrac{\sin2\omega t\sin\phi\,dt}{2}=0\qquad....(4)$ f. Substitute equations (3) and (4) in equation (2), we get, $P_{av}=\dfrac{e_0i_0}{T}\times\dfrac{T}{2}\cos\phi$ $=\dfrac{e_0i_0}{2}\cos\phi$ $=\dfrac{e_0}{\sqrt2}\cdot\dfrac{i_0}{\sqrt2}\cos\phi$ $=e_{rms}\times i_{rms}\times\cos\phi$ $\therefore P_{av}=e_{rms}\times i_{rms}\cos\phi\qquad....(5)$ Equation (5) represents required average power or true power dissipated in the AC circuit. ii. Power factor: a. Average power dissipated in LCR series circuit is given by, $P_{av}=e_{rms}\times i_{rms}\cos\phi$ b. This implies that, the average power dissipated in LCR series circuit depends on the rms values of current and emf as well as the phase difference $\phi$ between them. The term $\cos\phi$ is called power factor. c. It is given by, $\cos\phi=\dfrac{P_{av}}{e_{rms}\times i_{rms}}$ d. The average power $P_{av}$ is also termed as true power and product ( $e_{rms}\times I_{rms}$ ) is called apparent power. $\therefore\text{Power factor, }\cos\phi=\dfrac{\text{True power}}{\text{Apparent power}}$ e. Consider impedance triangle for series LCR circuit as shown in figure, $\therefore\cos\phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^2+(X_L-X_C)^2}}=\dfrac{\text{resistance}}{\text{impedance}}$
3.↑ back to the question
Expression for impedance of AC circuit containing LCR in series: i. Consider an alternating e.m.f applied to a series combination of pure inductor of inductance L, capacitor of capacitance C and resistor of resistance R as shown in figure (a). ii. Let $e_L$ , $e_C$ and $e_R$ be the r.m.s voltage across inductor, capacitor and resistor. As R, L and C are in series, the current at any instant through the three elements has the same amplitude and phase and given as, $i=i_0\sin\omega t$ . iii. As ' $e_R$ ' is in phase with current i, the vector $e_R$ is drawn in the same direction as that of i along positive direction of X axis. iv. $e_C$ lags the current i by $\dfrac{\pi}{2}$ radian (90°), hence the vector ' $e_C$ ' is drawn in the negative direction of the Y axis $e_C$ along OY'. v. $e_L$ leads the current by $\dfrac{\pi}{2}$ radian (90°), hence vector $e_L$ is drawn in the positive direction of Y axis. vi. The voltages through L, C and R are given by $e_L=iX_L$ , $e_C=iX_C$ and $e_R=iR\qquad....(1)$ vii. The voltages across L and C have a phase difference of 180°. Hence, the net reactive voltage is ( $e_L-e_C$ ) viii. Consider $e_L>e_C$ . The net reactive emf ( $e_L-e_C$ ) is in the direction of $e_L$ as shown in figure (b). Resultant of $e_R$ , $e_L$ and $e_C$ gives the r.m.s value (e) of applied e.m.f. which is represented by diagonal OP in figure (b). ix. From the figure (b), $OK=\sqrt{OA^2+OB'^2}$ $\therefore e^2=e_R^2+(e_L-e_C)^2$ From equation (1), $\therefore e^2=(iR)^2+(iX_L-iX_C)^2$ $\therefore e^2=i^2\left[R^2+(X_L-X_C)^2\right]$ $\therefore e=i\sqrt{R^2+(X_L-X_C)^2}\qquad....(2)$ $\therefore e=iZ$ where, $Z=\sqrt{R^2+(X_L-X_C)^2}$ = Impedance of circuit x. The quantity $\sqrt{R^2+(X_L-X_C)^2}$ represents the effective opposition offered by the inductor, capacitor and resistor connected in series to the flow of AC current. This total effective resistance of LCR circuit is called the impedance of the circuit and is represented by Z.
4.↑ back to the question
Given: $R=100\ \Omega$ , $V=220$ V, $f=50$ Hz To find: i. rms current ( $i_{rms}$ ) ii. Net power consumed ( $P_{av}$ ) Formulae: i. $i_{rms}=\dfrac{e_{rms}}{R}$ ii. $P_{av}=e_{rms}i_{rms}$ Calculation: From formula (i), $i_{rms}=\dfrac{220}{100}=2.2\text{ A}$ From formula (ii), $P_{av}=220\times2.2=484$ W Ans: i. The rms current in the circuit is 2.2 A. ii. Net power consumed over a full cycle is 484 W.