7.7 Power in AC Circuit: The Power Factor

We have seen that a voltage \(v=v_m \sin \omega t\) applied to a series \(R L C\) circuit drives a current in the circuit given by \(i=i_m \sin (\omega t+\phi)\) where
\(
i_m=\frac{v_m}{Z} \text { and } \phi=\tan ^{-1}\left(\frac{X_C-X_L}{R}\right)
\)
Therefore, the instantaneous power \(p\) supplied by the source is
\(
\begin{aligned}
& p=v i=\left(v_m \sin \omega t\right) \times\left[i_m \sin (\omega t+\phi)\right] \\
& =\frac{v_m i_m}{2}[\cos \phi-\cos (2 \omega t+\phi)] \dots(7.29)
\end{aligned}
\)
The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.37). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore,
\(
\begin{aligned}
& P=\frac{v_m i_m}{2} \cos \phi=\frac{v_m}{\sqrt{2}} \frac{i_m}{\sqrt{2}} \cos \phi \\
& =V I \cos \phi \dots[7.30(a)]
\end{aligned}
\)
This can also be written as,
\(
P=I^2 Z \cos \phi \dots[7.30(b)]
\)
So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle \(\phi\) between them. The quantity \(\cos \phi\) is called the power factor. Let us discuss the following cases:

Case (i) Resistive curcuit: If the circuit contains only pure \(R\), it is called resistive. In that case \(\phi=0, \cos \phi=1\). There is maximum power dissipation.

Case (ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is \(\pi / 2\). Therefore, \(\cos \phi=0\), and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current.

Case (iii) \(L C R\) series circuit: In an \(L C R\) series circuit, power dissipated is given by Eq. (7.30) where \(\phi=\tan ^{-1}\left(X_c-X_L\right) / R\). So, \(\phi\) may be non-zero in a \(R L\) or \(R C\) or \(R C L\) circuit. Even in such cases, power is dissipated only in the resistor.

Case (iv) Power dissipated at resonance in LCR circuit: At resonance \(X_c-X_L=0\), and \(\phi=0\). Therefore, \(\cos \phi=1\) and \(P=I^2 Z=I^2 R\). That is, maximum power is dissipated in a circuit (through \(R\) ) at resonance.

Example 7.7: (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
(b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Solution: (a) We know that \(P=I V \cos \phi\) where \(\cos \phi\) is the power factor. To supply a given power at a given voltage, if \(\cos \phi\) is small, we have to increase current accordingly. But this will lead to large power loss \(\left(I^2 R\right)\) in transmission.
(b)Suppose in a circuit, current \(I\) lags the voltage by an angle \(\phi\). Then power factor \(\cos \phi=R / Z\).
We can improve the power factor (tending to 1 ) by making \(Z\) tend to \(R\). Let us understand, with the help of a phasor diagram (Fig. 7.15)

how this can be achieved. Let us resolve I into two components. \(\mathbf{I}_p\) along the applied voltage \(\mathbf{V}\) and \(\mathbf{I}_q\) perpendicular to the applied voltage. \(\mathbf{I}_q\) as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. \(\mathbf{I}_{\mathrm{p}}\) is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current \(\mathbf{I}_{\mathrm{q}}\) by an equal leading wattless current \(\mathbf{I}_{\mathrm{q}}^{\prime}\). This can be done by connecting a capacitor of appropriate value in parallel so that \(\mathbf{I}_{\mathrm{q}}\) and \(\mathbf{I}_{\mathrm{q}}^{\prime}\) cancel each other and \(P\) is effectively \(I_{\mathrm{p}} V\).

Example 7.8: A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series \(L C R\) circuit in which \(\mathrm{R}=3 \Omega, L=25.48 \mathrm{mH}\), and \(\mathrm{C}=796 \mu \mathrm{~F}\). Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

Solution: 

\(
\begin{aligned}
&\text { (a) To find the impedance of the circuit, we first calculate } X_{\mathrm{L}} \text { and } X_{\mathrm{C}} \text {. }\\
&\begin{aligned}
X_L & =2 \pi \nu L \\
& =2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \Omega=8 \Omega \\
X_C & =\frac{1}{2 \pi \nu C}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&=\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}=4 \Omega\\
&\text { Therefore, }\\
&\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} \\
& =5 \Omega
\end{aligned}
\end{aligned}
\)

(b)
\(
\begin{aligned}
&\text { Phase difference, } \phi=\tan ^{-1} \frac{X_C-X_L}{R}\\
&=\tan ^{-1}\left(\frac{4-8}{3}\right)=-53.1^{\circ}
\end{aligned}
\)
Since \(\phi\) is negative, the current in the circuit lags the voltage across the source.

(c) The power dissipated in the circuit is
\(
P=I^2 R
\)
Now, \(I=\frac{i_m}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{283}{5}\right)=40 \mathrm{~A}\)
Therefore, \(P=(40 \mathrm{~A})^2 \times 3 \Omega=4800 \mathrm{~W}\)

(d) Power factor \(=\cos \phi=\cos \left(-53.1^{\circ}\right)=0.6\)

Example 7.9: Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.

Solution: (a) The frequency at which the resonance occurs is
\(
\begin{aligned}
\omega_0 & =\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}} \\
& =222.1 \mathrm{rad} / \mathrm{s} \\
\nu_r & =\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14} \mathrm{~Hz}=35.4 \mathrm{~Hz}
\end{aligned}
\)

(b) The impedance \(Z\) at resonant condition is equal to the resistance:
\(
Z=R=3 \Omega
\)
The rms current at resonance is
\(
=\frac{V}{Z}=\frac{V}{R}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}=66.7 \mathrm{~A}
\)
The power dissipated at resonance is
\(
P=I^2 \times R=(66.7)^2 \times 3=13.35 \mathrm{~kW}
\)
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8.

Example 7.10:  At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work?

Solution: The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.