A series circuit has an impedance of 60.0 Ω and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

In a series LCR A/C circuit, the voltage across the inductor leads current by a right angle, the voltage across capacitor lags the current by a right angle. The voltage across the resistor is in same phase with the current. The overall effect of all the components introduces a net phase shift between current and voltage.

If the inductive reactance is more than the capacitive reactance then the phase difference lies in the first quadrant. While if the capacitive reactance is more than the inductive reactance then the phase difference lies in the fourth quadrant.

Let the phase shift be $\varphi$. Then, power factor is given by $cos\left(\varphi \right)$. Power factor is the ratio of the power across the resistor and the total power across the circuit, which in nothing but the ratio of resistance and the total impedance of the circuit. Hence, $\cos \left(\varphi \right)=\frac{R}{Z}$, where, is the resistance in the circuit and z is the total impedance of the circuit.

To increase the power factor,$\cos \left(\varphi \right)$ needs to be increased which implies that$\varphi$ needs to be reduced. In the question it is specified that the voltage lags current. Which means$\varphi$ is in the fourth quadrant. That is capacitive reactance is more than the inductive reactance. To increase $\varphi$, therefore we need to increase the value of inductive reactance, which can be done by increasing the value of the inductance in the circuit. In other words an inductor has to be added to the circuit in series.

Given that at 50Hz frequency, the impedance of the circuit is$60\mathrm{\Omega }$ and power factor is 0.72. Hence,role="math" localid="1664192902249" $\cos \left(\varphi \right)=0.72$ ( $\varphi =-0.77\mathrm{rad}$, negative sign represents that$\varphi$ is in fourth quadrant). Therefore,

$$\begin{gathered} \frac{R}{Z}=0.72 \\ R=0.72\times Z \\ =0.72\times 60 \\ =43.2\Omega \end{gathered}$$

Now, we know that $\tan \left(\varphi \right)=\frac{X_L-X_c}{R}$, where,$X_L$ is the inductive reactance,$X_c$ is the capacitive reactance. Hence,

$$\begin{gathered} \tan \left(-0.77\right)=\frac{X_L-X_c}{43.2} \\ {X}_L-X_c=-41.89 \end{gathered}$$

For power factor to be unity, $\cos \left(\varphi \right)=1$. Hence, $\varphi =0$. Let$L\text{'}$ be the inductor that has to be added in series and$X_{L\text{'}}$ be its inductive reactance. Hence,

$$\begin{gathered} \tan \left(\varphi \right)=\frac{\left(X_{L\text{'}}+X_L\right)-X_C}{R} \\ 0=\frac{X_{L\text{'}}+\left(X_L-X_C\right)}{R} \\ 0=X_{L\text{'}}+\left(X_L-X_C\right) \\ {X}_{L\text{'}}=-\left(X_L-X_C\right) \\ {X}_{L\text{'}}=41.89 \end{gathered}$$

Now, we know that$X_{L\text{'}}=\omega L\text{'}$ where,$\omega$ is the angular frequency of the A/C signal. Therefore,

$$\begin{gathered} 41.89=2\mathrm{\pi }\times 50\times \mathrm{L}\text{'} \\ \mathrm{L}\text{'}=0.133\mathrm{H} \end{gathered}$$

Hence, an inductor of inductance 0.133 H has to be added to the circuit.