Question

In series $L-R$ circuit, $X_L=R$. Now a capacitor with $X_C=R$ added in series. New power factor:

• A. Same as initial
• B. $\frac{1}{\sqrt{2}}$ times the initial
• C. $\frac{1}{2}$ times the initial
• D. $\sqrt{2}$ times the initial

Answer 1

The correct option is D $\sqrt{2}$ times the initial

Given that,

$X_L=R$

$X_C=R$

We know that, the impedance is

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

When, $X_L=R$

Then,

$Z=R\sqrt{2}$

Now, the power factor is

$P_1=\frac{R}{Z}$

$P_1=\frac{R}{R\sqrt{2}}$

$P_1=\frac{1}{\sqrt{2}}$

Now, $X_c=R$

So, $Z=R$

Now, the new power factor is

$P_2=\frac{R}{Z}$

$P_2=\frac{R}{R}$

$P_2=1$

Now, the new power factor is

$P_1=\frac{1}{\sqrt{2}}$

$P_1\sqrt{2}=1$

$P_2=P_1\sqrt{2}$

Hence, the new power factor is $\sqrt{2}$ initial power factor