In a series LR circuit $X_L=R$ and power factor of the circuit is $\mathrm{P}_1$. When capacitor with capacitance C such that $X_L=X_C$is put in series, the power factor becomes $P_2$.. The ratio $\frac{P_1}{P_2}$ is
Select the correct option:
A
$\frac{1}{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{\sqrt{3}}{\sqrt{2}}$
D
$2: 1$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
In case of L-R circuit
$Z=\sqrt{X_L^2+R^2}$ \& power factor
$$
P_1=\cos \phi=\frac{R}{Z}
$$
As $\mathrm{X}_{\mathrm{L}}=\mathrm{R}$
$$
\begin{aligned}
& \Rightarrow Z=\sqrt{2} R \\
& \Rightarrow P_1=\frac{R}{\sqrt{2} R} \Rightarrow P_1=\frac{1}{\sqrt{2}}
\end{aligned}
$$
In case of L-C-R circuit
$$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$$
As $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}}$
$$
\Rightarrow Z=R
$$
$\Rightarrow \mathrm{P}_2=\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$
$$
\Rightarrow \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}
$$
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