In a series LR circuit, $X_L=R$, power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C=X_L$ is…
In a series LR circuit, $X_L=R$, power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C=X_L$ is added to the circuit the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be
$1: 3$
$1: \sqrt{2}$
$1: 1$
$1: 2$
Solution
$\begin{array}{ll} & \text { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}} \\ & \text { Given: } \mathrm{X}_{\mathrm{L}}=\mathrm{R} \\ \therefore \quad & \mathrm{P}_1=\frac{\mathrm{R}}{\sqrt{2 \mathrm{R}^2}}=\frac{1}{\sqrt{2}}\end{array}$
When $\mathrm{C}$ is connected, the circuit becomes a series LCR circuit.
$\begin{array}{ll}
\therefore & \text { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}} \\
& \text { Given: } \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}} \\
\therefore & \mathrm{P}_2=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2}}=1 \\
\therefore & \mathrm{P}_1: \mathrm{P}_2=1: \sqrt{2}
\end{array}$