Question

Power factor of an $L-R$ series circuit is $0.6$ and that of a $C–R$ series circuit is $0.5$. If the elements $(L,C,$ and $R)$ of the two circuits are joined in series, the power factor of this circuit is found to be $1$. The ratio of the resistance in the $L-R$ circuit to the resistance in the $C-R$ circuit is

• A. $\frac{6}{5}$
• B. $\frac{5}{6}$
• C. $\frac{4}{3\sqrt{3}}$
• D. $\frac{3\sqrt{3}}{4}$

Answer 1

The correct option is D $\frac{3\sqrt{3}}{4}$

For $L-R$ circuit:

Given $\cos \varphi =0.6=\frac{3}{5}$
$\Rightarrow \varphi ={53}^{\circ }$
$\tan {53}^{\circ }=\frac{X_L}{R_1}$
$\Rightarrow X_L=\frac{4R_1}{3}$


Given $\cos \varphi =0.5$
$\Rightarrow \varphi ={60}^{\circ }$
$\tan {60}^{\circ }=\frac{X_C}{R_2}$
$\Rightarrow \sqrt{3}=\frac{X_C}{R_2}$
$\Rightarrow X_C=\sqrt{3}{R}_2$

For $L-C-R$ combination:
Given that power factor is $1$ so,$X_L=X_C$
$\Rightarrow \frac{4R_1}{3}=\sqrt{3}{R}_2$
$\Rightarrow \frac{R_1}{R_2}=\frac{3\sqrt{3}}{4}$