Question

In the circuit shown in figure power factor of box is $0.5$ and power factor of circuit is $\sqrt{3}/2$. Current leading the voltage. Find the effective resistance (in ohms) of the box.

Answer 1

$\cos {\varphi }_1=0.5$
$\therefore {\varphi }_1={60}^o$
$\cos {\varphi }_2=\frac{\sqrt{3}}{2}$
$\therefore {\varphi }_2={30}^o$
Let $R$ be the effective resistance of the box. Then
$\tan {\varphi }_1=\frac{X_C}{R}$ or $\sqrt{3}=\frac{X_C}{R}$ ....(i)
$\tan {\varphi }_2=\frac{X_C}{R+10}$ or $\frac{1}{\sqrt{3}}=\frac{X_C}{R+10}$ .....(ii)
From these two equations we get $:R=5\Omega$