Question

In the circuit shown in the figure, power factor of the box is $0.5$ and power factor of the circuit is $\frac{\sqrt{3}}{2}$. If current leads the voltage, the effective resistance $\left(\text{in}\Omega \right)$ of the box will be

Answer 1

Since, the current leads the voltage, there is a capacitive element in the box.

$\therefore$ The power factor of the box

$\frac{R}{\sqrt{R^2+X_C^2}}=\frac{1}{2}$

$\Rightarrow X_C^2=3R^2$

Also, the power factor of the circuit

$\therefore \frac{R+10}{\sqrt{(R+10{)}^2+X_c^2}}=\frac{\sqrt{3}}{2}$

$\Rightarrow 4(R+10{)}^2=3(R+10{)}^2+9R^2$

$\Rightarrow (R+10{)}^2=9R^2$

$\Rightarrow R+10=3R$

$\Rightarrow R=5\Omega$

$\text{Correct ans :}5$