Power, Power Factor & Wattless Current question from JEE Main 2015 (11 Apr, Shift 1)

Answer: (d)

1. Current leads, so $X_C > X_L$: $C$ is too small.
2. To make $X_C = X_L$, total capacitance must increase, so $C'$ goes in parallel: $C_{eq} = C + C'$.
3. Unity power factor: $\frac{1}{\omega C_{eq}} = \omega L \Rightarrow C_{eq} = \frac{1}{\omega^2 L}$
4. $C' = \frac{1}{\omega^2 L} - C = \frac{1-\omega^2 LC}{\omega^2 L}$, so the answer is (d).