In a series L, C and R circuit with an ac source of frequency ω, the current leads the voltage by π/4. The value of capacitance C will be

$(ω^2 L+ωR)^{-1}$

The correct answer is Option (4) → $(ω^2 L+ωR)^{-1}$

Given that in an LCR series circuit the current leads the voltage by $\frac{\pi}{4}$:

Hence, $\tan \phi = \frac{X_C - X_L}{R} = 1$

⟹ $X_C - X_L = R$

Substitute $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$:

$\frac{1}{\omega C} - \omega L = R$

⟹ $\frac{1}{\omega C} = \omega L + R$

⟹ $C = \frac{1}{\omega(\omega L + R)}$

Final Answer: $C = (\omega^2 L + \omega R)^{-1}$