An a.c. e.m.f. of peak value 230 V and frequency 50 Hz is connected to a circuit with $\mathrm{R}=11.5 \Omega, \mathrm{~L}=2.5 \mathrm{H}$ and a capacitor all in series. The value of capacitance is ' $C$ ' for the current in the circuit to be maximum. The value of ' $C$ ' and maximum current are respectively ( $\pi^2=10$ )

Maximum current in an RLC series circuit occurs at resonance when $X_L = X_C$, making the impedance purely resistive with $Z = R$ and current $I_0 = \frac{V_0}{R}$.

Given $f = 50 \text{Hz}$ and $\pi^2 = 10$, angular frequency is $\omega = 2\pi f = 100\pi$ rad/s.

From $\omega L = \frac{1}{\omega C}$, capacitance is $C = \frac{1}{\omega^2 L} = \frac{1}{(100\pi)^2 (2.5)}$.
Substitute $\pi^2 = 10$ to find $C = \frac{1}{1000000 \times 2.5} = \frac{1}{250000} = 4\mu\text{F}$.

Maximum current is $I_0 = \frac{V_0}{R} = \frac{230}{11.5} = 20\text{A}$.
Thus, option A is correct.