1. Condition for maximum average power in a series LCR circuit

For a series LCR circuit, the average power is $P_{avg}=V_{rms}I_{rms}\cos\phi$ with $I_{rms}=\frac{V_{rms}}{Z}, \qquad Z=\sqrt{R^2+(X_L-X_C)^2}.$

The average power is maximum when the impedance is minimum, i.e. at resonance: $X_L=X_C.$

So we must choose the capacitor such that $X_C = X_L = 250\pi\ \Omega.$

2. Use the capacitive reactance formula

$X_C=\frac{1}{\omega C}$ where $\omega=2\pi f=2\pi(50)=100\pi\ \text{rad/s}.$

Thus, $\frac{1}{\omega C}=250\pi$ $\frac{1}{(100\pi)C}=250\pi.$

So, C=\frac{1}{100\pi\cdot 250\pi}=rac{1}{25000\pi^2}\ \text{F}.

Given: $\pi^2=10$ therefore, C=\frac{1}{25000\times 10}=rac{1}{250000}=4\times 10^{-6}\ \text{F}.

Hence, $C=4\ \mu\text{F}.$

3. Check options

• A: $4\,\mu$F ✅
• B: $25\,\mu$F ❌
• C: $400\,\mu$F ❌
• D: $40\,\mu$F ❌

So the correct answer is Option A.