A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k$$\Omega$$, an inductor of inductive reactance $$X_L = 250\pi$$ $$\Omega$$ and an unknown capacitor. The value of capacitance to maximise the average power should be: (take $$\pi^2 = 10$$)

We begin with the expression for the impedance of a series $$LCR$$ circuit

$$Z=\sqrt{R^{2}+\left(X_L-X_C\right)^{2}}.$$

The rms current is $$I=\dfrac{V}{Z},$$ so the average (true) power delivered to the circuit is

$$P_{\text{avg}} = V_{\text{rms}}\,I_{\text{rms}}\cos\phi = V_{\text{rms}} \left(\dfrac{V_{\text{rms}}}{Z}\right)\left(\dfrac{R}{Z}\right) = \dfrac{V_{\text{rms}}^{2}\,R}{Z^{2}}.$$

Because $$R$$ is fixed, $$P_{\text{avg}}$$ becomes maximum when the impedance $$Z$$ is minimum. Looking at the formula for $$Z,$$ the minimum value occurs when the bracketed term vanishes, that is, when

$$X_L = X_C.$$

This condition is called resonance, and at resonance the circuit behaves as a pure resistance: the phase angle $$\phi$$ becomes zero and $$\cos\phi = 1,$$ giving the greatest possible average power.

We are given the inductive reactance

$$X_L = 250\pi\; \Omega.$$

To satisfy $$X_L = X_C,$$ we must have

$$X_C = 250\pi\; \Omega.$$

The capacitive reactance is related to the capacitance by the formula

$$X_C = \frac{1}{\omega C},$$

where the angular frequency $$\omega$$ equals $$2\pi f.$$ The supply frequency is $$f = 50\;\text{Hz},$$ so

$$\omega = 2\pi f = 2\pi \times 50 = 100\pi\;\text{rad s}^{-1}.$$

Substituting $$X_C = 250\pi$$ and $$\omega = 100\pi$$ into $$X_C = \dfrac{1}{\omega C},$$ we get

$$250\pi = \frac{1}{100\pi\,C}.$$

Rearranging for $$C$$ gives

$$C = \frac{1}{100\pi \times 250\pi} = \frac{1}{25000\pi^{2}}\;\text{farad}.$$

The problem states that $$\pi^{2} = 10,$$ so

$$C = \frac{1}{25000 \times 10} = \frac{1}{250000}\;\text{F}.$$

To convert to microfarads, recall that $$1\;\text{F} = 10^{6}\;\mu\text{F},$$ therefore

$$C = \frac{10^{6}}{250000}\;\mu\text{F} = 4\;\mu\text{F}.$$

Hence, the correct answer is Option B.