A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contain a resistance $$R = 100 \Omega$$ and an inductor of inductive reactance $$X_L = 79.6 \Omega$$. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be ______ $$\mu F$$.

We have a series LCR circuit connected to a 220 V, 50 Hz AC source with $$R = 100 \, \Omega$$ and $$X_L = 79.6 \, \Omega$$, and we need to find the capacitance for maximum average power.

Since the average power in a series LCR circuit is $$P = V_{rms}^2 R / Z^2$$, where $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$, the power is maximized when the impedance $$Z$$ is minimized, which occurs at resonance, i.e. when $$X_L = X_C$$.

At resonance we have $$X_C = X_L = 79.6 \, \Omega$$.

The capacitive reactance is given by $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$, so solving for $$C$$ gives $$C = \frac{1}{2\pi f \cdot X_C} = \frac{1}{2\pi \times 50 \times 79.6}$$.

This is $$\frac{1}{2 \times 3.14159 \times 50 \times 79.6} = \frac{1}{314.159 \times 79.6} = \frac{1}{25007.1}$$, which equals $$3.999 \times 10^{-5} \text{ F} \approx 40 \, \mu\text{F}$$.

Therefore, the correct answer is Option X.