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 an AC source of 220 V at 50 Hz, with $$R = 100$$ $$\Omega$$ and $$X_L = 79.6$$ $$\Omega$$.

The average rate of energy supplied (power) is maximized at resonance, when the impedance equals $$R$$. At resonance, $$X_C = X_L$$, so

$$\frac{1}{2\pi f C} = X_L$$

Solving for $$C$$,

$$C = \frac{1}{2\pi f X_L}$$

Now substituting the values,

$$C = \frac{1}{2\pi \times 50 \times 79.6} = \frac{1}{2 \times 3.14159 \times 50 \times 79.6} = \frac{1}{25000.5}$$

$$C \approx 4.0 \times 10^{-5} \text{ F} = 40 \text{ }\mu\text{F}$$

Hence, the correct answer is Option 2.