1. To maximize the average rate at which energy is supplied in a series LCR circuit, the circuit must be at resonance.

2. For resonance in a series LCR circuit: $X_L = X_C$ where $X_C = \frac{1}{\omega C}$

3. Given:

• Frequency: $f = 50\,\text{Hz}$
• Inductive reactance: $X_L = 79.6\,\Omega$

First compute angular frequency: $\omega = 2\pi f = 2\pi \times 50 = 100\pi\,\text{rad/s}$

4. At resonance: $X_C = X_L = 79.6$ So, $\frac{1}{\omega C} = 79.6$

Hence, $C = \frac{1}{\omega X_L}$

Substitute values: $C = \frac{1}{(100\pi)(79.6)}$

Using $\pi \approx 3.14$: $C = \frac{1}{100 \times 3.14 \times 79.6}$ $C = \frac{1}{24994.4}$ $C \approx 4.0 \times 10^{-5}\,\text{F}$

5. Convert into microfarads: $4.0 \times 10^{-5}\,\text{F} = 40 \times 10^{-6}\,\text{F} = 40\,\mu\text{F}$

6. Therefore, the required capacitance is: $\boxed{40\,\mu\text{F}}$

7. Comparison with stored answer:

• Derived answer = $40$
• Stored correct answer = $40$

They match.