The value of C is ___.

The circuit contains a metal-filament lamp (pure resistance) in series with a capacitor across a 200 V, 50 Hz a.c. supply. All given voltages and powers are rms values.

Step 1 Resistance of the lamp
The power absorbed by a pure resistor is $$P = \frac{V_r^{2}}{R}$$.
Given $$P = 500\;{\rm W}$$ and the rms voltage across the lamp $$V_r = 100\;{\rm V}$$,

$$R = \frac{V_r^{2}}{P} = \frac{100^{2}}{500} = 20\;\Omega.$$

Step 2 Current in the series circuit
For a resistor, $$I = \frac{V_r}{R}$$, hence

$$I = \frac{100}{20} = 5\;{\rm A\;(rms)}.$$

Step 3 Voltage across the capacitor
The resistor voltage is in phase with the current, while the capacitor voltage lags the current by $$90^{\circ}$$. Therefore, the supply-voltage phasor satisfies

$$V_s^{2} = V_r^{2} + V_c^{2}.$$

With $$V_s = 200\;{\rm V}$$ and $$V_r = 100\;{\rm V}$$,

$$V_c = \sqrt{200^{2} - 100^{2}} = \sqrt{40000 - 10000} = 100\sqrt{3}\;{\rm V}.$$

Step 4 Capacitive reactance
$$X_C = \frac{V_c}{I} = \frac{100\sqrt{3}}{5} = 20\sqrt{3}\;\Omega.$$

Step 5 Required capacitance
For a capacitor, $$X_C = \frac{1}{2\pi f C}$$, so

$$C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 20\sqrt{3}} = \frac{1}{2000\pi\sqrt{3}}\;{\rm F}.$$

The question allows the approximation $$\pi\sqrt{3} \approx 5$$; hence

$$C \approx \frac{1}{2000 \times 5} = \frac{1}{10000}\;{\rm F} = 1.0 \times 10^{-4}\;{\rm F} = 100\;\mu{\rm F}.$$

Even with a more accurate value of $$\pi\sqrt{3}$$ the result lies near $$92\;\mu{\rm F}$$, well inside the accepted range.

Final answer: $$C \approx 100\;\mu{\rm F}\; \bigl(\text{accepted range }90-105\;\mu{\rm F}\bigr).$$