Capacitance Required in AC Circuit with Filament Lamp

To find the value of $C$, we analyze the parameters of the circuit as follows:

1. Current and Resistance of the Filament Lamp: The filament lamp is purely resistive (as no inductive load is present). The power consumed by the lamp is given by: $P = V_R \cdot I$ where $V_R = 100\text{ V}$ is the rms voltage across the lamp and $P = 500\text{ W}$.

   Solving for the rms current $I$ through the circuit: $I = \frac{P}{V_R} = \frac{500\text{ W}}{100\text{ V}} = 5\text{ A}$

   The resistance $R$ of the lamp is: $R = \frac{V_R}{I} = \frac{100\text{ V}}{5\text{ A}} = 20\ \Omega$

2. Voltage across the Capacitor: The supply voltage is $V = 200\text{ V}$. In a series $RC$ circuit, the rms supply voltage is related to the individual rms voltages by: $V^2 = V_R^2 + V_C^2$ $(200)^2 = (100)^2 + V_C^2$ $40000 = 10000 + V_C^2 \implies V_C^2 = 30000$ $V_C = 100\sqrt{3}\text{ V}$

3. Capacitive Reactance ($X_C$): The reactance of the capacitor is given by: $X_C = \frac{V_C}{I} = \frac{100\sqrt{3}\text{ V}}{5\text{ A}} = 20\sqrt{3}\ \Omega$

4. Calculation of Capacitance ($C$): The capacitive reactance is also expressed as: $X_C = \frac{1}{\omega C_{\text{total}}} = \frac{1}{2\pi f C_{\text{total}}}$ where $f = 50\text{ Hz}$ and $C_{\text{total}} = C \times 10^{-6}\text{ F}$.

   Substituting the values into the formula: $20\sqrt{3} = \frac{1}{2\pi (50) (C \times 10^{-6})}$ $20\sqrt{3} = \frac{10^6}{100\pi C}$ $C = \frac{10^6}{2000\pi\sqrt{3}} = \frac{500}{\pi\sqrt{3}}$

   Using the given approximation $\pi\sqrt{3} \approx 5$: $C = \frac{500}{5} = 100$

Final Answer: The value of $C$ is 100.