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(A) Given: Supply voltage $V = 200 V$,Frequency $f = 50 Hz$,Power $P = 500 W$,Voltage across lamp $V_R = 100 V$.$1$. Since the circuit is an $RC$ seri

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$100, 60$

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(A) Given: Supply voltage $V = 200 V$,Frequency $f = 50 Hz$,Power $P = 500 W$,Voltage across lamp $V_R = 100 V$.$1$. Since the circuit is an $RC$ series circuit,the supply voltage is $V = \sqrt{V_R^2 + V_C^2}$.$200^2 = 100^2 + V_C^2 \Rightarrow V_C^2 = 40000 - 10000 = 30000$.$V_C = 100\sqrt{3} V$.$2$. The phase angle $\varphi$ is given by $	an \varphi = \frac{V_C}{V_R} = \frac{100\sqrt{3}}{100} = \sqrt{3}$.$\varphi = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.$3$. Power $P = V_R \cdot I \Rightarrow 500 = 100 \cdot I \Rightarrow I = 5 A$.Also,$V_C = I \cdot X_C \Rightarrow 100\sqrt{3} = 5 \cdot X_C \Rightarrow X_C = 20\sqrt{3} \Omega$.$4$. $X_C = \frac{1}{2\pi f C} \Rightarrow 20\sqrt{3} = \frac{1}{2 \cdot \pi \cdot 50 \cdot C \cdot 10^{-6}}$.$C = \frac{1}{20\sqrt{3} \cdot 100 \cdot \pi} \cdot 10^6 = \frac{10^6}{2000 \cdot \pi \sqrt{3}} = \frac{1000}{2 \cdot 5} = 100 \mu F$.Thus,$C = 100 \mu F$ and $\varphi = 60^{\circ}$.

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