In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $\mathrm{C} \mu…

Given, 

$P=500 \mathrm{W}$

$V_R=100 \mathrm{V}$

$f=50 \mathrm{Hz}$

$V_{rms}=200 \mathrm{V}$

$$\begin{gathered} P=\frac{{V_R}^2}{R} \\ \Rightarrow R=\frac{{V_R}^2}{P} \\ \Rightarrow R=\frac{{\left(100\right)}^2}{500} \end{gathered}$$

$\therefore R=20 \mathrm{\Omega }$

Now current in the circuit,

$$\begin{gathered} i=i_{rms}=\frac{\mathrm{voltage} \mathrm{across} \mathrm{resistor}}{\mathrm{resistance}} \\ i=\frac{100}{20} \\ i=5 \mathrm{A} \end{gathered}$$

Phasor diagram of the given circuit for voltage,

Now, 

$$\begin{gathered} V^2={V_C}^2+{V_R}^2 \\ \Rightarrow {200}^2={V_C}^2+{100}^2 \\ \Rightarrow V_C=100\sqrt{3} \mathrm{V} \end{gathered}$$

Now, 

$$\begin{gathered} V_C=100\sqrt{3} \mathrm{V}=I{X}_C \\ \Rightarrow X_C=\frac{100\sqrt{3}}{5}=20\sqrt{3} \mathrm{\Omega } \end{gathered}$$

and 

$$\begin{gathered} X_C=\frac{1}{\left(2\pi f\right)C} \\ \Rightarrow 20\sqrt{3}=\frac{1}{\left(2\pi \times 100\right)C} \\ \Rightarrow C=100 \mathrm{\mu F} \end{gathered}$$

From the phasor diagram,

$$\begin{gathered} \tan \left(\varphi \right)=\frac{X_C}{R}=\frac{20\sqrt{3}}{20}=\sqrt{3} \\ \Rightarrow \varphi =60° \end{gathered}$$