Question

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C\mu \text{F}$ across a $200\text{V},50\text{Hz}$ supply. The power consumed by the lamp is $500\text{W}$ while the voltage drop across it is $100\text{V}$. Assume that there is no inductive load in the circuit. Take $rms$ values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is $\varphi$.

Assume, $\pi \sqrt{3}\approx 5$.

The value of $C$ is -

Answer 1

$\sqrt{V_C^2+V_R^2}=E_{\text{rms}}$

$\Rightarrow V_C^2+{100}^2={200}^2$

$\Rightarrow V_C=100\sqrt{3}\text{V}$

We know that,

$i=\frac{P}{V_R}=\frac{500}{100}=5\text{A}$

Now,

$V_C=i{X}_C$

$C=\frac{i}{V_C\left(2\pi f\right)}=\frac{5}{100\sqrt{3}\times 2\pi \times 50}$

$\therefore C=1\times {10}^{-4}\text{F}=100\times {10}^{-6}\text{F}=100\mu \text{F}$