Physics - EMI/AC Question with Solution | TestHub

Electrical power is given by $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$

Resistance of lamp is $R=\frac{100\times 100}{50}\Rightarrow R=200\Omega$

Let $V_C$ and $V_R$ be the voltage across capacitor and resistor.

Here, $V_R^2+V_C^2={\mathrm{V}}^2$

Current through lamp is $i=\frac{100}{200}=\frac{1}{2}\mathrm{A}$

$\Rightarrow {\left(100\right)}^2+V_C^2={\left(200\right)}^2$

$\Rightarrow V_C^2=30000\Rightarrow V_C=100\sqrt{3}\mathrm{V}$

We know that, $V=i\times X_C$

Then, $X_C=200\sqrt{3}\Omega =\frac{1}{\omega C}$

So, capacitance of capacitor is $C=\frac{1}{20\times 50\times 20\sqrt{3}}=\frac{50\times 10^{-6}}{\sqrt{x}}$

$\sqrt{x}=50\times 10^{-6}\times 100\times 200\sqrt{3}$

Thus, value of $x=3$.