To light, a 50W, 100V lamp is connected, in series with a capacitor...

Electrical power is given by $P = \dfrac{{V}^{2}}{R} ⇒ R = \dfrac{{V}^{2}}{P}$

Resistance of lamp is $R = \dfrac{100 \times 100}{50} ⇒ R = 200Ω$

Let ${V}_{C}$ and ${V}_{R}$ be the voltage across capacitor and resistor.

Here, ${V}_{R}^{2} + {V}_{C}^{2} = {V}^{2}$

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

$⇒ {\left(100\right)}^{2} + {V}_{C}^{2} = {\left(200\right)}^{2}$

$⇒ {V}_{C}^{2} = 30000 ⇒ {V}_{C} = 100\sqrt{3}V$

We know that, $V = i \times {X}_{C}$

Then, ${X}_{C} = 200\sqrt{3}Ω = \dfrac{1}{\omega C}$

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

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

Thus, value of $x = 3$.