To light, a $$50 \text{ W}$$, $$100 \text{ V}$$ lamp is connected, in series with a capacitor of capacitance $$\dfrac{50}{\pi\sqrt{x}} \mu F$$, with $$200 \text{ V}$$, $$50 \text{ Hz}$$ AC source. The value of $$x$$ will be ______.

We have a $$50 \text{ W}$$, $$100 \text{ V}$$ lamp connected in series with a capacitor of capacitance $$C = \dfrac{50}{\pi\sqrt{x}} \text{ }\mu\text{F}$$ to a $$200 \text{ V}$$, $$50 \text{ Hz}$$ AC source. We need to find $$x$$.

The lamp is rated $$50 \text{ W}$$ at $$100 \text{ V}$$:

$$R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \text{ }\Omega$$

For the lamp to glow at full brightness, the voltage across it must be $$100 \text{ V}$$:

$$I = \frac{P}{V} = \frac{50}{100} = 0.5 \text{ A}$$

The total voltage is $$200 \text{ V}$$, so the impedance is:

$$Z = \frac{V_s}{I} = \frac{200}{0.5} = 400 \text{ }\Omega$$

For a series R-C circuit:

$$Z = \sqrt{R^2 + X_C^2}$$

$$400 = \sqrt{200^2 + X_C^2}$$

$$160000 = 40000 + X_C^2$$

$$X_C^2 = 120000$$

$$X_C = 200\sqrt{3} \text{ }\Omega$$

$$X_C = \frac{1}{2\pi f C}$$

$$200\sqrt{3} = \frac{1}{2\pi \times 50 \times C}$$

$$C = \frac{1}{200\sqrt{3} \times 100\pi} = \frac{1}{20000\sqrt{3}\pi}$$

Converting to $$\mu\text{F}$$:

$$C = \frac{10^6}{20000\sqrt{3}\pi} = \frac{50}{\sqrt{3}\pi} \text{ }\mu\text{F}$$

Comparing with $$C = \dfrac{50}{\pi\sqrt{x}} \text{ }\mu\text{F}$$:

$$\frac{50}{\pi\sqrt{x}} = \frac{50}{\sqrt{3}\pi}$$

$$\sqrt{x} = \sqrt{3}$$

$$x = 3$$

The value of $$x$$ is $$3$$.