A 20-watt, 50 V lamp is to be connected to AC mains of 250 V, 50 Hz. Calculate the value of series capacitor to run the lamp.

The correct option is 4.9 µF.

To understand why this is the correct answer, let us break down the calculations step-by-step.

Step 1: Determine the properties of the lamp
The lamp is rated at a power (P) of 20 W and a voltage (V_L) of 50 V.
We can find the rated current (I) flowing through the lamp using the formula:
$$P=V_L\times I$$
Rearranging for I:
$$I=\frac{P}{V_L}=\frac{20}{50}=0.4\text{A}$$
The resistance of the lamp (R) is:
$$R=\frac{V_L}{I}=\frac{50}{0.4}=125\Omega$$

Step 2: Set up the AC circuit impedance
The lamp (modeled as a pure resistance R) is connected in series with a capacitor of capacitance C. The circuit is connected to an AC supply voltage (V) of 250 V at a frequency (f) of 50 Hz.
The total impedance (Z) of this series R-C circuit is given by:
$$Z=\sqrt{R^2+X_C^2}$$
where X_C is the capacitive reactance of the series capacitor.

Step 3: Calculate the required impedance
To run the lamp at its rated current of 0.4 A when connected to the 250 V AC mains, the total impedance must satisfy:
$$Z=\frac{V}{I}=\frac{250}{0.4}=625\Omega$$

Step 4: Calculate the required capacitive reactance (X_C)
Using the relation for Z:
$$Z^2=R^2+X_C^2$$
Substitute the values Z = 625 Ω and R = 125 Ω:
$$X_C^2={625}^2-{125}^2$$
Using the algebraic identity a² - b² = (a - b)(a + b):
$$X_C^2=(625-125)\times (625+125)=500\times 750=375000$$
Taking the square root:
$$X_C=\sqrt{375000}\approx 612.37\Omega$$

Step 5: Calculate the value of the capacitor (C)
The relation between capacitive reactance and capacitance is:
$$X_C=\frac{1}{2\pi fC}$$
Rearranging the formula to solve for C:
$$C=\frac{1}{2\pi f{X}_C}$$
Substitute f = 50 Hz and X_C ≈ 612.37 Ω:
$$C=\frac{1}{2\times 3.14159\times 50\times 612.37}$$