Question

A bulb is rated at 100 V, 100 W. It can be treated as a resistor. If the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 Hz

is √x/π H. Find X

Answer 1

The inductance of the choke coil required to operate the bulb at its rated power with a 200 V, 50 Hz AC source is 1/π H, which means x in the given formula √x/π H equals 1.

Step-by-step explanation:

To determine the inductance of the choke coil required to operate a 100 W, 100 V bulb at its rated power using a 200 V, 50 Hz AC source, we first need to find the resistance of the bulb. The power rating of the bulb (P) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance. Therefore, R = V^2 / P = 100^2 / 100 = 100 ohms.

Next, we need to find the current drawn by the bulb at its rated power, which is I = P / V = 100 / 100 = 1 A. The RMS voltage of the AC source is 200 V, and the bulb needs to drop 100 V across it to operate at 100 W. Therefore, we need the remaining 100 V to be dropped across the choke coil.

Using Ohm's law V = I * R and the inductive reactance formula XL = 2 π f L, we have:

V = I * XL

100 = 1 * 2 * π * 50 * L

L = 100 / (π * 100) = 1/π H

Therefore, x, in the equation sqrt(x)/π, is 1.