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

Explanation

To solve the problem of finding the inductance of the choke coil that should be connected in series with a bulb rated at 100V and 100W when connected to an AC source of 200V and 50Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Resistance of the Bulb**:
The power rating of the bulb is given as 100W, and the voltage rating is 100V. We can use the formula for power:
\[ P = \frac{V^2}{R} \]
Rearranging this gives:
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{100^2}{100} = \frac{10000}{100} = 100 \, \Omega \]

2. **Calculate the Current through the Bulb**:
The current can be calculated using the power formula:
\[ P = V \cdot I \]
Rearranging gives:
\[ I = \frac{P}{V} \]
Substituting the values:
\[ I = \frac{100}{100} = 1 \, A \]

3. **Determine the Impedance of the Circuit**:
The total voltage supplied is 200V, and we know the current is 1A. The impedance (Z) can be calculated as:
\[ Z = \frac{V}{I} \]
Substituting the values:
\[ Z = \frac{200}{1} = 200 \, \Omega \]

4. **Relate Impedance to Resistance and Reactance**:
The impedance in an R-L circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where \(X_L\) is the inductive reactance. We can square both sides:
\[ Z^2 = R^2 + X_L^2 \]
Substituting the known values:
\[ 200^2 = 100^2 + X_L^2 \]
This simplifies to:
\[ 40000 = 10000 + X_L^2 \]
Rearranging gives:
\[ X_L^2 = 40000 - 10000 = 30000 \]

5. **Calculate the Inductive Reactance**:
Taking the square root gives:
\[ X_L = \sqrt{30000} = 100\sqrt{3} \, \Omega \]

6. **Relate Inductive Reactance to Inductance**:
The inductive reactance is related to the inductance (L) and the frequency (f) by:
\[ X_L = 2\pi f L \]
Rearranging for L gives:
\[ L = \frac{X_L}{2\pi f} \]
Substituting the values:
\[ L = \frac{100\sqrt{3}}{2\pi \cdot 50} \]
Simplifying gives:
\[ L = \frac{100\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{\pi} \, H \]

### Final Answer:
The inductance of the choke coil is:
\[ L = \frac{\sqrt{3}}{\pi} \, H \]