An ekectric bulb of 100 W - 300 V is connected with an AC supply of 500 V and `(150)/(pi)` Hz. The required inductance to save the electric bulb is

Options

A. 2 H
B. `(1)/(2)` H
C. 4 H
D. `(1)/(4)` H

Answer

To solve the problem of finding the required inductance to protect a 100 W, 300 V electric bulb when connected to a 500 V AC supply at a frequency of \( \frac{150}{\pi} \) Hz, we will follow these steps:

### Step 1: Calculate the resistance of the bulb

The resistance \( R \) of the bulb can be calculated using the formula:

\[ R = \frac{V^2}{P} \]

Where:
- \( V = 300 \, \text{V} \) (rated voltage)
- \( P = 100 \, \text{W} \) (rated power)

Substituting the values:

\[ R = \frac{300^2}{100} = \frac{90000}{100} = 900 \, \Omega \]

### Step 2: Calculate the maximum current through the bulb

The maximum current \( I_0 \) through the bulb can be calculated using the formula:

\[ I_0 = \frac{P}{V} \]

Substituting the values:

\[ I_0 = \frac{100}{300} = \frac{1}{3} \, \text{A} \]

### Step 3: Calculate the total impedance of the circuit

The total impedance \( Z \) of the circuit can be calculated using the maximum voltage \( V_0 \) and the maximum current \( I_0 \):

\[ Z = \frac{V_0}{I_0} \]

Where:
- \( V_0 = 500 \, \text{V} \)

Substituting the values:

\[ Z = \frac{500}{\frac{1}{3}} = 500 \times 3 = 1500 \, \Omega \]

### Step 4: Relate impedance to resistance and inductive reactance

The total impedance \( Z \) is related to the resistance \( R \) and the inductive reactance \( X_L \) as follows:

\[ Z = \sqrt{R^2 + X_L^2} \]

Substituting the known values:

\[ 1500 = \sqrt{900^2 + X_L^2} \]

### Step 5: Solve for inductive reactance \( X_L \)

Squaring both sides gives:

\[ 1500^2 = 900^2 + X_L^2 \]

Calculating \( 1500^2 \) and \( 900^2 \):

\[ 2250000 = 810000 + X_L^2 \]

Now, solving for \( X_L^2 \):

\[ X_L^2 = 2250000 - 810000 = 1440000 \]

Taking the square root:

\[ X_L = \sqrt{1440000} = 1200 \, \Omega \]

### Step 6: Calculate the inductance \( L \)

The inductive reactance \( X_L \) is related to the inductance \( L \) and the angular frequency \( \omega \):

\[ X_L = \omega L \]

Where \( \omega = 2 \pi f \) and \( f = \frac{150}{\pi} \):

Calculating \( \omega \):

\[ \omega = 2 \pi \left(\frac{150}{\pi}\right) = 300 \, \text{rad/s} \]

Now substituting \( X_L \) and \( \omega \) to find \( L \):

\[ 1200 = 300 L \]

Solving for \( L \):

\[ L = \frac{1200}{300} = 4 \, \text{H} \]

### Final Answer

The required inductance to save the electric bulb is:

\[ \boxed{4 \, \text{H}} \]