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 find the required inductance to save the electric bulb when connected to an AC supply, we can follow these steps:

### Step 1: Identify the given values
- Power of the bulb (P) = 100 W
- Voltage of the bulb (V_bulb) = 300 V
- Voltage of the AC supply (V_supply) = 500 V
- Frequency (f) = \( \frac{150}{\pi} \) Hz

### Step 2: Calculate the current through the bulb
Using the formula for power in an AC circuit:
\[ P = V \cdot I \]
We can rearrange this to find the current (I):
\[ I = \frac{P}{V_bulb} = \frac{100 \, \text{W}}{300 \, \text{V}} = \frac{1}{3} \, \text{A} \]

### Step 3: Calculate the voltage difference across the inductance
The voltage across the inductance (V_L) can be calculated as:
\[ V_L = V_{supply} - V_{bulb} = 500 \, \text{V} - 300 \, \text{V} = 200 \, \text{V} \]

### Step 4: Relate voltage across the inductance to inductance and current
Using the formula for inductive reactance:
\[ V_L = \omega L I \]
Where:
- \( \omega = 2 \pi f \)
- \( f = \frac{150}{\pi} \)

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

### Step 5: Substitute known values into the inductance formula
Now we can substitute \( V_L \), \( \omega \), and \( I \) into the equation:
\[ 200 = 300 L \left( \frac{1}{3} \right) \]

### Step 6: Solve for L
Rearranging the equation to solve for L:
\[ 200 = 100 L \]
\[ L = \frac{200}{100} = 2 \, \text{H} \]

### Final Answer
The required inductance to save the electric bulb is \( L = 2 \, \text{H} \).

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