A 60 W , 120 V bulb is connected to a 240 V , 60 Hz supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.

Options

A. `(2.3)/(pi)H`
B. `2sqrt(3)H`
C. `piH`
D. `(2sqrt(3))/(pi) H`

Answer

To solve the problem of finding the inductance required for a 60 W, 120 V bulb connected to a 240 V, 60 Hz supply, we can follow these steps:

### Step 1: Determine the resistance of the bulb
The power \( P \) of the bulb is given by the formula:
\[ P = \frac{V^2}{R} \]
Where:
- \( P = 60 \) W (power rating of the bulb)
- \( V = 120 \) V (rated voltage of the bulb)

Rearranging the formula to find the resistance \( R \):
\[ R = \frac{V^2}{P} = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]

### Step 2: Calculate the current through the bulb
Using Ohm's Law, the current \( I \) through the bulb can be calculated as:
\[ I = \frac{V}{R} = \frac{120}{240} = 0.5 \, \text{A} \]

### Step 3: Calculate the total impedance \( Z \)
The total supply voltage \( V_s \) is 240 V. The impedance \( Z \) can be calculated using:
\[ Z = \frac{V_s}{I} = \frac{240}{0.5} = 480 \, \Omega \]

### Step 4: Relate impedance to resistance and inductive reactance
For an LR circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where \( X_L \) is the inductive reactance, given by \( X_L = \omega L \) and \( \omega = 2 \pi f \).

### Step 5: Calculate \( \omega \)
Given the frequency \( f = 60 \) Hz:
\[ \omega = 2 \pi f = 2 \pi \times 60 = 120 \pi \, \text{rad/s} \]

### Step 6: Set up the equation for \( X_L \)
From the impedance equation:
\[ Z^2 = R^2 + X_L^2 \]
Substituting the known values:
\[ 480^2 = 240^2 + X_L^2 \]
Calculating:
\[ 230400 = 57600 + X_L^2 \]
\[ X_L^2 = 230400 - 57600 = 172800 \]
\[ X_L = \sqrt{172800} = 240 \sqrt{3} \, \Omega \]

### Step 7: Calculate the inductance \( L \)
Using the relationship \( X_L = \omega L \):
\[ 240 \sqrt{3} = 120 \pi L \]
Rearranging for \( L \):
\[ L = \frac{240 \sqrt{3}}{120 \pi} = \frac{2 \sqrt{3}}{\pi} \, \text{H} \]

### Final Answer
The value of the inductance \( L \) required is:
\[ L = \frac{2 \sqrt{3}}{\pi} \, \text{H} \]