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, we need to determine the value of inductance (L) required so that a 60 W, 120 V bulb connected to a 240 V, 60 Hz supply receives the correct voltage. Here’s the step-by-step solution:

### Step 1: Calculate the resistance of the bulb
The resistance of the bulb (R) can be calculated using the formula:

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

Where:
- \( V \) = rated voltage of the bulb = 120 V
- \( P \) = power of the bulb = 60 W

Substituting the values:

\[ R = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]

### Step 2: Determine the current flowing through the bulb
The current (I) flowing through the bulb can be calculated using Ohm's law:

\[ I = \frac{V}{R} \]

Where:
- \( V \) = rated voltage of the bulb = 120 V
- \( R \) = resistance of the bulb = 240 Ω

Substituting the values:

\[ I = \frac{120}{240} = 0.5 \, A \]

### Step 3: Calculate the total impedance (Z) of the circuit
The total impedance (Z) can be calculated using the formula:

\[ Z = \frac{V_{supply}}{I} \]

Where:
- \( V_{supply} \) = supply voltage = 240 V
- \( I \) = current = 0.5 A

Substituting the values:

\[ Z = \frac{240}{0.5} = 480 \, \Omega \]

### Step 4: Relate the impedance to resistance and reactance
The impedance (Z) in an LR circuit is given by:

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

Where:
- \( R \) = resistance of the bulb = 240 Ω
- \( X_L \) = inductive reactance = \( \omega L = 2 \pi f L \)

Rearranging the equation gives:

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

Substituting the values:

\[ X_L = \sqrt{480^2 - 240^2} \]

Calculating:

\[ X_L = \sqrt{230400 - 57600} = \sqrt{172800} = 240 \sqrt{3} \, \Omega \]

### Step 5: Calculate the inductance (L)
Using the relation \( X_L = \omega L \):

\[ X_L = 2 \pi f L \]

Substituting \( X_L \):

\[ 240 \sqrt{3} = 2 \pi (60) L \]

Solving for L:

\[ L = \frac{240 \sqrt{3}}{2 \pi (60)} = \frac{240 \sqrt{3}}{120 \pi} = \frac{2 \sqrt{3}}{\pi} \, H \]

### Final Answer:
The value of inductance \( L \) required is:

\[ L = \frac{2 \sqrt{3}}{\pi} \, H \]