One `10V, 60W` bulb is to be connected to `100V` line. The required inductance coil has self-inductance of value `(f=50Hz)`

Options

A. 0.052 H
B. 2.42 H
C. 16.2 H
D. 16.2 mH

Explanation

To solve the problem of finding the required inductance coil for a 10V, 60W bulb connected to a 100V line at a frequency of 50Hz, we will follow these steps:

### Step 1: Determine the Current through the Bulb
The power (P) consumed by the bulb is given as 60W, and the voltage (V) across the bulb is 10V. We can use the formula for power:

\[ P = V \cdot I \]

Rearranging this to find the current (I):

\[ I = \frac{P}{V} = \frac{60W}{10V} = 6A \]

### Step 2: Calculate the Total Voltage across the Inductor
The total voltage supplied by the line is 100V. The voltage across the bulb (V_B) is 10V, so the voltage across the inductor (V_L) can be calculated as follows:

\[ V_L = V_{total} - V_B = 100V - 10V = 90V \]

### Step 3: Use the Relationship between Voltage, Current, and Inductive Reactance
The voltage across the inductor can also be expressed in terms of the inductive reactance (X_L) and the current (I):

\[ V_L = I \cdot X_L \]

Substituting the known values:

\[ 90V = 6A \cdot X_L \]

From this, we can solve for X_L:

\[ X_L = \frac{90V}{6A} = 15 \, \Omega \]

### Step 4: Relate Inductive Reactance to Inductance
The inductive reactance (X_L) is related to the inductance (L) and the angular frequency (ω) by the formula:

\[ X_L = L \cdot \omega \]

Where:

\[ \omega = 2\pi f \]

Given that the frequency (f) is 50Hz, we can calculate ω:

\[ \omega = 2\pi \cdot 50 = 100\pi \, \text{rad/s} \]

### Step 5: Substitute to Find Inductance
Now we can substitute X_L and ω into the equation to find L:

\[ 15 = L \cdot (100\pi) \]

Rearranging gives:

\[ L = \frac{15}{100\pi} \]

Calculating this gives:

\[ L \approx \frac{15}{314.16} \approx 0.0477 \, \text{H} \]

### Step 6: Final Result
Thus, the required inductance L is approximately:

\[ L \approx 0.0477 \, \text{H} \text{ or } 47.7 \, \text{mH} \]

### Summary
The required inductance coil has a self-inductance of approximately 0.0477 H.

---