An electric lamp designed for operation on `10V AC` is connected to a `220V AC` supply, through a choke coil of inductance `2H` for proper operation. The angular frequency of the `AC` is `100sqrt(10)rad//s`. If a capacitor is to be used in place of the choke coil its cpacitance must be

Options

A. `1 mu F`
B. `2muF`
C. `5 muF`
D. `10 muF`

Answer

To solve the problem, we need to determine the capacitance required when replacing a choke coil with a capacitor in an AC circuit. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Voltage of the lamp, \( V = 10V \) (not directly needed for capacitance calculation)
- Supply voltage, \( V_s = 220V \) (not directly needed for capacitance calculation)
- Inductance of the choke coil, \( L = 2H \)
- Angular frequency, \( \omega = 100\sqrt{10} \, \text{rad/s} \)

### Step 2: Use the formula for resonant frequency
The formula for the resonant frequency in terms of inductance \( L \) and capacitance \( C \) is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 3: Rearrange the formula to solve for capacitance \( C \)
Squaring both sides of the equation gives:
\[ \omega^2 = \frac{1}{LC} \]
Rearranging this, we find:
\[ C = \frac{1}{\omega^2 L} \]

### Step 4: Substitute the known values into the equation
Now, we can substitute the values of \( \omega \) and \( L \) into the equation:
\[ C = \frac{1}{(100\sqrt{10})^2 \cdot 2} \]

### Step 5: Calculate \( (100\sqrt{10})^2 \)
Calculating \( (100\sqrt{10})^2 \):
\[ (100\sqrt{10})^2 = 10000 \cdot 10 = 100000 \]

### Step 6: Substitute back into the capacitance formula
Now substituting back:
\[ C = \frac{1}{100000 \cdot 2} = \frac{1}{200000} \]

### Step 7: Calculate the capacitance
Calculating \( C \):
\[ C = 5 \times 10^{-6} \, \text{F} \]

### Step 8: Convert to microfarads
Since \( 1 \, \text{F} = 10^6 \, \mu\text{F} \):
\[ C = 5 \, \mu\text{F} \]

### Final Answer
The required capacitance is:
\[ C = 5 \, \mu\text{F} \]

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