In resonating circuit value of inductance and capacitance is 0.1H and `200 mu`F. For same resonating frequency if value of inductance is 100H then necessary value of capacitance in `mu`F will be -

Options

A. 4
B. `0.2`
C. 2
D. `0.3`

Answer

To solve the problem, we will use the formula for the resonant frequency of an LC circuit, which is given by:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]

Where:
- \( f \) is the resonant frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.

### Step 1: Calculate the resonant frequency with the initial values.

Given:
- \( L_1 = 0.1 \, H \)
- \( C_1 = 200 \, \mu F = 200 \times 10^{-6} \, F \)

Substituting the values into the formula:

\[ f_1 = \frac{1}{2\pi\sqrt{L_1C_1}} = \frac{1}{2\pi\sqrt{0.1 \times 200 \times 10^{-6}}} \]

### Step 2: Simplify the expression.

Calculate \( L_1C_1 \):

\[ L_1C_1 = 0.1 \times 200 \times 10^{-6} = 0.1 \times 0.0002 = 2 \times 10^{-5} \]

Now, substitute this back into the frequency formula:

\[ f_1 = \frac{1}{2\pi\sqrt{2 \times 10^{-5}}} \]

### Step 3: Calculate the new resonant frequency with the new inductance.

Now, we need to find the necessary capacitance \( C_2 \) when \( L_2 = 100 \, H \) for the same resonant frequency \( f_1 \).

Using the resonant frequency formula again:

\[ f_2 = \frac{1}{2\pi\sqrt{L_2C_2}} = f_1 \]

Since \( f_1 = f_2 \), we can set the equations equal to each other:

\[ \frac{1}{2\pi\sqrt{L_1C_1}} = \frac{1}{2\pi\sqrt{L_2C_2}} \]

### Step 4: Rearranging the equation.

Squaring both sides and rearranging gives:

\[ L_1C_1 = L_2C_2 \]

### Step 5: Solve for \( C_2 \).

Substituting the known values:

\[ 0.1 \times 200 \times 10^{-6} = 100 \times C_2 \]

Now, we already calculated \( 0.1 \times 200 \times 10^{-6} = 2 \times 10^{-5} \):

\[ 2 \times 10^{-5} = 100 \times C_2 \]

### Step 6: Isolate \( C_2 \).

\[ C_2 = \frac{2 \times 10^{-5}}{100} = 2 \times 10^{-7} \, F \]

### Step 7: Convert \( C_2 \) to microfarads.

\[ C_2 = 2 \times 10^{-7} \, F = 0.2 \, \mu F \]

### Final Answer:

The necessary value of capacitance \( C_2 \) is \( 0.2 \, \mu F \).

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