A transmitter transmits at a wavelength of 300 m. A condenser of capacitance `2.4 (mu)F` is being used. The value of the inductance for the resonant circuit is approximatly

Options

A. (A) `10^(-4)H`
B. (B) `10^(-6)H`
C. (C) `10^(-8)H`
D. (D) `10^(-10)H`

Explanation

To solve the problem, we need to find the inductance \( L \) for a resonant circuit that includes a capacitor with a given capacitance and operates at a specific frequency determined by the wavelength of the transmitted signal.

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength \( \lambda = 300 \, \text{m} \)
- Capacitance \( C = 2.4 \, \mu\text{F} = 2.4 \times 10^{-6} \, \text{F} \)

2. **Calculate the frequency \( f \):**
The frequency can be calculated using the formula:
\[ f = \frac{c}{\lambda} \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ f = \frac{3 \times 10^8 \, \text{m/s}}{300 \, \text{m}} = 10^6 \, \text{Hz} \]

3. **Use the resonance frequency formula:**
The resonance frequency \( f_0 \) for an LC circuit is given by:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
Setting \( f_0 = f \):
\[ 10^6 = \frac{1}{2\pi\sqrt{LC}} \]

4. **Rearrange the equation to solve for \( L \):**
Squaring both sides gives:
\[ (10^6)^2 = \frac{1}{(2\pi)^2 LC} \]
Rearranging this gives:
\[ LC = \frac{1}{(2\pi)^2 (10^6)^2} \]
Thus,
\[ L = \frac{1}{(2\pi)^2 C (10^6)^2} \]

5. **Substitute the value of \( C \):**
Now substituting \( C = 2.4 \times 10^{-6} \, \text{F} \):
\[ L = \frac{1}{(2\pi)^2 (2.4 \times 10^{-6}) (10^6)^2} \]

6. **Calculate \( L \):**
First, calculate \( (2\pi)^2 \):
\[ (2\pi)^2 \approx 39.478 \]
Now substitute this value:
\[ L = \frac{1}{39.478 \times 2.4 \times 10^{-6} \times 10^{12}} \]
\[ L = \frac{1}{39.478 \times 2.4 \times 10^{6}} \approx \frac{1}{9.47472 \times 10^{6}} \approx 1.055 \times 10^{-8} \, \text{H} \]

7. **Final Result:**
Therefore, the value of the inductance \( L \) is approximately:
\[ L \approx 10^{-8} \, \text{H} \]