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 approximately

Options

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

Explanation

To find the value of the inductance \( L \) for the resonant circuit, we can follow these steps:

### Step 1: Calculate the frequency of the signal
The frequency \( f \) can be calculated using the formula:
\[ f = \frac{C}{\lambda} \]
where:
- \( C \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) is the wavelength, given as \( 300 \, \text{m} \)

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

### Step 2: Use the resonant frequency formula
At resonance, the frequency \( f \) is related to the inductance \( L \) and capacitance \( C \) by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]

### Step 3: Rearranging the formula to find \( L \)
Rearranging the formula to solve for \( L \):
\[ L = \frac{1}{(2\pi f)^2 C} \]

### Step 4: Substitute the values of \( f \) and \( C \)
We know:
- \( f = 10^6 \, \text{Hz} \)
- \( C = 2.4 \, \mu\text{F} = 2.4 \times 10^{-6} \, \text{F} \)

Substituting these values into the equation:
\[ L = \frac{1}{(2\pi \times 10^6)^2 \times (2.4 \times 10^{-6})} \]

### Step 5: Calculate \( L \)
Calculating \( (2\pi \times 10^6)^2 \):
\[ (2\pi \times 10^6)^2 \approx (6.2832 \times 10^6)^2 \approx 39.4784 \times 10^{12} \]

Now substituting this back into the equation for \( L \):
\[ L = \frac{1}{39.4784 \times 10^{12} \times 2.4 \times 10^{-6}} = \frac{1}{9.474576 \times 10^{6}} \approx 1.057 \times 10^{-8} \, \text{H} \]

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