A transmitter circuit used for transmission of EM waves having wavelength `960 m` If capacitor used in circuit was of `2.56 microF`, then the self inductance of the inductor coil used in the circuit such that resonance occurs, is `P*10^(-8)` Find `P`

Answer

To find the self-inductance \( L \) of the inductor coil used in the circuit, we can use the resonance condition for an LC circuit, which is given by:

\[ \omega = \frac{1}{\sqrt{LC}} \]

Where:
- \( \omega \) is the angular frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.

### Step 1: Calculate the frequency \( f \) from the wavelength \( \lambda \)

The speed of electromagnetic waves in a vacuum is given by \( c \approx 3 \times 10^8 \, \text{m/s} \). The relationship between speed, frequency, and wavelength is:

\[ c = f \lambda \]

Rearranging for frequency \( f \):

\[ f = \frac{c}{\lambda} \]

Substituting the values:

\[ f = \frac{3 \times 10^8 \, \text{m/s}}{960 \, \text{m}} = 3.125 \times 10^5 \, \text{Hz} \]

### Step 2: Calculate the angular frequency \( \omega \)

The angular frequency \( \omega \) is related to the frequency \( f \) by:

\[ \omega = 2 \pi f \]

Substituting the value of \( f \):

\[ \omega = 2 \pi \times 3.125 \times 10^5 \approx 1.964 \times 10^6 \, \text{rad/s} \]

### Step 3: Use the resonance condition to find \( L \)

From the resonance condition:

\[ \omega^2 = \frac{1}{LC} \]

Rearranging for \( L \):

\[ L = \frac{1}{\omega^2 C} \]

Substituting the values of \( \omega \) and \( C \):

Given \( C = 2.56 \, \mu F = 2.56 \times 10^{-6} \, F \):

\[ L = \frac{1}{(1.964 \times 10^6)^2 \times (2.56 \times 10^{-6})} \]

Calculating \( \omega^2 \):

\[ \omega^2 \approx (1.964 \times 10^6)^2 \approx 3.865 \times 10^{12} \]

Now substituting this back into the equation for \( L \):

\[ L = \frac{1}{3.865 \times 10^{12} \times 2.56 \times 10^{-6}} \approx \frac{1}{9.88 \times 10^6} \approx 1.011 \times 10^{-7} \, H \]

### Step 4: Express \( L \) in the form \( P \times 10^{-8} \)

To express \( L \) in the desired form:

\[ L \approx 1.011 \times 10^{-7} \, H = 10.11 \times 10^{-8} \, H \]

Thus, \( P \) is approximately \( 10.11 \).

### Final Answer

Since \( P \) is typically taken as an integer, we can round \( P \) to \( 10 \).

\[ \text{Therefore, } P = 10 \]