Calculate the power developed by an amplitude modulated wave in a load resistance of `100 Omega`, if the peak voltage of carrier wave is 100 V and modulation index is 0.4.

Options

A. 50 watt
B. 54 watt
C. 104 watt
D. 4 watt

Explanation

To calculate the power developed by an amplitude modulated (AM) wave in a load resistance of \(100 \, \Omega\), given the peak voltage of the carrier wave \(E_C = 100 \, V\) and the modulation index \(m = 0.4\), we can follow these steps:

### Step 1: Calculate the power of the carrier wave \(P_C\)

The power \(P_C\) of the carrier wave can be calculated using the formula:

\[ P_C = \frac{E_C^2}{2R} \]

Where:
- \(E_C\) is the peak voltage of the carrier wave.
- \(R\) is the load resistance.

### Step 2: Substitute the values into the formula

Given:
- \(E_C = 100 \, V\)
- \(R = 100 \, \Omega\)

Substituting these values into the formula:

\[ P_C = \frac{(100)^2}{2 \times 100} = \frac{10000}{200} = 50 \, W \]

### Step 3: Calculate the total power \(P\) of the amplitude modulated wave

The total power \(P\) of the amplitude modulated wave can be calculated using the formula:

\[ P = P_C \left(1 + \frac{m^2}{2}\right) \]

### Step 4: Substitute the modulation index into the formula

Given:
- \(m = 0.4\)

Substituting the values into the formula:

\[ P = 50 \left(1 + \frac{(0.4)^2}{2}\right) \]

Calculating \(m^2\):

\[ m^2 = (0.4)^2 = 0.16 \]

Now substituting this back into the power equation:

\[ P = 50 \left(1 + \frac{0.16}{2}\right) = 50 \left(1 + 0.08\right) = 50 \times 1.08 \]

### Step 5: Calculate the final power

Calculating the final power:

\[ P = 50 \times 1.08 = 54 \, W \]

### Final Answer

The power developed by the amplitude modulated wave in a load resistance of \(100 \, \Omega\) is **54 W**.

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