A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively then total power radiated by the modulator is

Options

A. 10.25 kW
B. 11.25 kW
C. 12.75 kW
D. 17kW

Explanation

To solve the problem of finding the total power radiated by the modulator when a 10 kW carrier is sinusoidally modulated by two carriers with modulation indices of 30% and 40%, we can follow these steps:

### Step 1: Understand the Modulation Indices
The modulation indices given are:
- \( m_1 = 0.30 \) (30%)
- \( m_2 = 0.40 \) (40%)

### Step 2: Calculate the Effective Modulation Index
The effective modulation index \( m \) can be calculated using the formula:
\[ m = \sqrt{m_1^2 + m_2^2} \]
Substituting the values:
\[ m = \sqrt{(0.30)^2 + (0.40)^2} \]
Calculating the squares:
\[ m = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \]

### Step 3: Calculate the Total Power Radiated
The total power radiated by the modulator is given by the formula:
\[ P = P_c \left(1 + \frac{m^2}{2}\right) \]
Where \( P_c \) is the carrier power, which is 10 kW. Substituting the values:
\[ P = 10 \text{ kW} \left(1 + \frac{(0.5)^2}{2}\right) \]
Calculating \( \frac{(0.5)^2}{2} \):
\[ \frac{(0.5)^2}{2} = \frac{0.25}{2} = 0.125 \]
Now substituting back:
\[ P = 10 \text{ kW} \left(1 + 0.125\right) = 10 \text{ kW} \times 1.125 = 11.25 \text{ kW} \]

### Final Answer
The total power radiated by the modulator is **11.25 kW**.

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