An AM transmitter radiates 70 kW of carrier power. The radiated power at 75% modulation would be

Options

A. `89.69` kW
B. 79.20 kW
C. 85.59 kW
D. 83.69 kW

Answer

To solve the problem of finding the radiated power of an AM transmitter at 75% modulation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Carrier power, \( P_c = 70 \, \text{kW} \)
- Modulation index, \( \mu = 75\% = 0.75 \)

2. **Understand the Formula**:
The formula for the radiated power \( P_r \) in an AM transmitter is given by:
\[ P_r = P_c \left(1 + \frac{\mu^2}{2}\right) \]

3. **Substitute the Values into the Formula**:
Now, substitute the values of \( P_c \) and \( \mu \) into the formula:
\[ P_r = 70 \, \text{kW} \left(1 + \frac{(0.75)^2}{2}\right) \]

4. **Calculate \( \mu^2 \)**:
First, calculate \( \mu^2 \):
\[ \mu^2 = (0.75)^2 = 0.5625 \]

5. **Calculate \( \frac{\mu^2}{2} \)**:
Now, calculate \( \frac{\mu^2}{2} \):
\[ \frac{\mu^2}{2} = \frac{0.5625}{2} = 0.28125 \]

6. **Add 1 to the Result**:
Next, add 1 to the result:
\[ 1 + 0.28125 = 1.28125 \]

7. **Calculate the Radiated Power \( P_r \)**:
Finally, multiply by the carrier power:
\[ P_r = 70 \, \text{kW} \times 1.28125 = 89.6875 \, \text{kW} \]

8. **Round the Result**:
Rounding this to two decimal places gives:
\[ P_r \approx 89.69 \, \text{kW} \]

### Final Answer:
The radiated power at 75% modulation is approximately **89.69 kW**.

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