A broadcast radio transmitter radiates 12 kW when percentage of modulation is 50% , then the unmodulated carrier power is

Options

A. 5.67 kW
B. 7.15 kW
C. 9.7 kW
D. 12 kW

Explanation

To find the unmodulated carrier power (\(P_c\)) when the transmitter radiates a total power (\(P_t\)) of 12 kW with a modulation percentage of 50%, we can use the formula relating the carrier power, transmitted power, and modulation index.

### Step-by-Step Solution:

1. **Understand the Given Values**:
- Transmitted Power (\(P_t\)) = 12 kW
- Percentage of Modulation = 50%

2. **Convert Percentage of Modulation to Modulation Index**:
- Modulation Index (\(m\)) = Percentage of Modulation / 100 = 50% / 100 = 0.5

3. **Use the Formula for Transmitted Power**:
The relationship between the carrier power, transmitted power, and modulation index is given by:
\[ P_t = P_c \left(1 + \frac{m^2}{2}\right) \]
Rearranging this gives us:
\[ P_c = \frac{P_t}{1 + \frac{m^2}{2}} \]

4. **Calculate \(m^2\)**:
- \(m^2 = (0.5)^2 = 0.25\)

5. **Substitute \(m^2\) into the Formula**:
\[ P_c = \frac{12 \text{ kW}}{1 + \frac{0.25}{2}} \]

6. **Calculate \(\frac{0.25}{2}\)**:
- \(\frac{0.25}{2} = 0.125\)

7. **Add 1 to the Result**:
\[ 1 + 0.125 = 1.125 \]

8. **Calculate \(P_c\)**:
\[ P_c = \frac{12 \text{ kW}}{1.125} \approx 10.67 \text{ kW} \]

9. **Final Result**:
The unmodulated carrier power is approximately \(10.67 \text{ kW}\).

### Answer:
The unmodulated carrier power is approximately \(10.67 \text{ kW}\).

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