The total power content of an AM signal is 3000 w. For `100%` modulation, the power of CW and that of each side band are,

Options

A. 2kW, `0.5` kW
B. 3 kw, `0.75` kW
C. 4 KW, 1 kW
D. 2 KW, 1 kW

Answer

To solve the problem, we need to find the power of the carrier wave and the power of each sideband for an AM signal with a total power of 3000 W and 100% modulation.

### Step-by-Step Solution:

1. **Understand the Total Power in AM Signal**:
The total power \( P_t \) in an AM signal is given as 3000 W.

2. **Define Modulation Index**:
For 100% modulation, the modulation index \( \mu \) is equal to 1.

3. **Use the Power Relation in AM**:
The total power \( P_t \) in an AM signal can be expressed as:
\[ P_t = P_c \left(1 + \frac{\mu^2}{2}\right) \]
where \( P_c \) is the power of the carrier wave.

4. **Substitute Known Values**:
Substitute \( P_t = 3000 \, \text{W} \) and \( \mu = 1 \):
\[ 3000 = P_c \left(1 + \frac{1^2}{2}\right) \]
This simplifies to:
\[ 3000 = P_c \left(1 + 0.5\right) = P_c \times 1.5 \]

5. **Solve for Carrier Power \( P_c \)**:
Rearranging gives:
\[ P_c = \frac{3000}{1.5} = 2000 \, \text{W} \]
Therefore, the power of the carrier wave \( P_c \) is 2000 W.

6. **Calculate Power of Sidebands**:
The total power in the sidebands \( P_{sb} \) can be calculated as:
\[ P_{sb} = P_t - P_c = 3000 - 2000 = 1000 \, \text{W} \]

7. **Determine Power of Each Sideband**:
Since there are two sidebands in AM, the power of each sideband \( P_{sb} \) is:
\[ P_{sb} = \frac{1000}{2} = 500 \, \text{W} \]

### Final Results:
- Power of Carrier Wave \( P_c = 2000 \, \text{W} \) (or 2 kW)
- Power of Each Sideband \( P_{sb} = 500 \, \text{W} \) (or 0.5 kW)

### Conclusion:
The correct answer is:
- Carrier Power: 2 kW
- Each Sideband Power: 0.5 kW