When m=1 ,power carried by side bands is:

Options

A. 11.1% of the total power of AM wave
B. 22.2% of the total power of AM wave
C. 33.3% of the total power of AM wave
D. 44.4% of the total power of AM wave

Answer

To solve the problem regarding the power carried by sidebands when the modulation index \( m = 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The modulation index \( m = 1 \).
- We need to find the power carried by the sidebands in an Amplitude Modulated (AM) wave.

2. **Recall the Formula for Total Power in AM Wave**:
- The total power \( P_t \) of an AM wave is given by:
\[ P_t = \frac{m^2 E_c^2}{4R} + \frac{E_c^2}{2R} \]
- Here, \( E_c \) is the peak value of the carrier wave and \( R \) is the resistance.

3. **Recall the Formula for Power in Sidebands**:
- The power carried by the sidebands \( P_s \) is given by:
\[ P_s = \frac{m^2 E_c^2}{4R} \]

4. **Substitute \( m = 1 \) into the Formulas**:
- For total power:
\[ P_t = \frac{1^2 E_c^2}{4R} + \frac{E_c^2}{2R} = \frac{E_c^2}{4R} + \frac{2E_c^2}{4R} = \frac{3E_c^2}{4R} \]
- For sideband power:
\[ P_s = \frac{1^2 E_c^2}{4R} = \frac{E_c^2}{4R} \]

5. **Calculate the Fraction of Power in Sidebands**:
- The fraction of power in the sidebands relative to the total power is given by:
\[ \frac{P_s}{P_t} = \frac{\frac{E_c^2}{4R}}{\frac{3E_c^2}{4R}} = \frac{1}{3} \]

6. **Convert the Fraction to Percentage**:
- To express this as a percentage:
\[ P_s = \frac{1}{3} P_t \implies P_s \text{ (in percentage)} = \frac{1}{3} \times 100\% = 33.33\% \]

7. **Final Answer**:
- The power carried by the sidebands when \( m = 1 \) is **33.33%** of the total power of the AM wave.