In AM wave total power of side bands is given by :

Options

A. `P_(S ) =(E_C ^(2 ))/(4R )`
B. `P_(s )=(4E _(C )^(2 ))/(R )`
C. `P_(s ) =(m^2 E _(c )^(2))/(4R )`
D. `P_(s )=(E _(c )^(2) )/(4m ^(2 R)`

Answer

To find the total power of sidebands in an Amplitude Modulated (AM) wave, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Power of Sidebands**:
The total power of sidebands in an AM wave can be derived from the modulation index and the RMS value of the carrier signal.

2. **Using the Formula**:
The power of the sidebands (P_s) is given by the formula:
\[ P_s = \frac{M E_{RMS}^2}{R} \]
where:
- \( M \) is the modulation index,
- \( E_{RMS} \) is the RMS value of the modulated signal,
- \( R \) is the resistance.

3. **Expressing RMS Values**:
The RMS value of the carrier signal \( E_{c} \) can be expressed as:
\[ E_{RMS} = \frac{E_c}{\sqrt{2}} \]
The modulation index can also be represented in terms of its RMS value:
\[ M = \frac{m}{\sqrt{2}} \]
where \( m \) is the peak modulation index.

4. **Substituting Values into the Formula**:
Now substituting these expressions into the power formula:
\[ P_s = \frac{\left(\frac{m}{\sqrt{2}}\right) \left(\frac{E_c}{\sqrt{2}}\right)^2}{R} \]

5. **Simplifying the Expression**:
Simplifying the above expression gives:
\[ P_s = \frac{m}{\sqrt{2}} \cdot \frac{E_c^2}{2} \cdot \frac{1}{R} \]
This simplifies to:
\[ P_s = \frac{m E_c^2}{4R} \]

6. **Final Expression**:
Thus, the total power of the sidebands in an AM wave is:
\[ P_s = \frac{m^2 E_c^2}{4R} \]

7. **Identifying the Correct Option**:
Now we can check the options provided in the question. The derived expression matches option C:
\[ P_s = \frac{m^2 E_c^2}{4R} \]

### Final Answer:
The total power of the sidebands in an AM wave is given by:
\[ P_s = \frac{m^2 E_c^2}{4R} \]