An AM wave has `1800` watt of total power content, For `100%` modulation the carrier should have power content equal to

Options

A. 1000 watt
B. 1200 watt
C. 1500 watt
D. 1600 watt

Explanation

To solve the problem of finding the carrier power content for an AM wave with a total power of 1800 watts at 100% modulation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Total Power (Pt) = 1800 watts
- Modulation Index (ma) for 100% modulation = 1

2. **Recall the Formula for Total Power in AM Waves:**
The total power in an AM wave can be expressed as:
\[ P_t = P_c \left(1 + \frac{m_a^2}{2}\right) \]
where \(P_t\) is the total power, \(P_c\) is the carrier power, and \(m_a\) is the modulation index.

3. **Substitute the Known Values:**
Since \(m_a = 1\) for 100% modulation, we can substitute this into the formula:
\[ P_t = P_c \left(1 + \frac{1^2}{2}\right) = P_c \left(1 + \frac{1}{2}\right) = P_c \left(\frac{3}{2}\right) \]

4. **Rearranging the Formula to Find Carrier Power (Pc):**
To find \(P_c\), we rearrange the equation:
\[ P_c = \frac{P_t}{\frac{3}{2}} = \frac{P_t \cdot 2}{3} \]

5. **Substituting the Total Power:**
Now substitute \(P_t = 1800\) watts into the equation:
\[ P_c = \frac{1800 \cdot 2}{3} = \frac{3600}{3} = 1200 \text{ watts} \]

6. **Conclusion:**
The power content of the carrier for 100% modulation is:
\[ P_c = 1200 \text{ watts} \]

### Final Answer:
The carrier power content should be equal to **1200 watts**.

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