For an amplitude modulated wave, the maximum amplitude is found to be 12 V and minimum amplitude is found to be 4V. The modulation index of this wave is ______ %

Options

A. 25
B. 50
C. 75
D. 20

Explanation

To find the modulation index (μ) of the amplitude modulated wave, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Maximum and Minimum Amplitudes**:
- Given maximum amplitude (V_max) = 12 V
- Given minimum amplitude (V_min) = 4 V

2. **Set Up the Equations**:
- The maximum amplitude can be expressed as:
\[ V_{max} = V_C + V_A \]
- The minimum amplitude can be expressed as:
\[ V_{min} = V_C - V_A \]
Where \( V_C \) is the carrier amplitude and \( V_A \) is the amplitude of the modulating signal.

3. **Write the Two Equations**:
- From the maximum amplitude:
\[ V_C + V_A = 12 \quad \text{(1)} \]
- From the minimum amplitude:
\[ V_C - V_A = 4 \quad \text{(2)} \]

4. **Add the Two Equations**:
- Adding equations (1) and (2):
\[ (V_C + V_A) + (V_C - V_A) = 12 + 4 \]
\[ 2V_C = 16 \]
\[ V_C = 8 \, \text{V} \]

5. **Substitute \( V_C \) Back to Find \( V_A \)**:
- Substitute \( V_C \) into equation (1):
\[ 8 + V_A = 12 \]
\[ V_A = 12 - 8 = 4 \, \text{V} \]

6. **Calculate the Modulation Index**:
- The modulation index (μ) is given by:
\[ \mu = \frac{V_A}{V_C} \]
- Substitute the values:
\[ \mu = \frac{4}{8} = \frac{1}{2} \]

7. **Convert to Percentage**:
- To express the modulation index as a percentage:
\[ \mu \times 100 = \frac{1}{2} \times 100 = 50\% \]

### Final Answer:
The modulation index of this wave is **50%**.

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