The modulation index for an A.M. wave having maximum and minimum peak-to-peak voltages of 14 mV and 6 mV respectively is-

Options

A. `0.6`
B. `1.4`
C. `0.4`
D. `0.2`

Answer

To find the modulation index (μ) for the given A.M. wave with maximum and minimum peak-to-peak voltages, we can follow these steps:

### Step 1: Understand the formula for modulation index
The modulation index (μ) for an amplitude modulated wave is given by the formula:
\[ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} \]
where \(A_{max}\) is the maximum amplitude and \(A_{min}\) is the minimum amplitude.

### Step 2: Identify the given values
From the question, we have:
- Maximum peak-to-peak voltage (\(A_{max}\)) = 14 mV
- Minimum peak-to-peak voltage (\(A_{min}\)) = 6 mV

### Step 3: Calculate the amplitudes
Since the values given are peak-to-peak voltages, we need to convert them to amplitudes:
- The amplitude corresponding to the maximum voltage is:
\[ A_{max} = \frac{14 \, \text{mV}}{2} = 7 \, \text{mV} \]
- The amplitude corresponding to the minimum voltage is:
\[ A_{min} = \frac{6 \, \text{mV}}{2} = 3 \, \text{mV} \]

### Step 4: Substitute the values into the formula
Now, we can substitute \(A_{max}\) and \(A_{min}\) into the modulation index formula:
\[ \mu = \frac{7 \, \text{mV} - 3 \, \text{mV}}{7 \, \text{mV} + 3 \, \text{mV}} = \frac{4 \, \text{mV}}{10 \, \text{mV}} \]

### Step 5: Simplify the expression
Now, simplify the fraction:
\[ \mu = \frac{4}{10} = 0.4 \]

### Step 6: Conclusion
Thus, the modulation index for the A.M. wave is:
\[ \mu = 0.4 \]

### Final Answer
The modulation index is 0.4.

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