Question

The maximum peak-to-peak voltage of an AM wave is $16mV$, and the minimum peak-to-peak voltage is $4mV$. The modulation factor is equal to

• A. 0.6
• B. 0.3
• C. 0.8
• D. 0.25

Answer 1

The correct option is A

0.6

Step 1: Given data

The maximum peak-to-peak voltage of the AM wave = $16mV$
The minimum peak-to-peak voltage of the AM wave = $4mV$

Step 2: Formula used

Modulation factor is given by ${\mu }_V=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}$.

Step 3: Calculate the maximum and minimum voltage

Maximum Voltage, $V_{max}=\frac{16}{2}=8mV$

Minimum Voltage, $V_{min}=\frac{4}{2}=2mV$

Step 4: Write the equation and substitute the values

Modulation factor for voltage,

${\mu }_V=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}$

$\Rightarrow {\mu }_V=\frac{8-2}{8+2}$

$\Rightarrow {\mu }_V=\frac{6}{10}$

$\Rightarrow {\mu }_V=0.6$

$\because$ The modulation factor is equal to $0.6$.

Hence, option A is correct.