Question

The maximum peak to peak voltage of an AM wire is$24mV$ and the minimum peak to peak voltage is$8mV$. The modulation factor is

• A. $10\%$
• B. $20\%$
• C. $25\%$
• D. $50\%$

Answer 1

The correct option is D

$50\%$

Step 1: Given data:

The maximum peak-to-peak voltage of an AM wire$V_{max}=24mV$

The minimum peak-to-peak voltage $V_{min}=8mV$

Step 2: Formula used:

The ratio of the change of amplitude in carrier wave after the modulation to the amplitude of the unmodulated carrier wave.$Modulationfactor=\frac{\left(V_{max}-V_{min}\right)}{\left(V_{max}+V_{min}\right)}$

Step 3: Calculating the modulation factor

The modulation factor can be calculated as,

$\begin{array}{rcl}{V}_{max}& =& \frac{24}{2}\\ & =& 12mV\\ V_{min}& =& \frac{8}{2}=4mV\\ Modulationfactor& =& \frac{\left(V_{max}-V_{min}\right)}{\left(V_{max}+V_{min}\right)}\\ & =& \frac{\left(12-4\right)}{\left(12+4\right)}\\ & =& \frac{8}{16}\\ & =& \frac{1}{2}\\ & =& 0.5\times 100\\ & =& 50\%\end{array}$

Thus, the modulation factor is $50\%$