Q. The maximum peak to peak voltage of an AM wire is $24\ mV$ and the minimum peak to peak voltage is $8\ mV$. The modulation factor is
A) 10%
B) 20%
C) 25%
D) 50%
Explanation
(d) Here, $V_{\max} = \frac{24}{2} = 12\ mV$ and $V_{\min} = \frac{8}{2} = 4\ mV$
Now, $m = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}} = \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.