The maximum peak to peak voltage of an AM wave is 24 mV and the minimum peak to peak voltage is 8 mV. The modulation factor is
Answer & explanation
Correct answer: option 4
Vmax = 24/2 = 12 mV and Vmin = 8/2 = 4 mV
The modulation factor $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} = \frac{12 - 4}{12 + 4} = \frac{1}{2} = 0.5 $ = 50%