Q 12-14-144JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
A carrier wave $V_C(t) = 160\sin(2\pi\times10^6t)$ volts is made to vary between $V_{max} = 200$ V and $V_{min} = 120$ V by a message signal $V_m(t) = A_m\sin(2\pi\times10^3t)$ volts. The peak voltage $A_m$ of the modulating signal is ______ V.
Numerical value type. Enter your answer.
Answer: 40
$V_{max} = A_c + A_m$ and $V_{min} = A_c - A_m$, so $A_m = \dfrac{200 - 120}{2} = 40$ V (and $A_c = 160$ V, as given).
Solution by Sreeraj P, M.Sc Physics