Q 12-08-150JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
The maximum amplitude for an amplitude modulated wave is found to be $12$ V while the minimum amplitude is found to be $3$ V. The modulation index is $0.6x$ where $x$ is ______.
Numerical value type. Enter your answer.
Answer: 1
$$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} = \frac{12 - 3}{12 + 3} = 0.6$$
So $0.6x = 0.6 \Rightarrow x = 1$.
Solution by Sreeraj P, M.Sc Physics