Q83 / 90 · JEE Main 2021 (27 Jul Shift 2) · Physics

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 $0.6 x$ where $x$ is _____.

Correct answer: 1

Solution

$A_{\max} = A_{c} + A_{m} = 12$

$A_{\min} = A_{c} - A_{m} = 3$

$\Rightarrow A_{c} = \frac{15}{2}$ and $A_{m} = \frac{9}{2}$

Modulation index $= \frac{A_{m}}{A_{c}} = \frac{\frac{9}{2}}{\frac{15}{2}} = 0.6$

$\Rightarrow x = 1$

Hence, the correct answer is 1.