QJEEM23Ph585c: JEE Main 2023 Physics - Class 12 JEEM23Ph585c Solution: For an amplitude modulated wave the minimum amplitude is 3…

In the case of Amplitude Modulation, the modulation index (µ) can also be represented as:

$$\mu =\frac{A_{max}-A_{min}}{A_{max}+A_{min}}$$

where:

• $A_{\max}$ is the maximum amplitude of the modulated wave,
• $A_{\min}$ is the minimum amplitude of the modulated wave.

Given that the modulation index is 60% or 0.6 and the minimum amplitude $A_{\min}$ is 3 V, we can rearrange this formula to solve for the maximum amplitude $A_{\max}$ :

$$0.6=\frac{A_{max}-3}{A_{max}+3}$$

Solving this equation gives:

$$A_{max}=12\text{V}$$

So, the correct answer is 12 V.