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(B) The modulation index $\mu$ is given by $\mu = \frac{A_m}{A_c} = 0.6$,where $A_m$ is the amplitude of the modulating signal and $A_c$ is the amplit

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$12$

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(B) The modulation index $\mu$ is given by $\mu = \frac{A_m}{A_c} = 0.6$,where $A_m$ is the amplitude of the modulating signal and $A_c$ is the amplitude of the carrier wave.The minimum amplitude of the modulated wave is given by $A_{min} = A_c - A_m = 3\,V$.Substituting $A_m = 0.6 A_c$ into the equation: $A_c - 0.6 A_c = 3\,V$.$0.4 A_c = 3\,V \Rightarrow A_c = \frac{3}{0.4} = 7.5\,V$.Now,calculate $A_m$: $A_m = 0.6 	imes 7.5 = 4.5\,V$.The maximum amplitude of the modulated wave is given by $A_{max} = A_c + A_m$.$A_{max} = 7.5 + 4.5 = 12\,V$.

For an amplitude modulated wave the minimum amplitude is $3\,V$,while the modulation index is $60\%$. The maximum amplitude of the modulated wave is $......\,V$ (in $,V$)

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