Ans. (2) Sol. Given, $\mathrm{A}_{\mathrm{c}}=15 \mathrm{~V}$ $$ \mathrm{A}_{\mathrm{m}}=3 \mathrm{~V} $$ Maximum amplitude of modulated wave $$ A_{\max }=A_c+A_m=15+3=18 \mathrm{~V} $$ Minimum amplitude of modulated wave $$ \begin{aligned} & A_{\min }=A_c-A_m=15-3=12 V \\ & \therefore \frac{A_c+A_m}{A_c-A_m}=\frac{18}{12}=\frac{3}{2} \end{aligned} $$