Correct Answer - Option 2 : 66 %

__Concept:__

The generalized AM expression is represented as:

s(t) = Ac [1 + μa mn (t)] cos ωc t

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

The power in a single sideband will be:

\({P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}\)

With \(P_c=\frac{A^2}{2}\), the above can be written as:

\({P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}\)

\({P_s} = \frac{{{A_c^2μ^2}}}{8}\)

\(Power Saved=\frac{P_c}{P_{total}}\) ---(1)

**Power Saved = Pc in DSB - SC**

\(Power \ Saved=\frac{2}{2 \ + \ μ^2}\)

__Analysis__:

When μ = 1, the transmitted power will be:

\({P_t} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)=\frac{3}{2}P_c\)

\(Power Saved=\frac{P_c}{P_c(1 \ + \ \frac{μ^2}{2})}\)

As μ = 1

\(Power \ Saved=\frac{2}{2 \ + \ 1^2}\times 100\)

**Power Saved = 66 %**