Correct Answer - Option 4 : 94.4%

__Concept__:

The power of a transmitted AM wave is given as:

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

\(P_t = {P_c} + P_c\frac{μ ^2}{2}\)

Power in the carrier = Pc

Power in both the sidebands is given by:

\(P_s= \frac{{{P_c}{μ^2}}}{2}\)

Since the power is distributed equally to the left and to the right side of the sideband, the power in one of the sidebands is given by:

\(P_{s1}= \frac{{{P_c}{u^2}}}{4}\)

__Calculation__:

When the carrier and 1 of the sideband are suppressed, the amount of power saving will be:

\( = {P_c} + \frac{{{P_c}{μ ^2}}}{4}\)

The percentage of power saved will be calculated as:

\(\frac{{{P_c}\left( {1 + \frac{{{μ ^2}}}{4}} \right)}}{{{P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)}} \times 100\)

\(\Rightarrow \frac{{4 + {μ ^2}}}{{4 + 2{μ ^2}}} \times 100\)

With μ = 0.50, the amount of power saving will be:

\(=\frac{{4 + {{\left( {0.5} \right)}^2}}}{{4 + 2{{\left( {0.5} \right)}^2}}} \times 100 \)

\(= \frac{{4.25}}{{4.5}} = 94.44\%\)