Correct Answer - Option 4 : 17.43 W

__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:

\(P_s= \frac{{{P_c}{μ^2}}}{2}\)

__Calculation__:

Given:

μ = 0.7

P_{c} = 71.14 W

\(P_s= \frac{{{P_c}{μ^2}}}{2}\)

\(P_s= \frac{{{71.14} \ \times \ 0.7{^2}}}{2}\)

**P**_{s} = 17.43 W