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
Pc = 71.14 W
\(P_s= \frac{{{P_c}{μ^2}}}{2}\)
\(P_s= \frac{{{71.14} \ \times \ 0.7{^2}}}{2}\)
Ps = 17.43 W