Correct Answer - Option 2 : 1.53 kW
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
Total Power in the sidebands is given by:
\(P_s= \frac{{{P_c}{μ^2}}}{2}\) ---(1)
Calculation:
With Pt = 10 kW and μ = 0.6, we can write:
\(10k= {P_c}\left( {1 + \frac{{{(0.6)^2}}}{2}} \right)\)
10k = Pc × 1.18
Pc = 8.47 kW
Using Equation (1), we get the power at the sidebands as:
\(P_s= \frac{{{P_c}{μ^2}}}{2}=\frac{8.47\times 0.36}{2}\)
PS ≈ 1.53 kW
