Correct Answer - Option 1 : 68.1 kW
Concept:
The transmitted power in AM is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)\)
Pc = Carrier Power
μ = Modulation index
Calculation:
With μ = 0.85 and PC = 50 kW, the radiation power will be:
\({P_t} = 50k\left( {1 + \frac{{{(0.85) ^2}}}{2}} \right)\)
Pt ≈ 68.1 kW
