Correct Answer - Option 2 : 15.2 kW and 20 kW

__Concept:__

The total transmitted power for an AM system is given by:

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

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

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

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

__Calculation__:

Given P_{s} = 4.85 kW and μ = 0.8.

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}=4.85k\)

\( P_c\frac{{{(0.8)^2}}}{2}=4850\)

\(P_c=\frac{4850\times 2}{0.64}\)

**P**_{c} ≈ 15.2 kW

Now, the total power radiated will be:

\({P_t} = {P_c} + P_s\)

P_{t} = 15.2 + 4.85

**P**_{t} = 20 kW