Correct Answer - Option 3 : 1%

__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

\(P_c=\frac{{A_m}^2}{2}\)

Power in both the sidebands is given by:

\(P_s= \frac{{{P_c}{μ^2}}}{2}\)

Since the power is distributed equally to the left and to the right side of the sideband, the power in one of the sidebands is given by:

\(P_{s1}= \frac{{{P_c}{u^2}}}{4}\)

__Calculation:__

Given μ = 20% = 0.2

\(P_c=\frac{{A_m}^2}{2}\)

Total Sideband power is:

\(P_{sb}=\frac{{A_m}^2\mu^2}{4}\)

Assume A_{m} = 1

\(P_{sb}=\frac{0.2^2}{4} \ \times \ 100\)

**P**_{sb} **= 1%**

**Hence option (3) is the correct answer.**