The total power content of an AM wave is `900` W. For `100%` modulation, the power transmitted by each side band is

Options

A. `50 W`
B. `100 W`
C. `150 W`
D. `200 W`

Answer

c

Explanation

`P_(c)=P_(t)[(2)/(2+m^(2))]=900[(2)/(2+1)]=600 W`
Now, `P_(LSB)=(m^(2))/(4)xxP_(c)=(1)/(4)xx600=150 W`