Question

In AM wave, total power of side bands is given by:

• A. $P_s=\frac{{E_c}^2}{4R}$
• B. $P_s=\frac{{{4E}_c}^2}{R}$
• C. $P_s=\frac{{{m^{2}E}_c}^2}{4R}$
• D. $P_s=\frac{{E_c}^2}{4m^{2}R}$

Answer 1

The correct option is C $P_s=\frac{{{m^{2}E}_c}^2}{4R}$

power of side band $P_s$ is given by

$P_s=m_e\frac{E_{rms}^2}{R}$

Here $E_{rms}=\frac{E_c}{\sqrt{2}}$ and $m_e=\frac{m}{\sqrt{2}}$

Putting value we get :

$P_s=\frac{m^2{E}_c^2}{4R}$

Hence option C is correct