The total power content of an AM wave is 900 | vedclass

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(C) The total power of an $AM$ wave is given by $P_t = P_c \left(1 + \frac{m^2}{2}\right)$,where $P_c$ is the carrier power and $m$ is the modulation

Vedclass · Accepted Answer

$150$

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(C) The total power of an $AM$ wave is given by $P_t = P_c \left(1 + \frac{m^2}{2}ight)$,where $P_c$ is the carrier power and $m$ is the modulation index.Given $P_t = 900 \, W$ and $m = 1$ (for $100\%$ modulation).$900 = P_c \left(1 + \frac{1^2}{2}ight) = P_c \left(1 + 0.5ight) = 1.5 P_c$.Therefore,$P_c = \frac{900}{1.5} = 600 \, W$.The power in each sideband is given by $P_{SB} = \frac{m^2 P_c}{4}$.Substituting the values: $P_{SB} = \frac{1^2 	imes 600}{4} = \frac{600}{4} = 150 \, W$.

Column $A$	Column $B$
$A$. $D$-layer	$I$. $250-400 \ km$
$B$. $E$-layer	$II$. $170-190 \ km$
$C$. $F_1$-layer	$III$. $95-120 \ km$
$D$. $F_2$-layer	$IV$. $65-75 \ km$

The total power content of an $AM$ wave is $900 \, W$. For $100\%$ modulation,the power transmitted by each sideband is......$W$

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