Q.

The total power content of AM wave is 1320 W. The percentage modulation if each side band contains 160 W is

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a

50 %

b

60 %

c

70 %

d

80 %

answer is D.

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Detailed Solution

PT=1320 W  We know, 12PSB=PLSB   and  12PSB=PUSB ∴PSB=2PLSB=2×160 = 320 W  Also , PT=Pc+PSB ⇒1320=Pc+320 ⇒Pc=1000 W  Now, PTPc=1+m22 ⇒13201000=1+m22 ⇒1.32=1+m22 ⇒0.32=m22 ⇒m2 = 0.64 ⇒m=0.8 i.em%=m×100=0.8×100= 80%
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