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%