When the percentage Modulation is 75, an AM transmitter produces 10 kW Power. The percentage power saving if carrier and one of the side band is suppressed is

detailed solution

Correct option is B

Given   :  % of modulation = 75 %                      ∴ m = 0.75  and  Transmitter power : Pt= 10 KW  Use, Pc = Pt1+m22 = 101.28=7.8 kW  We know that, Pt = Pc + PLSB +PUSB ⇒10 = 7.8 +2 PLSB           ∵ PLSB =PUSB ⇒PLSB = 1.1 kW  If we suppress  Pc  and  PLSB , then new Pt' = PUSB = 1.1 kW  So, saving in transmitted power  = 10 - 1.1 = 8.9 kW ∴Percentage saving = 8.910×100 = 89% II method %=Pc1+m24Pc(1+m22)×100=4+m24(2+m22)×100= 2+0.5625×100=89%