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Question

Physics Question on work, energy and power

The antenna current of an AMAM transmitter is 8A8\,A when only the carrier is sent, but it increases to 8.93A8.93\,A when the carrier is modulated by a single sine wave. Find the percentage modulation.

A

60.10%

B

70.10%

C

80.10%

D

50.10%

Answer

70.10%

Explanation

Solution

The power dissipated in any circuit is a function of the square of voltage across the circuit and the effective resistance of the circuit. Equation of AM wave reveals that it has three components of amplitude Ec,mEc/2E_{c}, m E_{c} / 2 and mEc/2.m E_{c} / 2 . Clearly, power output must be distributed among these components. Carrier Power, PC=(Ec/2)2R=Ec22RP_{C}=\frac{\left(E_{c} / \sqrt{2}\right)^{2}}{R}=\frac{E_{c}^{2}}{2 R} Total power of side bands, PS=(mEc/22)2R+(mEc/22)2RP_{S}=\frac{\left(m E_{c} / 2 \sqrt{2}\right)^{2}}{R}+\frac{\left(m E_{c} / 2 \sqrt{2}\right)^{2}}{R} =m2Ec24R=\frac{m^{2} E_{c}^{2}}{4 R} PSPC=12m2\therefore \frac{P_{S}}{P_{C}}=\frac{1}{2} m^{2} and PT=PC+PS=PC(1+m22)P_{T}=P_{C}+P_{S}=P_{C}\left(1+\frac{m^{2}}{2}\right) PTPC=1+m22\therefore \frac{P_{T}}{P_{C}}=1+\frac{m^{2}}{2} or (ITIC)2=1+m22\left(\frac{I_{T}}{I_{C}}\right)^{2}=1+\frac{m^{2}}{2} Given that, IT=8.93A,IC=8A,m=I_{T}=8.93 A , I_{C}=8 A , m= ? (8.938)2=1+m22\therefore \left(\frac{8.93}{8}\right)^{2}=1+\frac{m^{2}}{2} or 1.246=1+m221.246=1+\frac{m^{2}}{2} or m22=0.246\frac{m^{2}}{2}=0.246 or m=2×0.246=0.701m=\sqrt{2 \times 0.246}=0.701 =70.1%=70.1 \% The situation is summarised in figure. BC=AD=3cm,AB=DC=4cm,B C=A D=3 \,cm , A B=D C=4 cm , so AC=5cmA C=5 \,cm