A) 60.1%
B) 70.1%
C) 80.1%
D) 50.1%
Correct Answer: B
Solution :
The power dissipated in any circuit is a function of the square of voltage across the circuit and the e*ffective resistance of the circuit. Equation of AM wave reveals that it has three components of amplitude \[{{E}_{c}},\] \[m{{E}_{c}}/2\] and \[m{{E}_{c}}/2\]. Clearly, power output must be distributed among these components. Carrier Power, \[{{P}_{C}}=\frac{{{({{E}_{c}}/\sqrt{2})}^{2}}}{R}=\frac{E_{c}^{2}}{2R}\] Total power of side bands, \[{{P}_{S}}=\frac{{{(m{{E}_{c}}/2\sqrt{2})}^{2}}}{R}+\frac{{{(m{{E}_{c}}/2\sqrt{2})}^{2}}}{R}\] \[=\frac{{{m}^{2}}E_{c}^{2}}{4R}\] \[\therefore \] s\[\frac{{{P}_{S}}}{{{P}_{C}}}=\frac{1}{2}{{m}^{2}}\] and \[{{P}_{T}}={{P}_{C}}+{{P}_{S}}={{P}_{C}}\left( 1+\frac{{{m}^{2}}}{2} \right)\] \[\therefore \] \[\frac{{{P}_{T}}}{{{P}_{C}}}=1+\frac{{{m}^{2}}}{2}\] or \[\left( \frac{{{I}_{T}}}{{{I}_{C}}} \right)=1+\frac{{{m}^{2}}}{2}\] Given that, it \[{{I}_{T}}=8.93A,\,{{I}_{C}}=8A,m=?\] \[\therefore \] \[{{\left( \frac{8.93}{8} \right)}^{2}}=1+\frac{{{m}^{2}}}{2}\] or \[1.246=1+\frac{{{m}^{2}}}{2}\] or \[\frac{{{m}^{2}}}{2}=0.246\] or \[m=\sqrt{2\times 0.246}=0.701\] \[=70.1%\]
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