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An AM wave has \(1800 \mathrm{~W}\) of total power content for \(100 \%\) modulation the carrier should have power content equal to
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The correct answer is:
\(1200 \mathrm{~W}\)
Total power of AM wave, \(P_t=1800 \mathrm{~W}\)
For \(100 \%\) modulation, depth of modulation, \(m=1\) If \(P_c\) be the carrier power, then we know that,
\(\begin{aligned}
& & P_t=P_c\left(1+\frac{m^2}{2}\right) \\
\Rightarrow & P_t & =P_c\left(1+\frac{1}{2}\right) \Rightarrow P_t=\frac{3}{2} P_c \\
\Rightarrow & P_c & =\frac{2}{3} P_t \quad=\frac{2}{3} \times 1800=1200 \mathrm{~W}
\end{aligned}\)
For \(100 \%\) modulation, depth of modulation, \(m=1\) If \(P_c\) be the carrier power, then we know that,
\(\begin{aligned}
& & P_t=P_c\left(1+\frac{m^2}{2}\right) \\
\Rightarrow & P_t & =P_c\left(1+\frac{1}{2}\right) \Rightarrow P_t=\frac{3}{2} P_c \\
\Rightarrow & P_c & =\frac{2}{3} P_t \quad=\frac{2}{3} \times 1800=1200 \mathrm{~W}
\end{aligned}\)
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