An AM wave has 1800 ,W of total power content | vedclass

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(B) The total power of an $AM$ wave is given by the formula: $P_t = P_c(1 + \frac{m^2}{2})$, where $P_t$ is the total power, $P_c$ is the carrier powe

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$1200$

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(B) The total power of an $AM$ wave is given by the formula: $P_t = P_c(1 + \frac{m^2}{2})$, where $P_t$ is the total power, $P_c$ is the carrier power, and $m$ is the modulation index.Given: $P_t = 1800 \,W$ and $m = 1$ (for $100 \%$ modulation).Substituting the values into the formula:$1800 = P_c(1 + \frac{1^2}{2})$$1800 = P_c(1 + 0.5)$$1800 = 1.5 P_c$$P_c = \frac{1800}{1.5} = 1200 \,W$.Therefore, the carrier power is $1200 \,W$.

An $AM$ wave has $1800 \,W$ of total power content. For $100 \%$ modulation, the carrier should have power content equal to: (in $\,W$)

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