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

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Question

(B) The total power $P_t$ in an $AM$ wave is given by the formula: $P_t = P_c \left( 1 + \frac{m_a^2}{2} \right)$,where $P_c$ is the carrier power and

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$1200$

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

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

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