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A broadcast `AM` transmitter radiates ` 50 kW` of carrier power . What will be the radiated power at `50%` modulation ? |
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Answer» Correct Answer - 56.25 kW `P_(t) = P_( c ) ( 1 + (mu^(2))/(2)) = 50 ( 1 + (1)/(8))` `= (450)/(8) = 56.25 kW` |
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