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(C) Given: Primary voltage $V_{P} = 2300\,V$,Secondary voltage $V_{S} = 230\,V$,Primary current $I_{P} = 5\,A$,Efficiency $\eta = 90\% = 0.9$.The effi

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$45$

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(C) Given: Primary voltage $V_{P} = 2300\,V$,Secondary voltage $V_{S} = 230\,V$,Primary current $I_{P} = 5\,A$,Efficiency $\eta = 90\% = 0.9$.The efficiency of a transformer is defined as the ratio of output power $(P_{S})$ to input power $(P_{P})$:$\eta = \frac{P_{S}}{P_{P}} \Rightarrow P_{S} = \eta 	imes P_{P}$.Since power $P = V 	imes I$,we can write:$V_{S} 	imes I_{S} = 0.9 	imes (V_{P} 	imes I_{P})$.Substituting the given values:$230 	imes I_{S} = 0.9 	imes 2300 	imes 5$.Solving for $I_{S}$:$I_{S} = \frac{0.9 	imes 2300 	imes 5}{230} = 0.9 	imes 10 	imes 5 = 45\,A$.Therefore,the output current is $45\,A$.

$A$ power transmission line feeds input power at $2300\,V$ to a step-down transformer with its primary windings having $4000$ turns,giving the output power at $230\,V.$ If the current in the primary of the transformer is $5\,A,$ and its efficiency is $90\%,$ the output current would be......$A$

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