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Question

(A) The input power $P_{	ext{in}}$ is given by the product of primary voltage $V_p$ and primary current $I_p$.$P_{	ext{in}} = V_p 	imes I_p = 2300

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The input power $P_{	ext{in}}$ is given by the product of primary voltage $V_p$ and primary current $I_p$.$P_{	ext{in}} = V_p 	imes I_p = 2300 	ext{ V} 	imes 5 	ext{ A} = 11500 	ext{ W}$.Efficiency $\eta$ is defined as the ratio of output power $P_{	ext{out}}$ to input power $P_{	ext{in}}$.Given $\eta = 90\% = 0.9$,so $P_{	ext{out}} = \eta 	imes P_{	ext{in}} = 0.9 	imes 11500 	ext{ W} = 10350 	ext{ W}$.The output power is also given by $P_{	ext{out}} = V_s 	imes I_s$,where $V_s = 230 	ext{ V}$ is the output voltage.$10350 	ext{ W} = 230 	ext{ V} 	imes I_s$.$I_s = \frac{10350}{230} 	ext{ A} = 45 	ext{ A}$.

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