A step-down transformer is used on a 1000,V l | vedclass

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(A) The efficiency $\eta$ of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}}$.Given

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$3$

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(A) The efficiency $\eta$ of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}}$.Given: Input voltage $V_p = 1000\,V$,Output current $I_s = 20\,A$,Output voltage $V_s = 120\,V$,and Efficiency $\eta = 80\% = 0.8$.Output power $P_{	ext{out}} = V_s 	imes I_s = 120\,V 	imes 20\,A = 2400\,W$.Input power $P_{	ext{in}} = V_p 	imes I_p = 1000\,V 	imes I_p$.Using the efficiency formula: $0.8 = \frac{2400}{1000 	imes I_p}$.Solving for $I_p$: $I_p = \frac{2400}{1000 	imes 0.8} = \frac{2400}{800} = 3\,A$.Therefore,the current drawn from the line is $3\,A$.

$A$ step-down transformer is used on a $1000\,V$ line to deliver $20\,A$ at $120\,V$ at the secondary coil. If the efficiency of the transformer is $80\%$,the current drawn from the line is ......$A$.

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