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(B) For an ideal (loss-free) transformer,the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac

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$40 \, V, \, 40 \, A$

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(B) For an ideal (loss-free) transformer,the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{500}{2500} = \frac{1}{5}$Given $V_s = 200 \, V$,we find the primary voltage:$V_p = V_s 	imes \frac{1}{5} = 200 	imes \frac{1}{5} = 40 \, V$For a loss-free transformer,the input power equals the output power:$P_p = P_s \Rightarrow V_p \cdot I_p = V_s \cdot I_s$Given $I_s = 8 \, A$,we find the primary current:$I_p = I_s 	imes \frac{V_s}{V_p} = 8 	imes \frac{200}{40} = 8 	imes 5 = 40 \, A$Thus,the primary voltage is $40 \, V$ and the primary current is $40 \, A$.

$A$ loss-free transformer has $500$ turns on its primary winding and $2500$ turns in its secondary winding. The meters of the secondary indicate $200 \, V$ at $8 \, A$ under these conditions. The voltage and current in the primary are:

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