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(A) For an ideal transformer,the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_s}{V_p} = \frac{N_s}{N_p}$Substituting the gi

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$240 \,V, \,5 \,A$

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(A) For an ideal transformer,the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_s}{V_p} = \frac{N_s}{N_p}$Substituting the given values:$\frac{V_s}{120} = \frac{200}{100}$$V_s = 120 	imes 2 = 240 \,V$For an ideal transformer,the power input equals the power output $(V_p I_p = V_s I_s)$:$I_s = \frac{V_p I_p}{V_s} = \frac{120 	imes 10}{240}$$I_s = \frac{1200}{240} = 5 \,A$Thus,the secondary voltage is $240 \,V$ and the secondary current is $5 \,A$.

The primary winding of a transformer has $100$ turns and its secondary winding has $200$ turns. The primary is connected to an ac supply of $120 \,V$ and the current flowing in it is $10 \,A$. The voltage and the current in the secondary are

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