For a transformer,the turns ratio is 3 and it | vedclass

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(B) Given: Turns ratio $\frac{n_{s}}{n_{p}} = 3$,Efficiency $\eta = 0.75$,Primary current $I_{p} = 2 \,A$,Primary voltage $V_{p} = 100 \,V$. For a tra

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$300 \,V, 0.5 \,A$

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(B) Given: Turns ratio $\frac{n_{s}}{n_{p}} = 3$,Efficiency $\eta = 0.75$,Primary current $I_{p} = 2 \,A$,Primary voltage $V_{p} = 100 \,V$. For a transformer,the voltage ratio is equal to the turns ratio: $\frac{V_{s}}{V_{p}} = \frac{n_{s}}{n_{p}} = 3$. Therefore,$V_{s} = 3 	imes V_{p} = 3 	imes 100 \,V = 300 \,V$. Efficiency $\eta$ is defined as the ratio of output power to input power: $\eta = \frac{V_{s} I_{s}}{V_{p} I_{p}}$. Substituting the values: $0.75 = \frac{300 	imes I_{s}}{100 	imes 2}$. $0.75 = \frac{300 	imes I_{s}}{200} = 1.5 	imes I_{s}$. $I_{s} = \frac{0.75}{1.5} = 0.5 \,A$. Thus,the secondary voltage is $300 \,V$ and the secondary current is $0.5 \,A$.

For a transformer,the turns ratio is $3$ and its efficiency is $0.75$. The current flowing in the primary coil is $2 \,A$ and the voltage applied to it is $100 \,V$. Then the voltage and the current flowing in the secondary coil are ... respectively.

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