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(B) Given:Efficiency of the transformer,$\eta = 90\% = 0.9$Input power,$P_{in} = 3\, kW = 3000\, W$Primary voltage,$V_{p} = 200\, V$Secondary current,

Vedclass · Accepted Answer

$450\, V, 15\, A$

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(B) Given:Efficiency of the transformer,$\eta = 90\% = 0.9$Input power,$P_{in} = 3\, kW = 3000\, W$Primary voltage,$V_{p} = 200\, V$Secondary current,$I_{s} = 6\, A$$1$. To find the current in the primary coil $(I_{p})$:Since $P_{in} = V_{p} 	imes I_{p}$,we have:$I_{p} = \frac{P_{in}}{V_{p}} = \frac{3000\, W}{200\, V} = 15\, A$$2$. To find the voltage across the secondary coil $(V_{s})$:Efficiency $\eta = \frac{P_{out}}{P_{in}} = \frac{V_{s} 	imes I_{s}}{P_{in}}$$0.9 = \frac{V_{s} 	imes 6\, A}{3000\, W}$$V_{s} = \frac{0.9 	imes 3000}{6} = \frac{2700}{6} = 450\, V$Thus,the secondary voltage is $450\, V$ and the primary current is $15\, A$.

$A$ transformer having an efficiency of $90\%$ is working on a $200\, V$ and $3\, kW$ power supply. If the current in the secondary coil is $6\, A$,the voltage across the secondary coil and the current in the primary coil respectively are:

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