A transformer with efficiency 80% works at 4, | vedclass

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(A) Given:Efficiency $\eta = 80\% = 0.8$Input Power $P_{in} = 4\, kW = 4000\, W$Primary Voltage $V_p = 100\, V$Secondary Voltage $V_s = 200\, V$$1$. C

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$40\, A, 16\, A$

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(A) Given:Efficiency $\eta = 80\% = 0.8$Input Power $P_{in} = 4\, kW = 4000\, W$Primary Voltage $V_p = 100\, V$Secondary Voltage $V_s = 200\, V$$1$. Calculate Primary Current $(I_p)$:Since $P_{in} = V_p 	imes I_p$, we have:$4000 = 100 	imes I_p$$I_p = 40\, A$$2$. Calculate Secondary Current $(I_s)$:Efficiency $\eta = \frac{P_{out}}{P_{in}} = \frac{V_s 	imes I_s}{P_{in}}$$0.8 = \frac{200 	imes I_s}{4000}$$0.8 = \frac{I_s}{20}$$I_s = 0.8 	imes 20 = 16\, A$Thus, the primary and secondary currents are $40\, A$ and $16\, A$ respectively.

$A$ transformer with efficiency $80\%$ works at $4\, kW$ and $100\, V$. If the secondary voltage is $200\, V$, then the primary and secondary currents are respectively:

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