A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the second

Question

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

Accepted Answer

1. **Primary Current ($I_p$):** The input power ($P_{in}$) is given by $P_{in} = V_p I_p$. Given $P_{in} = 3000$ W and $V_p = 200$ V, we calculate $I_p = P_{in} / V_p = 3000 / 200 = 15$ A .
2. **Secondary Voltage ($V_s$):** The efficiency ($\eta$) is defined as the ratio of output power to input power: $\eta = P_{out} / P_{in}$. Given $\eta = 90\% = 0.9$, the output power is $P_{out} = 0.9 	imes 3000 = 2700$ W. 
3. Since $P_{out} = V_s I_s$ and $I_s = 6$ A, we have $2700 = V_s 	imes 6$, which gives $V_s = 2700 / 6 = 450$ V.
Thus, $V_s = 450$ V and $I_p = 15$ A.

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