Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

300 V, 15 A

450 V, 15 A

450 V, 13.5 A

600 V, 15 A

Step-by-Step Solution

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 .
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 $P_{out} = 0.9 \times 3000 = 2700$ W.
Since $P_{out} = V_s I_s$ and $I_s = 6$ A, we have $2700 = V_s \times 6$ , which gives $V_s = 2700 / 6 = 450$ V. Thus, $V_s = 450$ V and $I_p = 15$ A.

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