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MHTCET 20 Apr 2026 Shift 2 Paper

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Question : 40 of 150

Marks: +1, -0

A current of 4 A is flowing at 230 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2300 V and 40 \% of power is lost, then the current in the secondary will be

0.24 A
0.2 A
0.12 A
0.4 A

Solution:

Concept:

The transformer power relation is

P = VI

. Since

40\%

power is lost, the secondary receives only

60\%

of the primary power.

Explanation:

Primary power is calculated as:

P_p = V_p I_p = 230 \times 4 = 920\ \mathrm{W}

After a

40\%

loss, useful secondary power is:

P_s = 60\% \times 920 = \frac{60}{100} \times 920 = 552\ \mathrm{W}

Using the secondary voltage

V_s = 2300\ \mathrm{V}

:

P_s = V_s I_s \Rightarrow 552 = 2300 \times I_s

Therefore:

I_s = \frac{552}{2300} = 0.24\ \mathrm{A}

Answer:

The current in the secondary coil is

0.24\ \mathrm{A}

, which is option A.

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