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MCQEasyJEE 2024AC Generator & Transformer

JEE Physics 2024 Question with Solution

The primary side of a transformer is connected to a 230 V230 \, \text{V}, 50 Hz50 \, \text{Hz} supply. Turns ratio of primary to secondary winding is 10:110:1. Load resistance on the secondary side is 46 Ω46 \, \Omega. The power consumed in it is:

  • A

    12.5 W12.5 \, \text{W}

  • B

    10.0 W10.0 \, \text{W}

  • C

    11.5 W11.5 \, \text{W}

  • D

    12.0 W12.0 \, \text{W}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Primary voltage is 230 V230 \, \text{V}, turns ratio is NpNs=10\frac{N_p}{N_s} = 10, and load resistance on the secondary side is 46 Ω46 \, \Omega.

Find: The power consumed by the load resistance.

For an ideal transformer,

VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}

So,

Vs=Vp10=23010=23 VV_s = \frac{V_p}{10} = \frac{230}{10} = 23 \, \text{V}

Using Ohm's law for the secondary load,

Is=VsR=2346=0.5 AI_s = \frac{V_s}{R} = \frac{23}{46} = 0.5 \, \text{A}

Now the power consumed by the load is,

P=Is2R=(0.5)2×46=0.25×46=11.5 WP = I_s^2 R = (0.5)^2 \times 46 = 0.25 \times 46 = 11.5 \, \text{W}

Therefore, the power consumed in the load resistance is 11.5 W11.5 \, \text{W}. The correct option is C.

Direct Voltage Method

Given: V1V2=N1N2=10\frac{V_1}{V_2} = \frac{N_1}{N_2} = 10 and R=46 ΩR = 46 \, \Omega.

Find: Power consumed by the secondary load.

First find the secondary voltage:

V2=23010=23 VV_2 = \frac{230}{10} = 23 \, \text{V}

Then use the direct relation for resistor power,

P=V22R=23×2346=11.5 WP = \frac{V_2^2}{R} = \frac{23 \times 23}{46} = 11.5 \, \text{W}

This method works because the load is purely resistive, so once the secondary voltage is known, power can be obtained directly without separately calculating current. The correct option is C.

Common mistakes

  • Using the turns ratio in the wrong direction. Since NpNs=10\frac{N_p}{N_s} = 10, the secondary voltage is smaller, not larger. Use VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} carefully.

  • Calculating power with the primary voltage instead of the secondary voltage. The resistor is connected on the secondary side, so power must be found using VsV_s or IsI_s, not 230 V230 \, \text{V} directly.

  • Forgetting that the load resistance is 46 Ω46 \, \Omega and substituting a wrong value or skipping Ohm's law. First find VsV_s, then use Is=VsRI_s = \frac{V_s}{R} or P=Vs2RP = \frac{V_s^2}{R}.

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