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NVAEasyJEE 2023AC Generator & Transformer

JEE Physics 2023 Question with Solution

An ideal transformer with a purely resistive load operates at 12 kV12 \, \text{kV} on the primary side. It supplies electrical energy to a number of nearby houses at 120 V120 \, \text{V}. The average rate of energy consumption in the houses served by the transformer is 60 kW60 \, \text{kW}. The value of resistive load (RsR_s) required in the secondary circuit will be in mΩ:

Answer

Correct answer:240

Step-by-step solution

Standard Method

Given: Primary voltage = 12 kV=12000 V12 \, \text{kV} = 12000 \, \text{V}, secondary voltage = 120 V120 \, \text{V}, output power = 60 kW=60000 W60 \, \text{kW} = 60000 \, \text{W}.

Find: The resistive load RsR_s in the secondary circuit.

For an ideal transformer, the voltage ratio is

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

So,

12012000=NsNp=1100\frac{120}{12000} = \frac{N_s}{N_p} = \frac{1}{100}

Also, for an ideal transformer, input power equals output power:

IpVp=IsVs=60000 WI_p V_p = I_s V_s = 60000 \, \text{W}

Using the secondary side values,

Is=60000120=500 AI_s = \frac{60000}{120} = 500 \, \text{A}

Now using Ohm's law for the purely resistive load,

Rs=VsIs=120500=0.24 ΩR_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.24 \, \Omega

Thus,

Rs=240 mΩR_s = 240 \, \text{m}\Omega

Therefore, the value of the resistive load in the secondary circuit is 240 mΩ240 \, \text{m}\Omega.

Using Power Conservation Step by Step

Given: Vp=12000 VV_p = 12000 \, \text{V}, Vs=120 VV_s = 120 \, \text{V}, and power consumed = 60000 W60000 \, \text{W}.

Find: The value of RsR_s.

The transformer relation is

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

Hence,

12012000=NsNp\frac{120}{12000} = \frac{N_s}{N_p}

which gives

NsNp=1100\frac{N_s}{N_p} = \frac{1}{100}

For an ideal transformer,

Pin=PoutP_{in} = P_{out}

and power is given by

P=IVP = IV

Therefore,

IpVp=IsVs=60000 WI_p V_p = I_s V_s = 60000 \, \text{W}

Primary current is

Ip=6000012000=5 AI_p = \frac{60000}{12000} = 5 \, \text{A}

Secondary current is

Is=60000120=500 AI_s = \frac{60000}{120} = 500 \, \text{A}

Now the secondary resistance is

Rs=VsIs=120500=0.240 Ω=240 mΩR_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.240 \, \Omega = 240 \, \text{m}\Omega

Therefore, the required resistive load is 240 mΩ240 \, \text{m}\Omega.

Common mistakes

  • Using the primary voltage instead of the secondary voltage while calculating the load resistance is incorrect because RsR_s is the resistance connected in the secondary circuit. Use Rs=VsIsR_s = \frac{V_s}{I_s} with secondary-side quantities only.

  • Forgetting that the transformer is ideal and not using power conservation leads to a wrong current value. Since input power equals output power, first use IsVs=60000 WI_s V_s = 60000 \, \text{W} to find the secondary current.

  • Leaving the answer in Ω\Omega instead of converting to mΩ gives an incomplete final response. After finding 0.24 Ω0.24 \, \Omega, convert it as 0.24 Ω=240 mΩ0.24 \, \Omega = 240 \, \text{m}\Omega.

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