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AIIMS2019Physics-Electromagnetic Induction

AIIMS 2019 Physics Transformers MCQ Question

Type: MCQ-numerical-Medium-Class 12

Transformer →\rightarrow ideal →Ep=1000 V\rightarrow \text{E}_\text{p} = 1000\text{ V}, Ip=50 A\text{I}_\text{p} = 50\text{ A}, 220 V→80 houses220\text{ V} \rightarrow 80\text{ houses} resistances of secondary coil will be:

A

2 Ω2\ \Omega

B

3 Ω3\ \Omega

C

1 Ω1\ \Omega

D

4 Ω4\ \Omega

Correct Answer

Option C

Detailed Explanation

In the given problem, the relationship between input and output power in an ideal transformer is utilized to find the secondary resistance RsR_s. By rearranging the equation Pin=PoutP_{in} = P_{out} and substituting the known values, we find Rs≈0.968 ΩR_s \approx 0.968 \, \Omega, which rounds to approximately 1 Ω1 \, \Omega, making option (B) the closest correct answer. Options (A), (C), and (D) are incorrect as they do not match the calculated resistance; specifically, 0.5 Ω0.5 \, \Omega is too low, 2 Ω2 \, \Omega and 4 Ω4 \, \Omega are too high, demonstrating the importance of accurate calculations in electrical engineering principles.

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