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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2019 · 10 April · Morning Shift · Physics · Alternating Current

A transformer consisting of 300 turns in the primary and 150 turns...

A transformer consisting of 300300 turns in the primary and 150150 turns in the secondary gives output power of 2.2kW2.2kW . If the current in the secondary coil is 10A10A , then the input voltage and current in the primary coil are:

Options

  1. A440V440V and 20A20A
  2. B220V220V and 20A20A
  3. C440V440V and 5A5A
  4. D220V220V and 10A10A

Correct answer

C. 440V440V and 5A5A

Step-by-step solution

Assuming 100% efficiency,

V2V1=i1i2=N2N1\dfrac{{V}_{2}}{{V}_{1}} = \dfrac{{i}_{1}}{{i}_{2}} = \dfrac{{N}_{2}}{{N}_{1}}

i1i2=N2N1i110=150300i1=5A\dfrac{{i}_{1}}{{i}_{2}} = \dfrac{{N}_{2}}{{N}_{1}}\dfrac{{i}_{1}}{10} = \dfrac{150}{300}{i}_{1} = 5A

P=VI2200=5×V1V1=440VP = VI2200 = 5 \times {V}_{1}{V}_{1} = 440V

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