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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Physics
Chapter
Alternating Current

JEE Advanced 2013 · Paper 2 · Physics · Alternating Current

Paragraph: A thermal power plant produces electric power of 600 ~kW...

Paragraph: A thermal power plant produces electric power of 600 kW600 \mathrm{~kW} at 4000 V4000 \mathrm{~V}, which is to be transported to a place 20 km20 \mathrm{~km} away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.
Question: In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1:101: 10. If the power to the consumers has to be supplied at 200 V200 \mathrm{~V}, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

Options

  1. A200:1200: 1
  2. B150:1150: 1
  3. C100:1100: 1
  4. D50:150: 1

Correct answer

A. 200:1200: 1

Step-by-step solution

NpNs=40,000200=2001\dfrac{{\text{N}}_{\text{p}}}{{\text{N}}_{\text{s}}} = \dfrac{\text{40,000}}{2 0 0} = \dfrac{2 0 0}{1}
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