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

JEE Advanced 2013 · Paper 2 · Physics · Current Electricity

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: If the direct transmission method with a cable of resistance 0.4Ωkm−10.4 \Omega \mathrm{km}^{-1} is used, the power dissipation (in \%) during transmission is

Options

  1. A2020
  2. B3030
  3. C4040
  4. D5050

Correct answer

B. 3030

Step-by-step solution

P=600×1000=4000×I⇒I=150A\text{P} = 6 0 0 \times 1 0 0 0 = 4 0 0 0 \times \text{I} ⇒ \text{I} = 1 5 0 \text{A}
dHdt=(150)2×0.4×20\dfrac{\text{dH}}{\text{dt}} = {\left(1 5 0\right)}^{2} \times \text{0.4} \times 2 0
Which is 3030 % of 600kW600kW.
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