JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Current Electricity

Paragraph for Questions 15 and 16

A thermal power plant produces electric power of

600 k W

4000 V

, which is to be transported to a place 20 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.

If the direct transmission method with a cable of resistance

0.4Ω km^{- 1}

is used, the power dissipation (in %) during transmission is

(A)20
(B)30✓ Correct
(C)40
(D)50

Answer: (B)

Solution

Current in direct transmission:

I = \frac{P}{V} = \frac{600 \times 1 0 ^{3}}{4000} = 150 A

Total cable resistance:

R = 0.4 \times 20 = 8 Ω

Power loss:

P_{loss} = I^{2} R = (150)^{2} \times 8 = 180 kW

Percentage loss:

% = \frac{180}{600} \times 100 = 30%

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Current Electricity

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