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Current Electricity question

2013 · Shift 2 · Q52
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  5. /2013 · Shift 2 · Q52

Current Electricity question

2013 · Shift 2 · Q52

JEE AdvancedPhysicsCurrent ElectricityMCQ+3 / −1
A thermal power plant produces electric power of 600 kW at 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 power factor unity. All the currents and voltages mentioned are rms values.If the direct transmission method with a cable of resistance 0.4 Ω\OmegaΩ km −-− 1 is used, the power dissipation (in %) during transmission is
  1. A
    20
  2. B
    30
  3. C
    40
  4. D
    50
View written solutionFree

Correct answer: B

  1. Given data
  • Power generated: P=600 kW=6×105 WP = 600\,\text{kW} = 6\times 10^5\,\text{W}P=600kW=6×105W
  • Transmission voltage: V=4000 VV = 4000\,\text{V}V=4000V
  • Distance: 20 km20\,\text{km}20km
  • Cable resistance: 0.4 Ω km−10.4\,\Omega\,\text{km}^{-1}0.4Ωkm−1

For direct transmission, the power is sent at 4000 V4000\,\text{V}4000V itself.

  1. Current in the transmission line

Using

P=VIP = VIP=VI

so

I=PV=6×1054000=150 AI = \frac{P}{V} = \frac{6\times 10^5}{4000} = 150\,\text{A}I=VP​=40006×105​=150A
  1. Total resistance of the cable

The place is 20 km20\,\text{km}20km away, so the transmission length is 20 km20\,\text{km}20km.

Given resistance per km is 0.4 Ω/km0.4\,\Omega/\text{km}0.4Ω/km, hence

R=0.4×20=8 ΩR = 0.4 \times 20 = 8\,\OmegaR=0.4×20=8Ω
  1. Power dissipated in the cable

Since the cable is purely resistive,

Ploss=I2RP_{\text{loss}} = I^2RPloss​=I2R

Thus,

Ploss=(150)2×8P_{\text{loss}} = (150)^2 \times 8Ploss​=(150)2×8 =22500×8=180000 W=180 kW= 22500 \times 8 = 180000\,\text{W} = 180\,\text{kW}=22500×8=180000W=180kW
  1. Percentage power dissipation
% loss=PlossP×100\%\text{ loss} = \frac{P_{\text{loss}}}{P} \times 100% loss=PPloss​​×100 =180600×100=30%= \frac{180}{600} \times 100 = 30\%=600180​×100=30%
  1. Checking options
  • A: 20%20\%20% ❌
  • B: 30%30\%30% ✅
  • C: 40%40\%40% ❌
  • D: 50%50\%50% ❌

Therefore, the correct option is B.

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