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Current Electricity question

2020 · 5 Sep · Shift 1 · Q61
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  5. /2020 · 5 Sep · Shift 1 · Q61

Current Electricity question

2020 · 5 Sep · Shift 1 · Q61

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
An electrical power line, having a total resistance of 2 Ω\OmegaΩ, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately :
  1. A
    85%
  2. B
    96%
  3. C
    72%
  4. D
    91%
View written solutionFree

Correct answer: B

  1. Given data
  • Resistance of transmission line: R=2 ΩR = 2\,\OmegaR=2Ω
  • Power delivered to the consumer: Pload=1 kW=1000 WP_{\text{load}} = 1\text{ kW} = 1000\text{ W}Pload​=1 kW=1000 W
  • Voltage at the consumer end: V=220 VV = 220\text{ V}V=220 V

We need the efficiency of transmission:

η=power delivered to loadtotal power sent×100%\eta = \frac{\text{power delivered to load}}{\text{total power sent}} \times 100\%η=total power sentpower delivered to load​×100%
  1. Find the current in the line

Since the load receives 1000 W1000\text{ W}1000 W at 220 V220\text{ V}220 V,

I=PloadV=1000220≈4.545 AI = \frac{P_{\text{load}}}{V} = \frac{1000}{220} \approx 4.545\text{ A}I=VPload​​=2201000​≈4.545 A
  1. Find power loss in the transmission line

Power lost in the line is

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

So,

Ploss=(4.545)2×2P_{\text{loss}} = (4.545)^2 \times 2Ploss​=(4.545)2×2 (4.545)2≈20.66(4.545)^2 \approx 20.66(4.545)2≈20.66 Ploss≈20.66×2=41.32 WP_{\text{loss}} \approx 20.66 \times 2 = 41.32\text{ W}Ploss​≈20.66×2=41.32 W
  1. Find total power sent
Pinput=Pload+Ploss=1000+41.32=1041.32 WP_{\text{input}} = P_{\text{load}} + P_{\text{loss}} = 1000 + 41.32 = 1041.32\text{ W}Pinput​=Pload​+Ploss​=1000+41.32=1041.32 W
  1. Calculate efficiency
η=10001041.32×100%\eta = \frac{1000}{1041.32} \times 100\%η=1041.321000​×100% η≈96%\eta \approx 96\%η≈96%
  1. Match with options

The closest option is:

  • B: 96%

Therefore, the correct answer is B.

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