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1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted

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प्रश्न

1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

  1. power is transmitted at 220 V. Comment on the feasibility of doing this.
  2. a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transfomer is used to bring voltage to 220 V. (ρCu = 1.7 × 10–8 SI unit)
दीर्घउत्तर
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उत्तर

i. When power is transmitted at 220 V.

Power lost in transmission P = I2R

P = VI

P = 1 MV = 106 W

V = 220 Volt

`I = P/V = 1000000/220`

`I = 50000/11 A`

R = `ρ I/A`

⇒ R = `(ρl)/(pir^2)` 

I = 10 km × 2 = 20,000 M

∴ A = πr2

r = 0.5 cm = 0.5 cm × 10–2 = 5 × 10–3 cm

ρCu = 1.7 × 10–8 Ωm

R = `(ρI)/A`

∴ R = `(1.7 xx 10^-8 xx 20,000)/(3.14 xx 5 xx 10^-3 xx 5 xx 10^-3)`

= `(170 xx 20000 xx 10^(-8+6))/(314 xx 5 xx 5)`

= `(170 xx 20000)/(314 xx 25 xx 100)` Ω

R = `(170 xx 800)/(314 xx 100) = (170 xx 4)/157 = 680/157` ≅ 4Ω

∴ Power loss = I2R

= `50000/11 xx 50000/11 xx 4 = (100 xx 10^8)/121 = 8.26 xx 10^7`

Power loss in heating = 82.6 MW

As 82.6 MW > 1 MW

So this method cannot be used to transmit the power.

ii. When power is transmitted at 11000 V

P = 106 W

VI = 1000000

11000 I = 1000000

`I = 1000000/11000 = 1000/11`

∴ RCu = 4Ω as already calculated in part (i)

∴ Power loss = P = I2R

P = `1000/11 xx 1000/11 xx 4 = 4000/121 xx 4^4`

P = `3.3 xx 10^4` Watt

Fractional power loss = `(3.3 xx 10^4)/10^6 = 3.3/1000` = 0.033

Power loss in 3.3%

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अध्याय 7: Alternating Current - MCQ I [पृष्ठ ४५]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
अध्याय 7 Alternating Current
MCQ I | Q 7.28 | पृष्ठ ४५

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