A town situated $20\,\mathrm{km}$ away from a power plant generating power at $440\,\mathrm{V}$ requires $600\,\mathrm{kW}$ of electric power at $200\,\mathrm{V}$. The resistance of the two-wire line carrying power is $0.4\,\Omega$ per km. The town gets power from the line through a $3000$–$220\,\mathrm{V}$ step-down transformer at a substation. (i) Find the line power loss in the form of heat. (ii) How much power must the plant supply, assuming negligible leakage loss?

(i) Line loss $= 640\,\mathrm{kW}$. (ii) Plant must supply $1240\,\mathrm{kW}$.

CBSE Class 12 · 2003 · Alternating CurrentShort Answer

A town situated

20 km

away from a power plant generating power at

440 V

requires

600 kW

of electric power at

200 V

. The resistance of the two-wire line carrying power is

0.4 Ω

per km. The town gets power from the line through a

3000

–

220 V

step-down transformer at a substation.
(i) Find the line power loss in the form of heat.
(ii) How much power must the plant supply, assuming negligible leakage loss?

Answer: (i) Line loss $= 640 kW$ .
(ii) Plant must supply $1240 kW$ .

Solution

The two-wire line has total length

2 \times 20 = 40 km

, so its resistance is

R = 40 \times 0.4 = 16 Ω

The line current (on the

3000 V

transmission side) is

I = \frac{P}{V} = \frac{600 \times 1 0 ^{3}}{3000} = 200 A

(i) Heat loss in the line:

P_{loss} = I^{2} R = (200)^{2} (16) = 640 \times 1 0^{3} W = 640 kW

(ii) Power the plant must supply = power delivered + line loss:

P_{plant} = 600 + 640 = 1240 kW

P_{loss} = 640 kW, P_{plant} = 1240 kW

CBSE Class 12 Board · 2003 · Subjective

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CBSE Class 12 · 2003 · Alternating CurrentShort Answer

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