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Alternating Current Physics Ncert Solutions Class 12th Physics Class 12th Alternating Current Overview

7.25 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

(a) Estimate the line power loss in the form of heat.

(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(c) Characterize the step up transformer at the plant.

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Payal Gupta | Contributor-Level 10

4 months ago

7.25 Total electric power required, P = 800 kW = 800 $\times 10^{3}$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V’ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $\times 15 \times 0.5 Ω$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer, $V_{1}$ = 4000 V

Output voltage from the transformer, $V_{2}$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 \times 10^{3}}{4000}$ = 200 A

Line power loss = $I^{2} R$ = $200^{2} \times$ 15 = 600 $\times 10^{3}$ W = 600 kW

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7.25 Total electric power required, P = 800 kW = 800 $\times 10^{3}$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V’ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $\times 15 \times 0.5 Ω$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer, $V_{1}$ = 4000 V

Output voltage from the transformer, $V_{2}$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 \times 10^{3}}{4000}$ = 200 A

Line power loss = $I^{2} R$ = $200^{2} \times$ 15 = 600 $\times 10^{3}$ W = 600 kW

Since the leakage power loss is negligible, the total power to be supplied by plat = 800 + 600 = 1400 kW

Voltage drop in the transmission line = IR = 200 $\times 15$ = 3000 V. Hence, the total voltage transmitted from plant = 3000 + 4000 = 7000 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 7000 V

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7.25 Total electric power required, P = 800 kW = 800 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> WSupply voltage, V = 220 VElectric plant generating voltage, V’ = 440 VDistance between the town and power generating station, d = 15 kmResistance of the two wires lines carrying power = 0.5 Ω /kmTotal resistance of the wire, R = 2 <math><mo>×</mo><mn>15</mn><mi></mi><mo>×</mo><mn>0.5</mn><mi></mi><mi>Ω</mi></math> = 15 ΩStep-down transformer rating 4000 – 220 V, henceInput voltage to the transformer,  <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 4000 VOutput voltage from the transformer,  <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 220 Vrms current in the wire lines is given asI = <math><mfrac><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>800</mn><mi mathvariant="normal"></mi><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4000</mn></mrow></mrow></mfrac></math> = 200 ALine power loss = <math><msup><mrow><mrow><mi>I</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mi>R</mi><mi></mi></math> = <math><msup><mrow><mrow><mn>200</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>×</mo></math> 15 = 600 <math><mo>×</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn><mi></mi></mrow></mrow></msup></math> W = 600 kWSince the leakage power loss is negligible, the total power to be supplied by plat = 800 + 600 = 1400 kWVoltage drop in the transmission line = IR = 200 <math><mo>×</mo><mn>15</mn></math> = 3000 V. Hence, the total voltage transmitted from plant = 3000 + 4000 = 7000 VSince the power generated is 440V, the transformer rating at the plant should be 440 V – 7000 V

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