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

7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

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Payal Gupta | Contributor-Level 10

4 months ago

7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage, $V_{1}$ = 40000 V

Output voltage, $V_{2}$ = 220 V

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

Source potential, V = 220 V

Voltage at which electric plant generates power, 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

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 \times 10^{3}}{40000}$ = 20 A

Line power loss = $I^{2} R$ = $20^{2} \times$ 15 = 6 $\times 10^{3}$ W = 6 kW

Since the le

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7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage, $V_{1}$ = 40000 V

Output voltage, $V_{2}$ = 220 V

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

Source potential, V = 220 V

Voltage at which electric plant generates power, 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

rms current in the wire lines is given as

I = $\frac{P}{V_{1}}$ = $\frac{800 \times 10^{3}}{40000}$ = 20 A

Line power loss = $I^{2} R$ = $20^{2} \times$ 15 = 6 $\times 10^{3}$ W = 6 kW

Since the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kW

Voltage drop in the transmission line = IR = 20 $\times 15$ = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 V

Hence, power loss during transmission = $\frac{6}{806}$ $\times 100$ = 0.744 %

In the previous exercise, power loss during transmission = $\frac{600}{1400}$ $\times 100$ = 42.86 %

Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

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7.26 The rating of step-down transformer = 40000 V – 220 VHence, the input voltage,  <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 40000 VOutput voltage,  <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 220 VTotal 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> WSource potential, V = 220 VVoltage at which electric plant generates power, 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 Ωrms 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>40000</mn></mrow></mrow></mfrac></math> = 20 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>20</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>×</mo></math> 15 = 6 <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 = 6 kWSince the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kWVoltage drop in the transmission line = IR = 20 <math><mo>×</mo><mn>15</mn></math> = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 VSince the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 VHence, power loss during transmission = <math><mfrac><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>806</mn></mrow></mrow></mfrac></math> <math><mo>×</mo><mn>100</mn><mi></mi></math> = 0.744 %In the previous exercise, power loss during transmission = <math><mfrac><mrow><mrow><mn>600</mn></mrow></mrow><mrow><mrow><mn>1400</mn></mrow></mrow></mfrac></math> <math><mo>×</mo><mn>100</mn></math> = 42.86 %Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

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