Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.25
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 25
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Total electric power required, P = 800 kW = 800 x 103 W
Supply voltage, V = 220 V
Voltage at which electric plant is generating power, V’ = 440 V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power =0.5 Ω/km
Total resistance of the wires, R = (15 + 15)0.5 =15Ω
A step-down transformer of rating 4000 – 220 V is used in the sub-station.
Input voltage, V1 = 4000 V
Output voltage, V2 = 220 V
rms current in the wire lines is given as:
I = P V1 = 800 x 10³/4000 = 200 A
Ans (a).
Line power loss = I2R = (200)2 x 15 =600 x103W = 600 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of the current:
Total power supplied by the plant = 800 kW + 600 kW = 1400 kW
Ans (c).
Voltage drop in the power line = IR = 200 x 15 = 3000 V
Hence, total voltage transmitted from the plant
= 3000 + 4000 =7000 V
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V — 7000 V.