Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.26
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 26
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The rating of a step-down transformer is 40000 V-220 V.
Input voltage,V, = 40000 V
Output voltage,V2 = 220 V
Total electric power required,
P = 800 kW = 800 x 103 W
Source potential, V = 220 V
Voltage at which the electric plant generates 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 wire lines,
R = (15 + 15)0.5 = 15 Ω
P= V1I
Rms current in the wire line is given as:
=> I = P/V1
= 800 x 103/40000= 20 A
Ans (a).
Line power loss = I2R = (20)2 x 15 = 6 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of current.
Hence, power supplied by the plant = 800 kW + 6kW = 806 kW
Ans (c).
Voltage drop in the power line = IR = 20 x 15 = 300 V
Hence, voltage that is transmitted by the power plant
= 300 + 40000 = 40300 V
The power is being generated in the plant at 440 V.
Hence, the rating of the step-up transformer needed at the plant is 440 V – 40300 V.
Hence, power loss during transmission
= (600/1400 )x 100 = 42.8%
1400
In the previous exercise, the power loss due to the same reason is (6/806 )x 100 = 0.744%
Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.