Given:

Supply voltage, V = 220 V

Total electric power required, P = 800 kW = \( 800 × 10^{3} W \)

The voltage at which the electric plant is generating power, V’ = 440 V

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

The resistance of the two wireline carrying powers = 0.5\( \omega /km \)

Total resistance of the wires, R = (15 + 15)0.5 = 15 \( \omega \)

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 coil, \( I = \frac{P}{V_{1}}\) \( \Rightarrow 800 * \frac{10^{3}}{4000}\) \( \Rightarrow 200 A\)

(A) Estimate The Line Power Loss In The Form Of Heat.

\( Power Loss = I^{2}R\) \( \Rightarrow 200^{2} * 15 \) \( \Rightarrow 600 KW\)

(B) How Much Power Must The Plant Supply, Assuming There Is Negligible Power Loss Due To Leakage?

Total power supplied by the path – P = 800 + 600

\( \Rightarrow 1400 KW\)

(C) Characterise The Step Up Transformer At The Plant.

Voltage drop is IR = 3000 V

The secondary voltage is 3000 + 4000

\( \Rightarrow 7000 V\)

Power generated is at a primary voltage of 440 V

Hence the transformer rating is 440 V − 7000 V

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