# Q7.25 (b)  A small town with a demand of $800 kW$ of electric power at $220V$ is situated $15km$ away from an electric plant generating power at $440V$. The resistance of the two wire line carrying power is $0.5\Omega$  per km.The town gets power from the line through a $4000-220V$ step-down transformer at a sub-station in the town.How much power must the plant supply, assuming there is negligible power loss due to leakage?

P Pankaj Sanodiya

Power required

$P=800kW=800*10^3kW$

The total resistance of the two-wire line

$R=2*15*0.5=15\Omega$

Input voltage

$V_{input}=4000V$

Output voltage:

$V_{output}=220V$

RMS current in the wireline

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

Now,

Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

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