Question

A power station delivers 690 kW of power at 12,000 V to a factory through wires with total resistance 5.0 Ω. How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V? Express your answer using two significant figures.

Answer 1

Answer:

15.579 kW

Explanation:

current passing through the wire

$I = \dfrac{P}{V} = \dfrac{690 \times 10^3}{12000} = 57.5 A$

power loss

$P_{loss} = I^2R = (57.5)^2\times 5 = 16531.25 W$

the current transmission for 50,000 V is

$I' = \dfrac{P}{V'} = \dfrac{690 \times 10^3}{50000} = 13.8 A$

power loss

$P_{loss} = I^2R = (13.8)^2\times 5 = 952.2 W$

wasted power

   =  16531.25 W - 952.2 W

   = 15579.05 W = 15.579 kW

Hence, the power wasted is equal to 15.579 kW

Answer 2

Electric power delivers from power station is the rate of energy transfer per unit time. The less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

What is electric power?

The electric power is the amount of electric energy transferred per unit time. It can be given as,

$P=IV$

Here, (I) is the current and (V) is the electric potential difference.

Given information-

The power delivers by the power station is 690 kW.

The voltage of the power is 12,000 V.

The resistance of the wire is 5.0 Ω.

As the power delivers by the power station is 690 kW and the voltage of the power is 12,000 V. Thus the current flowing through the wire is,

$I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{12000}\\I=57.5 \rm A$

The power loss due to the resistance of the wire is 5.0 Ω is,

$P_{loss}=I^2R\\P_{loss}=(57.5)^2\times5\\P_{loss}=16531.25\rm W$

Now the electricity is delivered at 50,000 V rather than 12,000 V.

The power delivers by the power station is 690 kW and the The voltage of the power is 50,000 V. Thus the current flowing through the wire is,

$I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{50000}\\I=13.8 \rm A$

The power loss due to the resistance of the wire is 5.0 Ω is,

$P_{loss}=I^2R\\P_{loss}=(13.8)^2\times5\\P_{loss}=952.9\rm W$

As when the voltage of the power is 12,000 V the power loss is 16531.25 W and when the voltage of the power is 50,000 V the power loss is 952.9 W.

Thus less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is,

$P_{saved}=16531.25-952.2\\P_{saved}=15579\rm W\\P_{saved}=15.579\rm kW$

Hence, the amount of less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

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