## What percentage of the original power is lost help please?

A power transmission line 50km long has a total resistance of 0.60 ohm. A generator produces 7 kilowatts at 100V. In order to reduce the power loss due to heating of the transmission line, the voltage is stepped up to 100kV using a transformer.

What percentage of the original power would be lost if the transformer were not used?

What percentage of the original power is lost when the transformer is used?

How would I figure this out? Any help is highly appreciated thanks!

### One Response to “What percentage of the original power is lost help please?”

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Physics Masteron January 18th, 2012That is some kinda tricky but we are gonna work it out

Thermal power consumed by the transmission line due to its resistance:

P(c)=RI^2

Where P(c) is power consumed

I= produced power divided by produced voltage: I=P/V

(1) In the case before voltage was raised

I=P/V=7000/100=70 A

Thermal power consumed in this case:

P(c)=RI^2=0.6*(70)^2=2940 Watt

Lost power percentage=2940/7000= 42.0%

(2) In the case after voltage has been raised:

P=IV(produced power)

When P is constant then we have

IV=constant ⇒ I1V1=I2V2 ⇒ I2/I1=V1/V2

I2/I1=100/100K=1/1000

And since the lost power percentage is directly proportional to (I^2)

then the ratio between the lost power after and before the voltage raising is equal to (I2/I1)^2 =(1/1000)^2=1/10^6

In math language: P(c)2/P(c)1=(I2/I1)^2=1/10^6

This implies that: P(c)2=P(c)1/10^6

then lost power percentage after voltage raising = 42/10^6 = 42*10^-6%

The second part can be obtained the same way the first part was done as follows:

I=P/V=7000/100000=0.07A

P(c)=RI^2=0.6*(0.07)^2= 0.00294 watt

Lost power percentage in this case=0.00294/7000=42*10^-6%

Comments:

1- This problem would be a little bit more complicated but even more realistic if the efficiency of the transformer was considered.

2- One should differentiate between Power produced and Power consumed (big difference).