Question

A power line 50km long has a total resistance of 0.60ohm. A generator produces 100V at 70A. in order to reduce energy loss due to heating of the transmission line, the voltage is stepped up with a transformer with a primary to secondary turns ratio of 1 to 100. What percentage of the original energy is lost when the transformer is not used?

Answer 1

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%