# Homework Help: Voltage drop and transformer improvement

1. Jan 26, 2016

### AstroJMT42

1. The problem statement, all variables and given/known data

Power station produces AC power at 2.5Mw, I = 100A.
Transferred along 100km line with a resistance of 50 ohms to a factory. Calculate power lost and % efficiency

2. Relevant equations

P=VI
P = I2R

3. The attempt at a solution

Initial power at the station is 2.5x106 . Using I2R you get new power at the factory of 1002x50 = 5x105W

This means a total loss of 2x105 and a percentage of 8%.

I'm pretty confident this is correct but a fellow student is adamant that I2R is power loss not total power out. His conclusion is that the line is actually 80% efficient, which sounds way to efficient to be correct, especially seeing as there are no transformers. Can anyone please confirm which of us is correct?

second part then takes the same original power and introduces a 1:5 step up transformer and 5:1 step down and asks to calculate the new power loss and new efficiency.

I worked initial V = 25,000 and after step up V = 125,000
New power after this using I2R is 2,480,000W which I calculated to be an efficiency of 99.2%
Again could you please check my logic is sound? There were no answers given.

Apologies it is actually 2 questions being asked but the second one leads directly from the first so thought it was best to keep it in one post.

Thanks!

2. Jan 26, 2016

### BvU

Check your calculations. $2.5 \ 10^6 - 5 \ 10^5 = 2 \ 10^5$ ?
Is the 50 $\Omega$ the resistance of the load in the plant or is it the resistance of the power cables ? So is it the useful power or the lost power ?

3. Jan 26, 2016

### AstroJMT42

50 Ω is the resistance in the cable.

2,500,000W originally - 500,000 = 2,000,000W sorry that was a silly arithmetic error.

So that makes 500,000 / 2,500,000 x 100 = 20%?

4. Jan 26, 2016

### BvU

Efficiency is useful power divided by input power. So: No.

Fellow student wins the argument.

Funny enouh you do have the right efficiency for the transformers in place case. Note the factor with which the loss is reduced!

5. Jan 26, 2016

### AstroJMT42

So I squared R is is the power loss, not the new power?

I think this is where my misunderstanding is coming from. I thought P=VI and P=I2R were interchangeable. We were given original power and had to use P=I2R to find what the power is after traveling down the line. divide this output power by the original and multiply by 100 to get the percentage.
Which part of that am I getting wrong? it's been a long day...

6. Jan 26, 2016

### CWatters

It depends which R you are talking about? The resistance of the cable or the load (aka the factory)?

If R is the resistance of the cable then I2R is the power dissipated (lost) in the cable.
If R is the resistance of the load (eg the equivalent resistance of the factory) then I2R is the power dissipated (used) by the factory.

In general they are. Provided you are talking about the right V, I and R.

Apply conservation of energy eg..

Power delivered to factory = Power generated - power lost in the cable.