P = VI = (281V)(3.75A) = 1054 W

[Tautomer22]

Homework Statement

A small power plant produces a voltage of 6.0 kV and 150 A. The voltage is stepped up to 240 kV by the transformer before it is transmitted to a substation. The resistance of the transmission line between the power plants and the substation is 75 Ohms.

What percentage of the power produced at the power plant is lost in the transmission of the substation?

a) 0.47%
b) 0.41%
c) 0.34%
d) 0.23%
e) 0.12% (This is correct answer)

Homework Equations

N_s/N_p=V_s/V_p=I_p/I_s

P = IV
P= I²R
P = V²/R
R = V/I

The Attempt at a Solution

I know something is wrong with my attempt. The first thing I did was solve for the current of the line when the voltage is stepped up to 240 kV, which is 3.75 A compared to the 150 amps we started with. The problem I have now is with the resistance... If I use R = V/I then I can't get the current and volts to line up properly, and, in any event, the volts or currents I could get by fixing one of the numbers doesn't come out in a reasonable order of magnitude. For example, if I use 240 kV as my volts, then I get a current of 3200 A which is outlandish, so I must not be using this relationship properly. So I guess what I'm looking for is how do I quantify the effect of resistance on the line?


Update: I found a way to get the answer, but I'm not sure why it should be this way... If I take the initial power at 9 * 10⁵ by multiplying 6000 V by 150 A, and then I get a value of power in the line from the 3.75 amps as follows: P = (3.75)²(75 Ohms) = 1054 watts. If I take this number (1054W) not as the power in the line, but the power lost in the line, then I can subtract it from 9 * 10⁵ and then simply generate a percent from the fierrence, giving me 0.12%

My new question: Why in the world is P = I²R equal to the power lost in the line, and not simply the power in the line?

 

[nil1996]

In the wire between the power plant and substation the current is same which is 3.75 A. Now you have given the resistance of the wire. From this you can calculate the power lost across the wire.Then for finding the percentage you can compare power created before transmitting and power lost. This will give you the answer

 

[Tautomer22]

In the wire between the power plant and substation the current is same which is 3.75 A. Now you have given the resistance of the wire. From this you can calculate the power lost across the wire.Then for finding the percentage you can compare power created before transmitting and power lost. This will give you the answer

I understand your logic, but it's the numbers I don't understand... With an resistance of 75 ohms and a current of 3.75, the voltage I calculate is = 281 V (which is very far off from the expected 240,000 V). Also, if I simply use a power equation, such as I²R then I get a power equal to 1054 W, compared to the 9 * 10⁵ at the power station... Clearly, if I take the percent of this, it's not going to less than a 1% power loss, which is the range of the answer...

I'm using one of the equations incorrectly, or making a wrong assumption somewhere. What answer do you get, and how exactly did you get it with respect to numbers?

 

[haruspex]

if I use 240 kV as my volts, then I get a current of 3200 A which is outlandish

You are overlooking that there is an unknown load somewhere out on the user side. Your calculation is as though there is a short between the wires.

 

[Tautomer22]

You are overlooking that there is an unknown load somewhere out on the user side. Your calculation is as though there is a short between the wires.

So you're saying I need to focus on the turns ratio of the transformer rather than the resistance in ohms? I understand what you're saying, but I suppose I attacked this problem under the assumtpion that I would solve it by

1) Finding the power at the power plant
2) Finding the power delivered to the substation

I didn't realize I would necessarily have to consider what was beyond the substation. I thought the P=I^2R would give me the power delivered to the substation, rather than the power lost in the line.

I'm in the 2nd semester of a non-calculus based college physics course, so I've only just begun to be exposed to the mathematical expressions of currents.

 

[haruspex]

So you're saying I need to focus on the turns ratio of the transformer rather than the resistance in ohms?

No.

I didn't realize I would necessarily have to consider what was beyond the substation. I thought the P=I^2R would give me the power delivered to the substation, rather than the power lost in the line.

You don't need to worry about what load there is on the line, merely be aware that most of the power is going to service that load and only a portion is lost in the line.
You can treat the line and the load as two resistors in series. You calculated the current in the high voltage line, and you know its resistance, so...

 

[CWatters]

I understand your logic, but it's the numbers I don't understand... With an resistance of 75 ohms and a current of 3.75, the voltage I calculate is = 281 V (which is very far off from the expected 240,000 V).

Why are you expecting the voltage LOSS to be 240,000V ?

Think about it. If the voltage lost in transmission was that big the voltage at the town/destination would be zero. Not very useful :-)

The voltage drop down the transmission line is 281V making the voltage at the destination town a much more handy 240,000 - 281 = 239719V. (A step down transformer is used in the town to make 110V/230V)

If the voltage lost is 281V and the current is 3.75A what's the power lost?