We had this question in physics class, and there are two formulas that would seem to fit the data given.
P=IV, and
V=IR
Where P is power, I is current, V is voltage, and R is resistance.
I asked the teacher, and he said that P=IV is the correct one to use, but couldn't (or wouldn't) explain why.
So why is it correct, or even is it? I know that they each seem to give different results.
A power station generates electricity at 250 MW and 400kV. The electricity I'd transmitted to a distant town along cables with a resistance of 5.0 ohms.
What is the current with in the power lines?
What is the power loss in the cable?
P=IV, and
V=IR
Where P is power, I is current, V is voltage, and R is resistance.
I asked the teacher, and he said that P=IV is the correct one to use, but couldn't (or wouldn't) explain why.
So why is it correct, or even is it? I know that they each seem to give different results.
A power station generates electricity at 250 MW and 400kV. The electricity I'd transmitted to a distant town along cables with a resistance of 5.0 ohms.
What is the current with in the power lines?
What is the power loss in the cable?
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They're both true. Used correctly, they can't give different results.
You can't use V = IR because you don't know what R is. The major part of the resistance is whatever is at the other end, actually consuming power. (The technical name is the "load").
If you treated the circuit as just consisting of the cables and divided 400 kV by 5 ohms to get the current, then you'd be saying that all of the power is being consumed in the cables. That wouldn't be a very useful power distribution system. The idea is to get the power to the consumer, not waste it all in the cables. The resistance of this circuit is more than 5 ohms.
So you can use P = VI because I is the only unknown.
You could also use P = V^2/R to find the actual total resistance in the circuit (including the load). That comes from using V = IR, substituting I = V/R.
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Edit: So actually doing it that way, P = V^2/R. R = V^2/P = (400 x 10^3 V)^2 / (250 x 10^6 W) = 640 ohms. That's the total resistance of the circuit (which means the unknown equivalent resistance of everybody consuming power is 635 ohms).
Now you could use V = IR with R = 640 ohms to find the current. It should give the same answer as doing it the other way.
You can't use V = IR because you don't know what R is. The major part of the resistance is whatever is at the other end, actually consuming power. (The technical name is the "load").
If you treated the circuit as just consisting of the cables and divided 400 kV by 5 ohms to get the current, then you'd be saying that all of the power is being consumed in the cables. That wouldn't be a very useful power distribution system. The idea is to get the power to the consumer, not waste it all in the cables. The resistance of this circuit is more than 5 ohms.
So you can use P = VI because I is the only unknown.
You could also use P = V^2/R to find the actual total resistance in the circuit (including the load). That comes from using V = IR, substituting I = V/R.
---------------------------------------…
Edit: So actually doing it that way, P = V^2/R. R = V^2/P = (400 x 10^3 V)^2 / (250 x 10^6 W) = 640 ohms. That's the total resistance of the circuit (which means the unknown equivalent resistance of everybody consuming power is 635 ohms).
Now you could use V = IR with R = 640 ohms to find the current. It should give the same answer as doing it the other way.