# Does P=I^2R = really P=V^2/R

#### JDee

I understand P = VI and this makes total sense. A coulomb is pushed or pulled along a wire at a certain Voltage and we can work out the amount of power produced.

Now correct me if I am wrong but P = I^2R is used to determine to amount of power lost in the wire during transmission from point A to point B. The Resistance in the wire is a fixed value. So using a transformer to step up the Voltage and in effect reduce the Ampres (P=VI) we can transfer the power with less loss due to resistance.

Let's say I want to transfer 1000W @ 100V over a wire with 2R. P=10^2 x 2 (P = 300W). However someone has tried to explain P=V^2/R but using the same example P=100^2/2 (P = 5000W). This doesn't make sense to me.

So I would like to ask:
Is P =VI used to determine the amount of total power being pushed along the wire
Is P = I^2R used to determine the amount of power being lost due to resistance while moving the power from point A to point B.
Therefore is P in the two equations the same value? I would assume one is total Power and the other is Power loss and therefore not interchangeable

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#### SpaceBuff12

5+ Year Member
They use P=I^2R for power loss because the current is the same throughout the circuit. If you wanted to find the power loss using P=V^2/R then you would have to know the voltage drop from point A to B.

Edit: I'm assuming a simple circuit in this explanation. In fact, I should not have addressed your question without specifying conditions. It does not help to discuss these values in generalized terms because you must be specific about the applied conditions.

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#### SpaceBuff12

5+ Year Member
I realized that I didn't fully answer your question.

First, the numbers that you used do not make sense. You cannot have 1000W @ 100V with a resistance of 2 ohms; using P=VI your current will be 10 A and your R = 10 ohms.
Therefore is P in the two equations the same value? I would assume one is total Power and the other is Power loss and therefore not interchangeable
By law P=I^(2)*R = V^(2)/ R.

Do you have a specific problem that you wanted to discuss?

Is P = I^2R used to determine the amount of power being lost due to resistance while moving the power from point A to point B.
Do you have a specific example tied to this question? Power loss is a term used to describe power that does not contribute toward the intended purpose, e.g. power dissipation in transmission lines. Obviously, the resistance in that example is not the total resistance of the power grid, it is the resistance along the transmission line.

OP
J

#### JDee

Thanks for the response and clarifying. I don't have a specific problem or question. I have just been reading some text books and it never specifically stated P = I^2R is used to determine the amount of power lost between point A and point B. The books and your second post have made the assumption that someone is always working with a closed circuit. This confused me for a little until I worked it out, with the assistance of your first post.

Whereas the numbers I provided 1000W @ 100V (10 Amps) with a R of 2 ohms can defiantly be possible, when not in a closed circuit ie transmission line - R of the line = Voltage drop over the line / I which also converts to Voltage drop over the line = I / resistance of the line which makes perfect sense as Power drop over the line = I2 x R of the line.

I would I therefore be right in assuming the Voltage will drop over the line (where no load is in play). However the Amps will always stay the same.

Also I have a quick question in regards to capacitors. I understand a capacitor is two conductive surfaces with a dielectric between to provide resistance. The Voltage builds up on one side and when it gets high enough to push through the resistance it does so. It is my understanding the toroid at the top of a Tesla coil is an over sized capacitor (correct me if I am wrong). The toroid is 1 surface, the atmosphere as the dielectric and the earth as the other surface. Now my question is what is the difference between a capacitor and a spark gap?

#### Next Step Tutor

##### MCAT Guru
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5+ Year Member
Now my question is what is the difference between a capacitor and a spark gap?
There's like a 99% chance that you won't see questions about a spark gap on the MCAT. Or if you did, they would explain it in the passage such that any questions you got would actually be critical reading questions (did you read the passage carefully) rather than actual physics knowledge questions.

Having said that, it sounds like your mental model of a capacitor is wrong. The point of a capacitor is to store charge. You mention the notion of the voltage "push[ing] through the resistance". Instead, in the circuits you're going to see on the MCAT, a capacitor will simply store up charge until the potential difference across the plates is high enough that it can't hold anymore. The current won't "push through", it will simply stop flowing to the capacitor.

Imagine a simple circuit with one battery, one resistor, and one capacitor. First, if the resistor and capacitor are hooked up in series: the current will "push through" the resistor, and the build up on the capacitor. Eventually the voltage drop across the capacitor will equal the voltage of the battery and current will stop flowing. At that point, the voltage drop across the resistor will be 0 (since V = IR for the resistor).

Next, if the resistor and capacitor are hooked up in parallel: the capacitor will slowly build up charge until it plateaus, at which point some current will still "leak" through the circuit by going through the resistor.

A spark gap would be sort of like a failed (maybe we could loosely say "shorted out" or "burned out") capacitor. A voltage difference builds up between the electrodes but then the air between breaks down into ions capable of carrying charge, and current starts flowing through the air. So instead of storing charge, the "plates" of the spark gap are now transmitting charge.

Hope this helps! 