# How is it that power can vary if the current stays constant?

1. Jul 28, 2013

### qtac

Hello everyone,

I'm trying to teach myself basic circuit design from an EE textbook, but I'm starting to think myself in circles here. I have no trouble with the basic algebra, but the logic behind the formulas is giving me trouble.

The thing I'm most confused about right now is the relation of current to power:

1) $P=I^2R$
2) and $P = VI$

From these, it's clear that if the current stays constant and the voltage increases, power will increase proportionally... but I just can't wrap my head around why that's so.

Let's say we have a 12V battery connected to a resistor such that it draws 1A. We have another setup that uses a 24V battery and a stronger resistor, such that the current is similarly 1A. The circuit draws the same current but uses double the power. How can that be?

2. Jul 28, 2013

### FOIWATER

Power is measured in watts, or Joules/second
P=V*I = volts*amperes = (Joules/Coulomb)*(Coloumb/sec) = Joules/Second = power

If you have a situation where a load draws the same amount of current as another load but at a different voltage, consider that more energy is required to "push" current through the higher resistance load - because of its higher resistance.

The higher resistance requires more energy to accept the same amount of amperes as a lower resistance.

3. Jul 28, 2013

What would the "R" be for the two different cases? The calculate formula #1.

4. Jul 28, 2013

### qtac

That's an interesting way to look at it. Certainly makes a lot more sense than I was making in my own head.

Thank you

5. Jul 31, 2013

### sophiecentaur

There is no surprise that different resistances can be used to vary the power drawn from a constant Voltage supply (normal electrical supply - battery or alternator). The same thing will apply for a constant Current supply but such generators are relatively rare.