# Need help understanding power consumption

#### peterscache12

Joined Apr 3, 2015
2
Now, just to let you guys know, I am a beginner relative to you all. I have read up on ohms law, and a lot of other things. I've even got to the point of making a three phase ac wind turbine generator (though I don't understand it all the way.) However, something is really bothering me, and I hope it doesn't sound dumb (keep in mind that I'm only 15.) Ohms law states that V=IxR, and this other law says that W=VxI (watts, volts, current.) However, when placed into a circuit it is very strange. Let's say we have a 6v battery, and for convenience sake we're going to use it on a load (resistor) with a constant resistance, and not take into account the wire's resistance. Now, a 6v battery going through a resistor of 100 ohms will give us 0.06 amps. 0.06 x 6 = 0.36 watts. Now, a circuit with the same battery, and a load of 30 ohms will give us 0.3 amps. 0.3 x 6 = 1.8 watts. Now, tell me why a circuit with less resistance is consuming more power. To me, I've always thought that big power hungry machines in the home like air conditioners have a higher resistance, because a higher resistance means more energy output, right? Now, I know that a short circuit will draw unlimited amount of current, (at least as much as the battery is capable of), and that that power consumption must be pretty high. But there's no resistance! (Apart from the internal resistance of the battery.) Tell me why, particularly with my 6v circuit example, please!

#### #12

Joined Nov 30, 2010
18,223
You simply have the wrong intuitive feel right now. For a constant voltage, the lower the resistance, the more current will flow. It is only when you get to zero resistance that the intuition breaks down.
Think, 6 volts and 6 ohms...1 amp
Think 6 volts and 3 ohms...2 amps
6 volts and 1 ohm...6 amps
You can actually buy a 0.05 ohm resistor!
6 volts and 0.05 ohms...120 amps
6 volts and zero ohms...infinity amps.
Bang. Intuition breaks down.
Just hold that zero out of your mind until you get a better feel for it.
You already know the voltage source and the wires have resistance in the real world. The zero never happens except in exotic cases like superconductors. For all practical purposes, zero never happens until you get a few years under your belt.

#### t_n_k

Joined Mar 6, 2009
5,455
Another approach is to consider what's happening with respect to the source supplying the power. For the case of an ideal DC source of EMF V [volts], delivering current I [amps] to the connected load, then the power delivered from the source to the load will be V*I [watts]. Clearly the source power (hence load power) increases as load current I increases. Load current I, can only increase if the load resistance decreases.

#### dl324

Joined Mar 30, 2015
13,814
Hi,

Another way to express power is to substitute $$I=V/R$$ and get $$W = V^2/R$$. That makes it easy to see that power dissipation is inversely proportional to resistance.

HTH
Dennis

#### #12

Joined Nov 30, 2010
18,223
Hi,

Another way to express power is to substitute $$I=V/R$$ and get $$W = V^2/R$$. That makes it easy to see that power dissipation is inversely proportional to resistance.

HTH
Dennis
and the R in the denominator shows that the answer set includes infinity, which is impossible in the real world.

#### upand_at_them

Joined May 15, 2010
766
It's usually helpful to think of the case of no resistor. With no resistor for a load (no wire, no anything) you say that it consumes no power, right? Well no resistor is the same thing as a resistor with infinite resistance.

So a larger resistance consumes less power than a smaller resistance (assuming a constant voltage source).