I built a hot wire cutter. My nichrome wire has 1.5 ohms of resistance. When I feed my hot wire from a 12 volt transformer controlled by a 120v variac, I can control the amount of voltage from the variac, therefore controlling the voltage from the transformer.

As I turn the voltage up, I notice that my hot wire cutter starts cutting foam at around 46 volts measured at the variac, which is about 4.6 volts measured coming out of the transformer.

So here are my questions.

If my wire has 1.5 ohms of resistance, what does that tell me?

Let's call the voltage from the 12v transformer 4.5 volts for simplicity's sake. Well, with 1.5 ohms and 4.5 volts, that means I've got a 3 amp load, or 13.5 watts, right?

Or does it mean that I've got 45 volts coming out of the variac, and with a 1.5 ohm resistance in the nichrome wire, that means I've got a 30 amp load, or 135 watts? That can't be right, my variac has a 10 amp fuse.

When I started, I thought that with 1.5 ohms of resistance at 12 volts, that would mean 8 amps and 96 watts. But the nichrome wire will blow apart at that amperage. That's why I have to control the voltage with a variac. So as I reduce the voltage, the amps are reduced, but the resistance stays the same...right?

I have a little problem figuring ohm's law from one side of the transformer to the other. Can somebody dial me in?

SMF

The resistance of nichrome wire changes as it heats up. You really have to measure the current and applied voltage to see what the resistance is.

Let's start at the nichrome wire. With 1.5 ohms and 4.5 V, the current is I=V/R=4.5 V/1.5 ohm=3 A. The power is P=I*V=3 A*4.5 V=13.5 W. (You can also do P=V^2/R=(4.5 V)^2/1.5 ohm = 13.5 W.)

What about the primary side of the 12V transformer? If we assume an ideal transformer to make the calculations easy, power in = power out, so the input is also 13.5 W. Since the voltage is 45 V, you can determine the current by I=P/V=13.5 W/45 V=0.3 A.

The same holds true for the variac. If the input voltage is 120 V, the current is I=P/V=13.5 W/120 V=0.1125 A.

All of this is ignoring transformer losses and assumes the power factor is 1.0 for the entire system.

Brian

So I could apply a 90 amp load on the secondary (12 volt) side and it wouldn't blow my 10 amp fuse on the variac...right?

Right. Of course, this assumes the transformers are capable of that current.

Thanks Larry, that helps a lot.

SMF