I can't seem to heat up some nichrome wire I got from a toaster. I don't really know what the gauge is, but the wire diameter is a little less than a millimeter, and it's flat.

I have 3 options of a power supply:

  • 9 volts, 1 amp
  • 19.5 volts, 3.34 amps
  • 3 D batteries (about 7.1 amps and 4.45 volts)

I've tried all of these, but none of them even get the wire hot. I cut off 3 cm of the wire. I don't know how long to make it for it to work with one of the power supply options.

All I'm trying to accomplish is to heat up the wire at least a little.

  • \$\begingroup\$ You can measure the wire's resistance to find out. Note that the wire might be coated with a high-temperature insulator. \$\endgroup\$ Commented Jun 11, 2023 at 23:04
  • \$\begingroup\$ Even if there is no insulator, it's hard to terminate to nichrome. Are you sure you're making contact to it? \$\endgroup\$
    – Hearth
    Commented Jun 11, 2023 at 23:07
  • \$\begingroup\$ I'm just setting the nichrome on top of copper wire. do you think that may be the problem? \$\endgroup\$ Commented Jun 11, 2023 at 23:08
  • 1
    \$\begingroup\$ @MarstheLimit That will absolutely be the problem. If you short the nichrome with copper, you won't get enough current through the nichrome to heat it. \$\endgroup\$
    – Hearth
    Commented Jun 11, 2023 at 23:20
  • 6
    \$\begingroup\$ Keep in mind that a typical toaster is rated at somewhere around 1000 to 1500 W, which at a line voltage of 120 V, means that there's about 10A flowing. So regardless of the length of the nichrome, you're going to have to put about that much current through it to get a similar temperature! \$\endgroup\$
    – Dave Tweed
    Commented Jun 11, 2023 at 23:39

1 Answer 1


There's a disclaimer at the end of this answer, but please forgive me if I start with one too: I don't accept any responsibility for injury or fire caused by doing anything I say here. There are hot things, and you can get burned, or start a fire. If that happens, it's on you.

Find out the resistance-per-metre of the wire. Then you can do some math to figure this out. First, whatever length of wire you have, measure its resistance end-to-end. Then divide this resistance by the total length (in metres), and you have resistance-per-meter, which I will call \$K\$.

I estimate (can't know for sure) that a reasonable target might be 0.5W of power per centimetre of wire. That's enough to get it hot enough to feel very hot (and maybe burn the skin), but not to set fire to anything (unless you coil up a long section, and concentrate all the energy into a small space - don't do that).

In SI units, 0.5W/cm is 50W/m, so that's your goal. This value I will call Q, which is therefore \$Q=50W/m\$.

From this point on, I'll assume your wire to have \$K=10\Omega /m\$. You should of course substitute your own measured value for \$K\$.

Using the power law \$P=I^2R\$ we can work out what current will be required to deliver \$Q\$ watts to resistance \$K\$, and determine which sources can actually do the job:

$$ I = \sqrt{\frac{P}{R}} = \sqrt{\frac{50W/m}{10\Omega /m}} = 2.2A $$

Immediately it's clear which of your three sources can work. The 9V source is the only one incapable of providing the necessary 2.2A, and I won't consider it any further. The other two should be fine.

Now we have another question: what length of wire would dissipate exactly \$Q\$ watts of power in every metre of that length, given a certain voltage \$V\$ across it?

Total power \$P\$ dissipated in the entire length \$L\$ of the wire will be:

$$ P = Q \times L $$

Total resistance \$R\$ of a wire of length \$L\$ is:

$$ R = K\times L $$

Using the voltage variant of the power law:

$$ \begin{aligned} P &= \frac{V^2}{R} \\ \\ &= \frac{V^2}{KL} \\ \\ QL &= \frac{V^2}{KL} \\ \\ L^2 &= \frac{V^2}{QK} \\ \\ L &= \frac{V}{\sqrt{QK}} \\ \\ \end{aligned} $$

Let's apply this equation practically, using the two voltage sources that are capable (2nd and 3rd), again assuming your wire has \$K=10\Omega /m\$. The appropriate lengths will be:

$$ L_2 = \frac{V_2}{\sqrt{QK}} = \frac{19.5V}{\sqrt{50W/m \times 10\Omega/m}} = 0.87m $$

$$ L_3 = \frac{V_3}{\sqrt{QK}} = \frac{4.45V}{\sqrt{50W/m \times 10\Omega/m}} = 0.20m $$

You don't need to cut the wire, just apply a voltage between two points separated by 0.87m (for the 19.5V source) or 0.2m (for the 4.5V source).

I reiterate that your own value for \$K\$ is going to be different, and the figures I calculated here are valid only for \$K=10\Omega /m\$.

Also, you don't have to use \$Q=50W/m\$. Setting \$Q=10W/m\$ will produce some heating, but it might be barely noticeable, especially considering that the wire will be cooled by any air currents, and radiating away a good deal of that energy as infra-red. I'm just guessing here, though. Probably you should start with \$Q=10W/m\$, and work your way up.

I consider \$Q=100W/m\$ (that's 1W per centimetre) to be on the very high side. I suspect that at this power, the wire will get hot enough that touching it will burn the skin.

You have been warned, and I don't accept any responsibility for injury or fire caused by following this guide.


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