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So today I was working on a DIY project, that involves me heating up a nichrome wire, and for some reason the nichrome wire is not heating up. I am using a 38 gauge wire, I calculated that I would only need 0.33 AMPS to heat up the wire, using 10inch of nichrome, 12V voltage source, as well as I have read 0.4 amp in my circuit. My circuit consists of the same nichrome wire, measured to be 31 ohms, and a 330 ohm resistor in series; so using current divider, I calculated 0.36 current should be going through the nichrome, but instead the resistor heats up and the nichrome is cold,any idea were I missed up. I can provide photo of my circuit if needed.

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  • \$\begingroup\$ Do you really mean "cold" wire ? If your 12 volts is holding up, that is 4.6 watts into the nichrome wire. It isn't going to glow red hot at that power. It does sound like your 12 volts is holding up if resistor gets hot (0.43 watts in resistor) \$\endgroup\$ – Marla Dec 13 '15 at 0:06
  • \$\begingroup\$ Cold as an not changing at all I have a temperature sensor and saw no changes and according to this nichrome calculator: jacobs-online.biz/nichrome/NichromeCalc.html I should be getting 440 degrees F with 3.8 Watts of power needed \$\endgroup\$ – AceSammy Dec 13 '15 at 0:11
  • \$\begingroup\$ So where is 4.6 watts going ? I guess that is the real question (if you are applying 12v @ 0.4 amps \$\endgroup\$ – Marla Dec 13 '15 at 0:12
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    \$\begingroup\$ Are you sure you haven't mixed up the concepts of serial and parallel? A resistor in parallel with your nichrome makes no sense at all, and should not give the results you saw for the given values unless your wire was open-circuit. One in series could make a tiny bit of sense but yours would be way too big for that, and it's probably not the ideal way to regulate current either. \$\endgroup\$ – Chris Stratton Dec 13 '15 at 0:24
  • \$\begingroup\$ Are you sure the resistor is in parallel with the wire? Why would you do that? If it were in series, it might explain things since the resistor dissipate over 10 times more than the wire. \$\endgroup\$ – Olin Lathrop Dec 13 '15 at 0:29
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$$V=IR$$

If your Nichrome wire is \$31\Omega\$, and you apply \$12\mathrm{V}\$, then you will have a current of \$387\mathrm{mA}\$ flowing through the wire.

If instead you place a \$330\Omega\$ resistor in series, you now have a resistance of \$330+31=361\Omega\$. So doing the calculation again, you now have only \$33.2\mathrm{mA}\$ flowing through the circuit. The vast majority of the voltage is now dropped over the resistor.

$$P=IV=I^2R$$

You have \$33.2\mathrm{mA}\$ flowing through your circuit, so the Nichrome wire is dissipating \$P=I^2R=33.2^2 \times 31= 34.2\mathrm{mW}\$.

On the other hand, the resistor is dissipating \$P=I^2R=33.2^2 \times 330= 364\mathrm{mW}\$. Basically 10 times as much.


So the question really is, why do you have the resistor?


If you instead place the resistor in parallel with the Nichrome wire, it would do nothing useful. The Nichrome branch of the circuit will still show \$387\mathrm{mA}\$ flowing. You will also have \$36\mathrm{mA}\$ flowing through the parallel branch of the resistor, but this wouldn't change the amount of heat generated in the Nichrome - for all intensive purposes the resistor is essentially a separate circuit sharing the same supply.


It seems that the issue is your supply cannot deliver enough power. The voltage of the supply is basically dropping away meaning that you are only getting \$\approx 160\mathrm{mA}\$ through the wire, less than half of what you need. Noting the square in the Power formula, this means that the power dissipated by the Nichrome is less than 20% of what it would be if the voltage had not dropped away. Try to find a supply which can deliver more power.

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  • \$\begingroup\$ thanks this really helped, got rid of the resistor and the wire still is cold so I'am thinking I am a transistor is going to be needed \$\endgroup\$ – AceSammy Dec 13 '15 at 1:25
  • \$\begingroup\$ @EngineerEE with the Nichrome wire connected, could you measure the voltage across it, make sure it is 12V (if the supply can't provide enough current, you'll see the voltage being lower than expected). \$\endgroup\$ – Tom Carpenter Dec 13 '15 at 1:28
  • \$\begingroup\$ Your right it measured only 5V across the nichrome wire \$\endgroup\$ – AceSammy Dec 13 '15 at 1:33
  • \$\begingroup\$ Okay my temperature sensor is reading a small temperature change, and I do read the voltage being lower so it definitely is a current issue, Thank you for your help you've been very helpful \$\endgroup\$ – AceSammy Dec 13 '15 at 1:39
  • \$\begingroup\$ Transistor? How are you wiring the circuit up? \$\endgroup\$ – Passerby Dec 13 '15 at 7:37
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nichrome wire not heating up

Without seeing your circuit connectivity, it isn't very easy to see if you messed up in any way. Your parallel circuit resistance is ~28.3Ω ((31Ω*330Ω)/(31Ω+330Ω)). At 12V, your intended circuit should draw ~.42A (which is close enough to what you stated that you measured).

According to numbers that you provided, current through your 31Ω nichrome wire should be ~.387A (i.e. 12V/31Ω).
Current through your 330Ω resistor should be ~.036A (12V/330Ω).

Assuming that you are using Nichrome 60 (Chromel C) wire (42.2Ω/ft), then your nichrome wire length should be ~22.4cm or 8.8" and it should produce a wire temp of ~293°C or ~560°F @ ~12V & ~.387A. The dissipated power through the wire should be ~4.65W (.387A^2*31Ω).
Ref: handy nichrome wire temp calculator here.

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