I have a faulty car (bad central locking engines) and a fuse is blowing every time I open/lock the it. I'm looking to do a quick fix, and just replace it with an automatic fuse. The car runs on ~12V, and the fuse is at 15 ampere.

Are fuses designed to operate on specific voltages, and have their ampere "tolerance" rated for the specific voltage?

As I understand it, a 15A fuse for 12V is designed to blow at 180W (the heat produced in the fuse wiring is what makes it melt, right?).

So, if I use an ordinary 16A/220V fuse in my car, it will blow at 3520W, and should really be rated 293A for a 12V circuit...

If my reasoning is right, that fuses blow at effect and not ampere, what I'm looking for is a 0.8A fuse that is designed for 220V (I=180/230), and it would work as a good replacement...?

  • 3
    \$\begingroup\$ Surely it would be better to get to the root of the fault and stop the fuse blowing in the first place...? A "quick" fix is seldom that quick in the long run. You could be causing more damage to the system by doing it. \$\endgroup\$
    – Majenko
    Commented Nov 4, 2011 at 10:17
  • \$\begingroup\$ True true, my plan right now is to reset the fuse every time I lock/unlock the car ;) \$\endgroup\$ Commented Nov 4, 2011 at 11:25

2 Answers 2


The 180W reasoning is wrong. You don't have the full 12V across the fuse, only a fraction of it. Say the fuse has a 0.1\$\Omega\$ resistance, then 15A will cause a power dissipation of

\$ P = I^2 \cdot R = (15A)^2 \cdot 0.1 \Omega = 22.5W \$.

And that 22.5W in the thin wire is just too much, and the fuse will blow.
Note that the 12V doesn't appear anywhere in this, and you could do the same calculation for 15A at 230V AC. There's a "but", however. Fuses do have a rated voltage. Not for the blowing, as we've seen, but to make sure the remains of the fuse won't cause sparks when at a high voltage. So you can't use a 12V car fuse in a mains appliance, but the other way around should be safe.

Automatic fuses also react on current only. They have a coil which releases a latch if the number of coil-windings exceeds the rated value. Also here it's independent of voltage.

  • \$\begingroup\$ Great answer, here is a fuse I found, but this is a thermic fuse: elfaelektronikk.no/elfa3~no_no/elfa/… . Does you answer still apply? \$\endgroup\$ Commented Nov 4, 2011 at 9:10
  • 1
    \$\begingroup\$ @dog - No these work differently from the automatic fuse I described. I'm not quite sure, but I think it uses a bimetal to release the latch, so working on the generated heat, rather than just the current. The power formula is still valid. \$\endgroup\$
    – stevenvh
    Commented Nov 4, 2011 at 9:15
  • \$\begingroup\$ Ok thanks, I will buy a thermic fuse and test it before I install it in my car. \$\endgroup\$ Commented Nov 4, 2011 at 9:47

Everything is wrong about this question. The purpose of fuses in automotive wiring harnesses is to prevent electric fires when some type of electrical fault causes abnormally high current to flow in the wire. If your fuse blows every time you lock/unlock the doors, then you have an electrical fault that is causing a high current.

You must fix the fault, not replace the fuse with a re-settable type. Also, thermic fuse is not what you want.

Normal fuses operate based on heating. High current causes the fuse to get hot, and eventually melt. Once it melts, it opens the circuit and current stops flowing. Make sure the fuse has a DC rating. If the fuse only has AC current ratings, then it may not be suitable for use in a DC system. DC systems are more prone to arcing when current is interrupted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.