Why do PT fuses have internal resistance?

[majesus]

Curiosity question: for PT primary fuses, say 4160V, current limiting, ferrule mounting similar to the fuses shown in the following specification sheet: Link
I noticed that when I measure the fuse resistance with an ohm meter I usually get around 30ohms for a good, non-blown fuse. For all other fuses, the resistance is measured at 0.1 ohm resistance. Why do PTs have a built in resistance? Does anyone know?

Thanks,
Majesus

 

[itsmoked]

Unless I'm way off base it's likely to limit inrush and fault current since a potential transformer should not draw much current in its normal operation. And since it doesn't draw much the 30 ohms is insignificant since there will be little voltage drop.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Skogsgurra]

It is more about Prospective Short Circuit Current I[sub]PSC[/sub] and the rupture capacity of the fuse.

It takes quite a fuse to break the short circuit current reliably. And that would mean a large fuse body. Which is impractical. So, by making the fuse 30 ohms, you can limit the short circuit current to less than 4200/30 = around 140 A (less actually since there will be two fuses in series if shorted line-line). And that can be handled by a small cartridge fuse.

Have a look at a fused security test lead. Same principle and about the same resistance range. That is why you should never try to measure low ohms using such cables. And never measure any substantial current with them, either.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

 

[pwrtran]

Another thought would be when differential protection is used it can limit the differential operating current below the minimum pickup value to avoid dumping the entire bus or transformer.

 

[Skogsgurra]

??? PT - not CT.

And, wouldn't any such manipulation be detrimental, inflexible, non-documented and very much against good pratice?

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[cranky108]

A fuse is melted by heat, and an internal resistance creates that heat. Because a PT normally draws so little current, the 30 ohms dosen't matter to the PT.

Most current limiting fuses that I know of, are full of sand, that when melted becomes glass. So I would assume it takes some heat to cause this melting.

Most other fuses that I know of are in air, and I would assume don't require much heat to melt.

 

[Skogsgurra]

Most fuses, current limiting or not, are sand-filled. It is only the small 5x20 mm and similar that are not.

The heat caused by a short-Circuit current is caused by I[sup]2[/sup]R and if R is low, I[sup]2[/sup] gets high. So rupture will always be initiated. If the I[sup]2[/sup] is too high, which it easily is in kV systems, the fuse will explode and initiate an arc.

The 30 ohms, keep current at a reasonable level so the rupture can take place without any explosion. And, as already said, 30 ohms don't make much difference in the primary circuit of a PT.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.