Evaluating terminal current capacity.

  • Thread starter Adder_Noir
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  • #1
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Hi!

Just wondered if someone could help me. I'm trying to see if the following terminal can handle a higher than stated current at a lower than stated voltage.

http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=4458964"

It's rated to I think 24A at 500V. What I would like to know is how can I work out what it is rated to at 240V AC. I'd appreciate some help thanks :wink:
 
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Answers and Replies

  • #2
Averagesupernova
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Let me ask you this: Can you expect 20 amp fuse rated at 240 volts to carry twice the current at 120 volts?
 
  • #3
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My answer:

20 x 240 = 4800W

120 x 40 = 4800W

Given that a fuse rating and current capacity rating is based on the heat dissipated (i.e. work done on) the element/conductor then I would presume yes one can expect such a fuse to carry twice the current at 120V.

Sooooo

What you're saying is it's ok to evaluate the total power a terminal can stand and then evaluate your projected power using the P=IV equation and determine whether or not it falls within the total power rating? If so why don't manufacturer's simply state power capacity rather than specify a particular current and voltage?
 
  • #4
Averagesupernova
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Nope. You are incorrect. To determine the power dissipated in a device you need to multiply the current through the device by the voltage across the device. The fuse example does NOT have 120 or 240 volts across it. Neither will a connector. The fuse or connector may have 120 or 240 or whatever voltage from itself to another conductor, or chassis ground or whatever, but that is irrelevant. The voltage rating on a fuse or connector specifies how much voltage the insulation surround the conductor can withstand before breaking down. The current rating in a fuse or connector has nothing to do with the voltage rating. So what I'm telling you is that if you have a connector rated at 24 amps at some voltage, that connector is good for 24 amps, no more.
 
  • #5
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Ah I see of course. The fuse or connector terminals are at the same potential and thus it is not such a simple matter. The way I worked it out before would suggest that at maximum demand the fuse would have nearly 5kW of work being done on it to make it rupture which is nonsense as this is almost the equivalent of two kettles on full boil acting on a single fuse element.

Surely though the connector terminals or fuse element will have some impedance. I appreciate what you're stating about current flow but surely the magnitude of the voltage driving the current through the connector will have some affect on the work each electron does when passing through the connector and thus the amount of work done on the element/conductor in the form of heat will vary according to the driving voltage. Lower voltages will surely have less of a heating effect and will thus require larger currents to perform enough work on the conductor to make it rupture.

Is this not correct?
 
  • #6
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The driving voltage is largely irrelevant for a terminal, so long as it's below the rated maximum voltage.

The voltage drop across the terminal, which is generated by the current through the terminal, is what counts.

If you double the current through the terminal you will double the voltage across the terminal...

Which means that since P = I * V.... you'll get 4 times the rated power dissipated in the terminal.

So it'll get hot.

Very hot....

Can anyone smell smoke?
 
  • #7
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The driving voltage is largely irrelevant for a terminal, so long as it's below the rated maximum voltage.

The voltage drop across the terminal, which is generated by the current through the terminal, is what counts.

If you double the current through the terminal you will double the voltage across the terminal...

Which means that since P = I * V.... you'll get 4 times the rated power dissipated in the terminal.

So it'll get hot.

Very hot....

Can anyone smell smoke?
Interesting, thanks for answering that.

As for the comment highlighted in bold unfortunately for you I am savvy enough to understand exactly what it means and I would ask you to keep such comments to yourself in future. I thought this forum was supposed to attract people who were interested in the subject and whom are likely to ask questions. So in your case on this occasion there really is smoke without fire (see it's not that hard to be unpleasant in a clever fashion is it?).

The reason I'm asking is I'm trying to source a suitable replacement (at greater expense to myself) to the poor (although legal) standard of terminal blocks sold for the domestic electric market in the United Kingdom as I wish to distinguish myself infront of my customers by using higher standard materials than those simply sold off the wholesaler's shelf.

I hope that is quite clear. If you're struggling with any of the concepts described in this post do drop me a line and I'll explain them more thoroughly for you if needs be :smile:
 
  • #8
Averagesupernova
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Ah I see of course. The fuse or connector terminals are at the same potential and thus it is not such a simple matter. The way I worked it out before would suggest that at maximum demand the fuse would have nearly 5kW of work being done on it to make it rupture which is nonsense as this is almost the equivalent of two kettles on full boil acting on a single fuse element.

Surely though the connector terminals or fuse element will have some impedance. I appreciate what you're stating about current flow but surely the magnitude of the voltage driving the current through the connector will have some affect on the work each electron does when passing through the connector and thus the amount of work done on the element/conductor in the form of heat will vary according to the driving voltage. Lower voltages will surely have less of a heating effect and will thus require larger currents to perform enough work on the conductor to make it rupture.

Is this not correct?
The driving voltage to the network has nothing to do with how much current the connector or fuse will pass. Turning up the voltage to the network which in turn inceases current will most certainly cause more heating in the connector but no more heating than if we left the voltage the same and reduced the resistance elsewhere in the network in order to increase current flow.
 
  • #9
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Interesting, thanks for answering that.

As for the comment highlighted in bold unfortunately for you I am savvy enough to understand exactly what it means and I would ask you to keep such comments to yourself in future. I thought this forum was supposed to attract people who were interested in the subject and whom are likely to ask questions. So in your case on this occasion there really is smoke without fire (see it's not that hard to be unpleasant in a clever fashion is it?).

The reason I'm asking is I'm trying to source a suitable replacement (at greater expense to myself) to the poor (although legal) standard of terminal blocks sold for the domestic electric market in the United Kingdom as I wish to distinguish myself infront of my customers by using higher standard materials than those simply sold off the wholesaler's shelf.

I hope that is quite clear. If you're struggling with any of the concepts described in this post do drop me a line and I'll explain them more thoroughly for you if needs be :smile:
Carry on mate... if you're struggling with the concept of manufacturer's current ratings versus operating voltage, then maybe the 17th edition regs might be of more assistance to you.

P.S. The "smoke" comment was not a joke... putting too much current through a terminal is a recipe for disaster.
 
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  • #10
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Further reading has made it apparent that the "24A" current rating is related to conductor size, in this case 2.5 sq mm.

I would assume (since I've never seen one of these things) that you will be unable to insert a thicker wire into the clamp.

If you want 50A, you'll need to use 10sq mm cable with a different terminal... (This seems very large to me, but that's what the specs appear to say...).

The current carrying capacity of the said cable is, in turn, based on the heating effect at the stated current.
 

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