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how i calculate current of cable.as example-1 sq mm cable current is about 10 amp.how i calculate current of 95 sq mm cale etc....

2006-11-09 23:02:39 · 7 answers · asked by niranjan s 1 in Science & Mathematics Engineering

7 answers

You can find table ratings of cables in National Electrical Codes Book.
You will need to consider other factors like temperature ratings and voltage drop for length of cable.

2006-11-09 23:12:44 · answer #1 · answered by java 4 · 0 0

Yes, the IEE Regulations in the UK have charts of all cable sizes available. The voltage drop is expresses in Milli volts / Ampere /Metre, so as the cable gets longer or the current rise, the volt drop increases. It is limited to 2.5% of the stated supply voltage on most circuits. For the EHV cables, manufacturers data should also be consulted as the IEE Regulations do not cover this area.

2016-05-22 02:23:48 · answer #2 · answered by Anonymous · 0 0

There are tables for finding this out it also depends on whether the cable is enclosed and ambient temperature. Also I would not consider putting 10 amp through a 1 mm cable you need a larger gauge.

2006-11-09 23:21:54 · answer #3 · answered by deflagrated 4 · 0 0

DC Current will be linear with area. Therefore, if 1 mm2 carries 10 amps, 95 mm2 will carry 950 amps.

This ignores special effects if the current is AC. High frequency currents tend to move to the outside of the wire and the relation of current to area is no longer linear. This is called the Hall Effect.

2006-11-09 23:08:41 · answer #4 · answered by cfpops 5 · 0 0

You need to know the impedance of the cables, length of cables, the voltage and the load impedance before you can calculate the current

2006-11-10 00:07:42 · answer #5 · answered by Forgettable 5 · 0 0

Use this formula i=el/r ,where i=current,e=voltage or potential difference,r=internal resistance of cable and l=length of cable.
its simple heh...........simple

2006-11-10 12:32:09 · answer #6 · answered by simply the best. 2 · 0 0

OHMs law

2006-11-09 23:10:48 · answer #7 · answered by Anonymous · 0 0

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