# Current through Mains Cable

Discussion in 'Electrical Engineering' started by David Skinner, Sep 30, 2006.

1. ### David SkinnerGuest

Hi, quick question if you guys don't mind...

If a piece of mains cable can carry 240VAC at 13A safely, could the same
cable carry 12VDC at 25A safely?

Just looking to muck around with a couple of bits of kit out of a car -
it's not for a permanent installation anywhere. I just don't want it to
catch fire.

2. ### woodglassGuest

No, the quoted voltage rating of a cable is the maximum voltage that can be
applied between it's cores (AC or DC).

The quoted current rating is the maximum current that each core can carry.

So, your 240V, 13A cable would be safe to use at 24V, but only up to a
maximum current of 13A

woodglass...

3. ### Andrew GabrielGuest

Cable may not be up to the degree of vibration, chaffing, heat,
and oil/fuel contamination that might be expected of cable
used in cars.

4. ### ehsjrGuest

Power dissipated in the wire: P = I^2R
where I is the current and R is the resistance
of the wire. Notice how you don't need voltage
to arrive at the answer, when you know I.

Therefore, a wire rated at no more than 13 amps
cannot carry 25 amps, regardless of voltage.

Ed

5. ### David SkinnerGuest

Thanks for the replies everyone - It's nice to drop-in on a newsgroup
which still values helpfulness and facts. Looks like I either need to
double-up the conductors or go out and buy fatter cable.

6. ### Guest

| David Skinner wrote:
|> Hi, quick question if you guys don't mind...
|>
|> If a piece of mains cable can carry 240VAC at 13A safely, could the same
|> cable carry 12VDC at 25A safely?
|>
|> Just looking to muck around with a couple of bits of kit out of a car -
|> it's not for a permanent installation anywhere. I just don't want it to
|> catch fire.
|
| Power dissipated in the wire: P = I^2R
| where I is the current and R is the resistance
| of the wire. Notice how you don't need voltage
| to arrive at the answer, when you know I.

Indeed. The conditions of resistance and current _define_ the voltage
that is across the wire over that distance.

7. ### Guest

| In article <[email protected]>, says...
|
|> Power dissipated in the wire: P = I^2R
|> where I is the current and R is the resistance
|> of the wire. Notice how you don't need voltage
|> to arrive at the answer, when you know I.
|>
|> Therefore, a wire rated at no more than 13 amps
|> cannot carry 25 amps, regardless of voltage.
|
| Thanks for the replies everyone - It's nice to drop-in on a newsgroup
| which still values helpfulness and facts. Looks like I either need to
| double-up the conductors or go out and buy fatter cable.

I don't know about UK, but in the US, doubling up the conductors is not
permitted unless the conductor size is already rather large (well beyond
the 13 amps level).

If ring circuits are still permitted in the UK, maybe you can set that
up if it's not aready so.

8. ### Andrew GabrielGuest

Not very relevant to OP's situation, but since you asked...
It is permitted in the UK, but the extra protection required to
ensure each conductor isn't overloaded (generally a multi-pole
ganged breaker at both ends of the cable) means it's rarely
economic to use it just because you have lots of undersized
cable around.
A ring circuit doesn't count under our Conductors in Parallel
regs, as the cable size used is significantly larger than
that required just to share the max load in a parallel cable
run.