Increasing wire gauge by twisting pairs

Question

I have a cable like this with 4 wires inside. Each wire is 20 AWG.

I've been told that I can twist the ends together (i.e. green and red, white and black) on both ends of the cable and this will effectively increase the wire gauge.

This would make perfect sense to me if the wires were naked (no green/red/white/black plastic coat around it) so it would be copper twisted the full length instead of just the ends... does the coat affect the current? I'm not an electrician or EE so I'm not too sure about this.

I want to use this for sprinkler valves.

Answer 1

If you twist two wires together, each would carry half the current, so you'd "effectively increase the gauge." American Wire Gauges go down by about 10 for every factor of ten in cross-sectional area. If you had ten #20 wires connected in parallel, they could carry as much power as one #10 wire. With two #20 wires, you'd have the equivalent of one #17 wire.

(A handy "rule of thumb" value: #40 copper wire has about an Ohm of resistance for each foot. By the rule above, #30 would have an Ohm for every ten feet, and #20 an Ohm for every 100 feet.)

Note that connecting wires in parallel may work at DC or low frequency AC. For audio, RF, or other purposes, you'd just mess up the wire characteristics, and cause yourself problems.

Answer 2

Doubling the conductors has the effect of reducing the equivalent AWG by a factor of 3 (as in 1x 20AWG = 20 AWG; 2x 20 AWG = 17 AWG equivalent).

In order for the effect to continue with additional cable conductors, doubling is required each time (eg 2x 20 AWG = 17 AWG equivalent, to go down (larger) another 3 AWG would require doubling your 17 AWG equivalent once more; ie 4x 20 AWG = 14 AWG equivalent, 8x 20 AWG = 11 AWG equivalent; to go down another 3 AWG equivalent now requires 16 conductors, then 32, and so on).

Whether it's advisable to create a substitute cable in this way is debatable, and dependent on application, (it affects certain peramaters such as cable capacitance, inductance, and in AC frequency {audio, Radio, digital} applications may cause smearing of the signal if each conductor is not the exact same length physically and electrically), but if you want to use an existing installed cable, it may be adequate.

Bundling of cable (what you are effectively doing) reduces the heat dissipation capacity of the cable, so some safety factor should be considered (ie don't use a 2x 20 AWG cable for DC power if the application is power and requires a minimum or specified 17 AWG, but if the requirement is greater than 17 AWG but less than 20 AWG it might be OK, or if heat is not normally an issue, such as loudspeaker cable, heat dissipation issues can be ignored). Again, all depending on application.

Answer 3

On a uniform wire the resistance is defined as: R=(p*L)/A

p = Resistivity of material

L = length of wire

A = area of cross section

Increasing the area by twisting several wires together makes area larger = the resistance smaller. As the power dissipated by a resistance (your wire in this case) can be calculated with:

Power=Current^2 * Resistance That means that lowering the resistance will make your wire less hot. If you have too high resistance the wire gets hot, it might burn off or burn soemthing else, in worst case cause fire. If the current is small, probably the most annoying effect will be that at the end of the wire the voltage you put in will have decreased because of the resistance.

I don't know if a sprinkler in this case means a fire extinguishing sprinkler or just a gardening sprinkler. If it is for safety equipment I would surely get the correct wire gauge from the start, to not risk that a twisted wire gets "untwisted" so the sprinkler won't be able to operate.

Answer 4

It'll effectively "increase the gauge" by lowering the resistance of each pair of two paralleled conductors to 1/2 the resistance of a single conductor.

From the table below, 20 AWG has a resistance of 10.13 ohms per thousand feet, so 1/2 of that would be 5.06 ohms per thousand feet, which corresponds, roughly, to 17 AWG at 5.054 ohms per thousand feet.