# Resistors in Parallel (Not that easy)

Discussion in 'General Electronics Chat' started by Guest3123, Sep 24, 2016.

1. ### Guest3123 Thread Starter Member

Oct 28, 2014
334
18
So it's Rt = 1 / ((1/R1) + (1/R2))

That's very easy.

So if Rt = 1 / ((1/2.8Ω) + (1/2.8Ω))

Solve R1 : 1/2.8Ω = 0.35714285714285714285714285714286
Solve R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1/ (R1 + R2) = 1 / 0.71428571428571428571428571428571 = 1.4Ω

Rt = 1.4Ω
Easy right?

Resistors in series : Rt = R1 + R2, Done. Even a child could do the math.

But here's my problem.

Each one of those resistors are two more resistors in parallel.

So R1 has two resistors that make it up, in parallel.

R1 : (R1a & R1b)
R2 : (R2a & R2b)

2.8 x 2 = 5.6Ω
2.8 x 2 = 5.6Ω

Rt = 1 / ((1/5.6Ω) + (1/5.6Ω))

Solve R1 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R2 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R1 + R2 : 0.35714285714285714285714285714286
Solve 1 / (R1 + R2) = 2.8Ω

Easy right?

*** Here's the problem ***

36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.

Target Resistance 0.7 Ω
Target Wattage (Power) : 100 Watts.
Supply Voltage (Vs) : 8.4Vdc

It's a dual coil build, with a supply voltage of 8.4V, total resistance is 0.7Ω
0.7Ω x 2 coils = 1.4Ω each.

Rt = 1 / ((1/1.4Ω) + (1/1.4Ω))
R1 : 1/1.4 = 0.71428571428571428571428571428571
R2 : 1/1.4 = 0.71428571428571428571428571428571
R1 + R2 = 1.4285714285714285714285714285714
1 / (R1 + R2) = 0.7Ω

Ohm's Law : Watts (Voltage & Resistance) P = E² / R
8.4 x 8.4 = 70.56 / 0.7Ω = 100 Watts.

But each coil (1.4Ω) each, has two wires that make up the coil.

2.8 Ω per wire.

Rt = 1 / ((1/2.8) + (1/2.8))
R1 : 1/2.8Ω = 0.35714285714285714285714285714286
R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1 / (R1 + R2) = 1.4Ω Single Coil (with two wires).

The 36ga. get's wrapped on a straight piece of 24ga. : Clapton Coil

36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.

The 24ga is very long, if I cut it at 2.8Ω
25.4mm * 12 In. = 304.8mm / Ft.

36ga : 17.80Ω / 304.8mm = 0.0583989501 Ω/mm
24ga : 1.101Ω / 304.8mm = 0.0036122047 Ω/mm

Solve the length of 24ga. needed to hit 2.8Ω
2.8Ω / 0.0036122047 = 775mm

That's a very long piece of 24ga.

The 36ga is fine I guess.
2.8Ω / 0.0583989501 Ω/mm = 47.9mm

It can't be 2.8Ω for the 24ga, and It has to be 24ga.
How do I shrink the length of 24ga. without missing 2.8Ω

I don't know the math, maybe I should just leave it alone. Because I'm also afraid that I wont be able to handle the math, as simple as it might be.

Doing the above math is easy when the numbers are identical. But if I need to shrink the length of wire, keeping the 24ga. shorter than the 36ga. I can't wrap my head around it.

People make these coils, and there's no math involved.

Resistors inside Resistors that are all parallel.
I guess it would look like something like this as a circuit drawing.

Dual Clapton Coils Build Video.

The people that make these, DO NOT KNOW THE MATH. I do.. but as you saw from the math, the 24ga. that the 36ga. get's wrapped on, is way too long for a single coil. The 36ga. would never wrap on the whole thing, and on top of that, again, the 24ga. is too long to begin with.

Last edited by a moderator: Sep 25, 2016
2. ### AlbertHall Well-Known Member

Jun 4, 2014
2,278
450
I suspect that the 24ga wire is removed from the centre of the coil - it's just being used as a coil former - all that is actually used is the 36ga coil.

3. ### wayneh Expert

Sep 9, 2010
12,394
3,246
I think you're confused how these are made. The overall resistance is only slightly lower than that of the thicker center wire. The outer coil of narrow gauge wire provides surface area for heat transfer, but it carries relatively little current. It's much longer and has more ohms per unit length, so it has a much higher resistance. Adding a 50Ω length of thin wire to the 2Ω length of thick wire drops the total resistance very little.

There's no requirement that the thin wire have the same resistance as the fat one.

4. ### AlbertHall Well-Known Member

Jun 4, 2014
2,278
450
More looking: This video shows a similar process, but the final resistance is shown as 0.14Ω so wayneh is right.

5. ### Guest3123 Thread Starter Member

Oct 28, 2014
334
18
No, I thought that it was at 2nd glance, but in reality, a piece of resistance wire then inserted back into the coil.

As this video clearly shows.

Oct 28, 2014
334
18