# Calculating Copper Wire Gauge for Resistance Thermometer Bridge Circuit

• calcphonup
In summary: It is correct to say α=0.0042. Δt=(202-200)=2. R_t = 112Ω at 20°C. Punching in the figures = 0.9408Ω. 0.9408Ω/250 = 0.0037632Ω/m^-1. From properties of resistance table, SWG12 is too low, so......you should be using SWG16 instead.

#### calcphonup

a. The resistance thermometer bridge circuit shown in FIGURE 1 has a designed maximum temperature of 200°C, ignoring the effects of connecting wire resistance. If the connecting loop is 250 m determine the smallest gauge (swg) of copper wire which must be used if the indicated maximum temperature is to be less than 202°C.

The temperature coefficient of resistance of the thermometer is 0.0042 °C–1 and the resistance of the thermometer is 112 Ω at 20°C. Assume the connecting leads are at 20°C.

b. Using the gauge of copper wire calculated in part (a), calculate the maximum indicated temperature using a 3-wire system (as shown in FIGURE 2) over a distance of 125 m.Relevant EquationsRt = R0(1 + at)

R1RT = R2RS

R = pL/A = Resistance = (Resistivity) x Length of wire / cross-sectional area

A = (pi x d2) / 4

d = squareroot(4A /pi)

R_t=R_20×(1+α×∆t)
α=0.0042
R20 = 112Ω
∆t=202-20=182
R_t=R_20×(1+α×∆t)
R_t=112×(1+0.0042×182)
R_t= 197.6128

As both resistors on figure 1 are 150Ω, they cancel each other out.
Find the resistivity of copper at 20°C

ρ=RA÷L
Where:
R = Resistance of copper at 20°C
A = area
L = length
From the properties of resistance wires table:
When diameter = 0.376mm, resistance = 0.155m-1
0.376mm diameter = 0.11103645mm2 for the area.
Converting resistance to per mm = 0.000155m-1
For length 1m = 1000mm.

Therefore, ρ=0.000155×0.111036450÷1000=1.72×10^(-8)
Resistivity of copper wire at 20°C = 1.72×10^(-8)Ωm-1
To find the area the equation would be A=(ρ×L)/R
A=(1.72×10^(-5) Ωmm×250000mm)/197.6128Ω
A = 0.02176mm2

Find diameter = 0.16645mm

Too small of a diameter.

Any help would be greatly appreciated. Ive been stuck for a week

Moderator note: Moved from a technical forum.

Last edited by a moderator:
Hello @calcphonup ,

For the bridge circuit you have a relationship , I suppose that's what you mean with R1RT = R2RS ?

Where in your essay is ##R_4## ? Is the value given ?

Your calculation gives me the impression that you start out with a diameter and then calculate another diameter. What is the intention ?

It's really worthwhile to learn some ##\LaTeX## (link at lower left of the edit box). ##R_1 R_t = R_2 R_S## looks a lot better than R1RT = R2RS But you want to at least use subscripts (toolbar) R1 Rt = R2RS. No clue what ##R_S## is ...
You sure this is the right equation for the circuit shown ? Shouldn't it have ##R_4## somewhere ?

##\ ##

In other words R_s is the position of the slidewire tapper – is a measure of the resistance (and therefore the temperature) of the resistance thermometer. So for figure 2, R_S = R_4. However, I have not made it passed part (a) of the question yet where I am only using figure 1. I am not sure which route I should be going down. Should I be trying to find a diameter and reference the table to find my SWG or should I be trying to find the Ωm-1 of copper at 20degrees and then using the table?

calcphonup said:
I have not made it passed part (a) of the question
Does this mean that you finished part (a) or that you are still stuck in part (a)
I am under the impression that you are still stuck, right ?

OK, so for figure 1 you have an equilibrium condition ##R_1\,R_t = R_2\,R_s## and with ##R_1=R_2## this becomes ##R_s = R_t##.

The information
calcphonup said:
has a designed maximum temperature of 200°C, ignoring the effects of connecting wire resistance
can then be used to find ##R_s = 200 \ \Omega## as you did. The maximum value of ##R_S##, when the slider is at the top end, that is.
(or ##198 \ \Omega## if you think 2.5 digit accuracy is justified -- I would certainly agree). Never ever omit units ! and do the calculations with ##197.6128 \ \Omega##).

I must admit that I find it hard to interpret the exact problem statement in part (a) (is this the exact formulation ?) because with such a value of ##R_S## the indicated temperature is never more than 200 °C.

The way I would compose this exercise would be something like 'determine the smallest gauge ... if the indicated temperature is less than 202 °C when the actual temperature of the thermistor is 200 °C'

Then the 250 m of connecting wire should have a resistance of less than ##\alpha \,\Delta T\, R_t = 0.9408 \ \Omega ## and that lets you find a gauge.

---

Moving on to figure 2 (and to complicate things ##R_s=R_4## ) is now in order. What is the equilibrium condition ?

##\ ##

BvU said:
Does this mean that you finished part (a) or that you are still stuck in part (a)
I am stuck on part (a) as you correctly assumed.

BvU said:
Then the 250 m of connecting wire should have a resistance of less than αΔTRt=0.9408 Ω and that lets you find a gauge.
So is it correct to say α=0.0042. Δt=(202-200)=2. R_t = 112Ω at 20°C. Punching in the figures = 0.9408Ω. 0.9408Ω/250 = 0.0037632Ω/m^-1. From properties of resistance table, SWG12 is too low, so SWG 14 must be used? Is that correct?

BvU said:
Moving on to figure 2 (and to complicate things Rs=R4 ) is now in order. What is the equilibrium condition ?
As far as part (b) goes, I am not sure where to start? Other than the total resistance of the wire is 0.00532Ω/m(from the table)*125 = 0.665Ω. However, that is at 20 degrees and the resistance will increase with the temperature.

BvU said:
hard to interpret the exact problem statement in part (a) (is this the exact formulation ?)
...
Moving on to figure 2 ... What is the equilibrium condition ?

Wrt #5: I'm not clearvoyant and not the exercise composer. It's your exercise, so your judgment prevails. I do agree with your calculations.

Wrt #6:
calcphonup said:
Assume the connecting leads are at 20°C.
##\ ##