Calculating the length of a wire

two wires A and B made from two different materials have temperature coefficient of resistance equal to 0.0025 and 0.0005 ohm per degree Celsius respectively. It is desired to make a coil of wire having a resistance of 1200 ohms with a temperature coefficient of 0.001, using a suitable length of the two given wires connected in series. Determine the required length of wire A. ?

please help me solve this problem

I know these relationships

\[\frac{R1}{R2} = \frac{(T+t1)}{(T+t2)}\] ----->

\[{\Delta}_{ t} = t2-t1\]

but they don't seem to be useful for this problem
regards!

We know that the resisto of a wire can be calculated from its resistivity "ρ", cross-section "S" and length "L" by the formula:

R=ρ*L/S

furthermore the temperature coefficient "k" is so that:

R=Ro*(1+k*ΔT) where Ro is the resistance calculated at the reference temperature To

In you case you know you want as a result of the series of two wires a resistance Rto=1200 ohm with coefficient kt=0.001 then:

Rt=Rto*(1+kt*ΔT)

the two wires in series will have a total resistance of:

Rs=ρa*La/Sa*(1+ka*ΔT)+ρb*Lb/Sb*(1+ka*ΔT)

then we want Rt=Rs that is: Rto*(1+kt*ΔT)=ρa*La/Sa*(1+ka*ΔT)+ρb*Lb/Sb*(1+ka*ΔT)

we can set ΔT=0 to obtain Rto=ρa*La/Sa+ρb*Lb/Sb

then developing the original equation:

Rto+Rto*kt*ΔT=ρa*La/Sa+ρa*La/Sa*ka*ΔT+ρb*Lb/Sb+ρb*Lb/Sb*kb*ΔT

using the previous one (that calculated for ΔT=0) we have:

Rto*kt*ΔT=ρa*La/Sa*ka*ΔT+ρb*Lb/Sb*kb*ΔT

from wich ΔT can be eliminated. At the end we have the two equation:

Rto=ρa*La/Sa+ρb*Lb/Sb
Rto*kt=ρa*La/Sa*ka+ρb*Lb/Sb*kb

from the first:

ρb*Lb/Sb=Rto-ρa*La/Sa

substituting:

Rto*kt=ρa*La/Sa*ka+(Rto-ρa*La/Sa)*kb

that is

Rto*(kt-kb)=ρa*La/Sa*(ka-kb)

then La= Rto*(kt-kb)*Sa/[ρa*(ka-kb)]

so you have to know also "Sa" and "ρa" not given in your original text.

Please check I didn't do mistakes in the calculations.

if you have 1 metre of of the .0025 and N metres of .0005, then to get .001, you need Z metres of .0025 and ZN metres of .005 and this equals (Z+ZN ) .001. So sorting out terms
Z.0025 -Z.001 = -ZN .0005 + ZN .001
or Z.0015 = ZN .0005
Therefore N = 3,
Lets check
1m of .0025 + 3m of 0005 = 4m at .0040 or .001/m
QED
Frank

chuckey said:

if you have 1 metre of of the .0025 and N metres of .0005, then to get .001, you need Z metres of .0025 and ZN metres of .005 and this equals (Z+ZN ) .001. So sorting out terms
Z.0025 -Z.001 = -ZN .0005 + ZN .001
or Z.0015 = ZN .0005
Therefore N = 3,
Lets check
1m of .0025 + 3m of 0005 = 4m at .0040 or .001/m
QED
Frank

Click to expand...

How can you be sure the resistance of the series is 1200 ohms ?

You can not, what I have found is the ratios of the lengths required. if you tell us what the absolute resistivity is of each material, then we can calculate the actual length. BUT remember the ratio of 3:1 to get the temperature coefficient correct i.e. replace metres with ohms, so you need 300 ohms worth of the .0025 and 900 ohms worth of the .0005.
Frank