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Resistance of a wire can be determined by the formula R = resistivity*(length/area). What if you have a wire composed of two parallel metals that are in contact with one another? Will the resistivity be a weighted average of the two values of each metal?

athedcha
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4 Answers4

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That one's a bit harder than it seems.

So, first of all, let's assume the conductivities of the two metals are in the same order of magnitude, so that a simplification that "all significant current is carried by the lower-resistance wire" isn't justified.

In a first approach, you could model the wires as sum of small resistors that are contacted regularly:

       x·₁   x·₁   x·₁  x·₁
A ---+-[===]-+-[===]-+-[===]-+-[===]-+--
     |       |       |       |       |
B ---+-[===]-+-[===]-+-[===]-+-[===]-+--
       x·₂   x·₂  x·₂   x·₂

With conductor A having the specific resistance (ohm per meter) ₁, and B having ₂. Each resistor than has a resistance x· ; x is the small wire length between contacts.

We set the total length \$X= N\cdot\Delta x\$. A single "parallel element" has the resistance

$$\begin{align} R_{\Delta x} &= (\Delta x\cdot\rho_1)||(\Delta x\cdot\rho_2)\\ &= \frac{\Delta x\cdot\rho_1\,\cdot\,\Delta x\cdot\rho_2}{\Delta x\cdot\rho_1+\Delta x\cdot\rho_2}\\ &=\frac{(\Delta x)^2\rho_1\rho_2}{\Delta x\cdot(\rho_1+\rho_2)}\\ &=\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2} \end{align}$$

The total resistance of A||B then becomes:

$$\begin{align} R_X &= N\cdot R_{\Delta x}\\ &= N\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2}&\text{with $N=\frac{X}{\Delta x}$:}\\ &= X\frac{\rho_1\rho_2}{\rho_1+\rho_2} \end{align}$$

In other words, it's the same formula as if you put two resistors in parallel, just applied to the specific resistances (times the length of the conductors).

Notice that this construct will also be subject to the Seebeck Effect, which means that if you have a temperature difference between the ends of your composite wire, you might see a current flowing – a small one.

Marcus Müller
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No. You have two parallel resistors.

$$R_1=\rho_1\cdot\frac{l_1}{A_1}$$

$$R_2=\rho_2\cdot\frac{l_2}{A_2}$$

Parallelizing resistors follows the formula

$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots$$

For two resistors, this can be rewritten as

$$R=\frac{R_1\cdot R_2}{R_1+R_2}$$

Putting in \$R_1\$ and \$R_2\$ from above, with \$l = l_1 = l_2\$ and \$A = A_1 = A_2\$

$$R=\frac{\rho_1\frac{l}{A}\cdot\rho_2\frac{l}{A}}{\rho_1\frac{l}{A}+\rho_2\frac{l}{A}}$$

$$\cdots$$

$$R=\frac{\rho_1\cdot\rho_2}{\rho_1+\rho_2}\cdot\frac{l}{A}$$

When \$A_1 \neq A_2\$ things get more complicated.

Janka
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  • Thanks for the help! What would change if A_1 does not equal A_2 as you said? – athedcha Mar 05 '17 at 23:55
  • You had to solve the second formula from below with A1 and A2, which makes it impossible to separate l/A from the rho part. You could only separate l. – Janka Mar 06 '17 at 00:31
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Assuming the two metals which make up the wire have a constant cross section, then you can calculate the resistance of the wire this way:

  • Work out the cross section of each metal.
  • Use the resistivity of that metal and the length of the wire to work out the resistance of the two metals as if they were separate wires.
  • The total resistance of the combined wire is the resistance of the two sub-wires in parallel.
Jack B
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Assuming you have two exactly parallel uniformly thick wires contacting each other:

There's no sideways current(* You can calculate the resistances separately and the combined resistance by R=R1*R2/(R1+R2). The equivalent resistivity is R*Area/Length

The result: The equivalent resistivity for the combination of geometrically equal wires is 2* r1*r2/(r1+r2) where r1 and r2 ase the individual resistivities. Note: we have doubled the area.

*) How can I know that? - Simply there's as many volts/millimeter in both wires. Theres in every point zero volts at the sideways direction.