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?
Electronic – Electrical resistance of two conductors of equal surface area in contact with another
conductorsresistanceresistivity
Related Solutions
4.2 microns, not 4.2 microns squared. Area is not a unit of thickness. Divide your answer by pi, take the square root, and multiply by 2 to yield diameter
26 AWG is 40 ohms / 1000 foot as per this table: -
A 50 foot length of this wire (2 ohms) with 8A will dissipate 128 watts i.e. it doesn't sound a sensible choice. The volt drop at 8A will be 16 volts.
As for your formula I can't tell whether it agrees or disagrees with whatever reference you have.
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Best Answer
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.