Calculate Diameter of Single Copper Wire for Same Resistance as 9 Wires

[Physicsisfun2005]

9 copper wires of length L and diameter d are connected in parallel to form a single composite conductor of resistance R. What must be the diameter of a single copper wire of length L if it is to the the same resistance?

obviously since copper is the only thing involved i know the resistivity is the same given the same distance

R=(resistivity)L/(Area)...i believe this formula may help me...not sure

 

[maverick280857]

For parallel combination, $$\frac{1}{R_{eq}} = \sum \frac{1}{R_{i}}$$ where $$R_{i} = \rho \frac{L}{A}$$. Express A as a function of diameter.

 

[wais]

how to use it or if its the right one...

To calculate the diameter of a single copper wire for the same resistance as 9 wires, we can use the formula for resistance in a parallel circuit: Rt = 1/(1/R1 + 1/R2 + 1/R3 + ... + 1/Rn), where Rt is the total resistance and R1-Rn are the individual resistances of the wires. Since we want the total resistance to be the same, we can set Rt = R and solve for R1-Rn. In this case, we have 9 wires so n = 9. We also know that the resistivity is the same for all the wires, so we can write R1 = R2 = ... = Rn = R. Substituting this into the formula, we get R = 1/(1/R + 1/R + ... + 1/R) = 1/(9/R), which simplifies to R = R/9. Now, we can use the formula for resistance (R = (resistivity)L/A) to solve for the area A. Plugging in the known values, we get R = (resistivity)L/A = (resistivity)L/(pi*d^2/4), where d is the diameter of the single wire. Rearranging for d, we get d = sqrt(4*R/(pi*L*resistivity)). Therefore, the diameter of the single copper wire should be sqrt(4*R/(pi*L*resistivity)) to have the same resistance as 9 wires.