Question

Question: Nine copper wires of length land diameter dare connected in parallel to form a single composite conductor of resistance R. What must be the diameter Dof a single copper wire of length lif it is to have the same resistance?

Short answer

Answer

The diameter for the single copper wire is D = 3d.

Step-by-step solution

Write the given data

1. The length of each resistance isl
2. The diameter of each resistance isd

Determine the concept for resistance

Use the formula for parallel arrangement of resistances to find the equivalent resistance and the equation of resistance related with the length, the resistivity and the area of cross-section of wire.

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} \\ R=\frac{\rho L}{A} \end{gathered}$$

Calculate the diameter D of a single copper wire of length l if it is to have the same resistance 

As all nine resistances are in parallel. So, the equivalent resistance is given as follow:

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \\ \frac{1}{R_{eq}}=\frac{9}{R} \end{gathered}$$

So,

$R_{eq}=\frac{R}{9}$

But we know that the resistance of copper wire is as follow:

$R=\frac{\rho L}{a}$

and

$R_{eq}=\frac{\rho L}{A}$

Rewrite the equation for resistance as:

$\frac{\rho L}{A}=\frac{\rho L}{9a}$

So,

$A=9a$

Since, the length and the resistivity both are same.

$\frac{\pi D^2}{4}=\frac{9\times \pi d^2}{4}$

So, resolve for the diameter as:

$$\begin{gathered} D^2=9d^2 \\ D=3d \end{gathered}$$