Question

Two copper wire A and B of length 30m and 10m have radii 2cm and 1cm respectively. Compare the resistance of the two wires. Which will have less resistance?

Answer 1

Step 1: Given that:

Length of copper wire A ($l_A$) = $\text{30m}$

Radius of wire A ($r_A$) = $2cm=2\times {10}^{-2}m$

length of copper wire B($l_B$) = $10m$

Radius of copper wire B($r_B$) = $1cm={10}^{-2}m$

Step 2: Comparison of the resistances of both wires A and B:

The resistivity of a conductor is given by;

$R=\frac{\text{ρ}l}{A}$

$\text{ρ}=\frac{RA}{l}$

Where R is the resistance of the conductor, l is the length and A is the area of the cross-section of the conductor.

Since both wires, A and B are made of copper, their resistivity will be the same.

${\text{ρ}}_A={\text{ρ}}_B$

$\frac{R_A{A}_A}{l_A}=\frac{R_B{A}_B}{l_B}$

Putting the given values we get;

$\frac{R_A\times \pi \times {(2\times {10}^{-2}m)}^2}{30m}=\frac{R_B\times \pi \times {\left({10}^{-2}m\right)}^2}{10m}$

$\frac{R_A}{R_B}=\frac{30}{10\times 4}$

$\frac{R_A}{R_B}=\frac{3}{4}$

Step 3: Determination of the maximum resistance between A and B:

Since; $\frac{R_A}{R_B}=\frac{3}{4}$

Thus,

$R_A=R_B\frac{3}{4}$

$R_B=\frac{4}{3}{R}_A$

$R_B=1.11R_A$

That is;

$R_B>R_A$

Thus, wire B has the maximum resistance.

Thus,

$\frac{R_A}{R_B}=\frac{3}{4}$

And, wire B has the maximum resistance.