Question

Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 times the diameter of wire 2. What is the ratio of the resistance of wire 1 to the resistance of wire 2

Answer 1

Answer: 1.63 : 1

Explanation:

let length of wire 2 be 10 m, so that length of wire 1 will be 13.5 m

Also, let the diameter of wire 2 be 10 m, so that diameter if wire 1 will be 9.1 m

Area, A = πd²/4, so that, area of wire 2 will be

A2 = π*10²/4

A2 = 25π

A1 = π*9.1²/4

A1 = 20.7025π

Since they are both made if the same material, we can agree that their resistivity is the same, thus,

ρ = R1*A1/L1 = R2*A2/L2

R1*A1*L2 = R2A2L1, so that

R1/R2 = A2L1/A1L2

R1/R2 = L1/L2 * A2/A1 when we substitute, we have

R1/R2 = 13.5/10 * 25π/20.7025π

R1/R2 = 337.5π / 207.025π

R1/R2 = 1.63 : 1

Answer 2

Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity $\rho$.

The cross-sectional area of the of a wire is given by;

$A=\pi\frac{d^2}{4}................(1)$

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

$\rho=\frac{RA}{l}..................(2)$

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

$\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)$

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

$\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)$

Given;

$l_1=1.35l_2$

$\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)$

We then use equation (1) to fine the ratio of the area $A_1$ to $A_2$

bearing in mind that $d_1=0.91d_2$

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

$\frac{R_1}{R_2}= 1.35*0.8281=1.117935$