P28_025.PDF

25. Let r be the resistance of each of the narrow wires. Since they are in parallel the resistance R of the

composite is given by

R = 9

or R = r/ 9. Now r =4 ρ/πd 2 and R =4 ρ/πD 2 ,where ρ is the resistivity of copper. A = πd 2 / 4was

used for the cross-sectional area of a single wire, and a similar expression was used for the cross-sectional

area of the thick wire. Since the single thick wire is to have the same resistance as the composite,

4 ρ

πD 2 =

4 ρ

9 πd 2

⇒

D =3 d.