Question 26.25: A wire with a resistance of 6.0 Ω is drawn out through a die......
A wire with a resistance of 6.0 Ω is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.
THINK The resistance of an object depends on its length and the cross-sectional area.
EXPRESS Since the mass and density of the material do not change, the volume remains the same. If L_0 is the original length, L is the new length, A_0 is the original cross-sectional area, and A is the new cross-sectional area, then L_0A_0 = LA and
A=L_0 A_0 / L=L_0 A_0 / 3 L_0=A_0 / 3 .
ANALYZE The new resistance is
R=\frac{\rho L}{A}=\frac{\rho 3 L_0}{A_0 / 3}=9 \frac{\rho L_0}{A_0}=9 R_0,
where R_0 is the original resistance. Thus, R = 9(6.0 Ω) = 54 Ω.
LEARN In general, the resistances of two objects made of the same material but different cross-sectional areas and lengths may be related by
R_2=R_1\left(\frac{A_1}{A_2}\right)\left(\frac{L_2}{L_1}\right) .