2. Your boss asks you to construct a special resistor made of two metals. Metal 1 is iridium and metal 2 is uranium. The iridium is 25 cm long with a square cross section of 3.0 mm on a side. The uranium is 40 cm long, also with a square cross section of 3.0 mm on a side. The two metals are fused together at one of their 3.0 x 3.0 $mm^2$ ends, making a conductor that has a

total length of 65 cm.

(a) What is the electrical resistance measured between the ends of the composite rod, under the assumption that no additional resistance occurs at the point where the rods are fused‘?

Answer

$R= \rho*(L/S)$,

R is resistance, L is length, S is section

$S(uranium) = S(iridium) = 3*3 =9 (mm^2) = 9*(10^{-6}) (m^2)$

$\rho(uranium) =0.280 *(10^{-6}) ohm*m$

$\rho(iridium) = 0.05*(10^{-6}) ohm*m$

$R(uranium) = (0.28*10^{-6})*0.4/(9*10^{-6}) = 12.44*10^{-3}$ ohms

$R(iridium) = (0.05*10^{-6})*0.25/(9*10^{-6}) =1.39*10^{-3}$ ohms

Because the wires are linked one after the other the two resistances are is series

$R = R(uranium)+R(iridium) = 13.83*10^{-3}$ ohms