Two rods of copper and aluminium of each of equal length 20 cm and equal cross-sectional area `2 mm^(2)`. They are joined (i) in series and (ii) in parallel as shown in figure. Find the resistance of the combination in each case. Resistivity of copper `=1.7 xx10^(-8) Omega m` and resistivity of aluminium `= 2.6 xx 10^(-8)Omega m`.

Explanation

For copper rod, `l=20 cm=0.20 m`,
`A=2 mm^(2)=2xx10^(-6) m^(2), rho =1.7 xx10^(-8)Omega m`
`R_(1) = rho l/A=((1.7 xx10^(-8)) xx 0.20)/(2xx10^(-6))= 1.7 xx10^(-3)Omega`
For aluminium rod, l =20 cm =0.20 m,
`A= 2mm^(2) = 2xx10^(-6) m^(2), rho= 2.6 xx 10^(-8)Omega m`
`R_(2)=((2.6 xx 10^(-8))xx0.20)/(2xx10^(-6)) = 2.6 xx 10^(-3) Omega`
(i) When two rods are joined in series, their equivalent resistance,
`R = R_(1) +R_(2) =1.7 xx 10^(-3) +2.6 xx10^(-3)`
`~~ 4.3 xx 10^(-3)Omega`
(ii) When rods are in parallel, the equicvalent resistance is
`R =(R_(1)R_(2))/(R_(1)+R_(2))=((1.7 xx 10^(-3)) xx(2.6 xx 10^(-3)))/((1.7 xx 10^(-3)) + (2.6 xx 10^(-3)))`
`=(1.7 xx 2.6 xx 10^(-3))/4.3 =1.028 xx 10^(-3)Omega`
`=1.028 mOmega`