Two wires of equal diameters of resistivities `rho_(1)` and `rho_(2)` and length `x_(1)` and `x_(2)` respectively are joined in series. Find the equivalent resistivity of the combination.

Explanation

Resistance `R_(1)=(rho_(1)l_(1))/(A_(1)),R_(2)=(rho_(2)l_(2))/(A_(2))`
`l_(1)=x_(1),l_(2)=x_(2)`
As the wires are of equal diameters `A_(1)=A_(2)=A`.
`R_(1)=(rhox_(1))/(A),R_(2)=(rhox_(2))/(A_(2)),R=(rhox)/(A)`
where `x=x_(1)+x_(2),R=R_(1)+R_(2)`
`(rhox)/(A)=(rho_(1)x_(1))/(A)+(rho_(2)x_(2))/(A),rhox=rho_(1)x_(1)+rho_(2)x_(2)`
`rho(x_(1)+x_(2))=rho_(1)x_(1)+rho_(2)x_(2)[becausex=x_(1)+x_(2)]`
`thereforerho=(rho_(1)x_(1)+rho_(2)x_(2))/(x_(1)+x_(2)) also (1)/(sigma)=((x_(1))/(sigma_(1))+(x_(2))/(sigma_(2)))/(x_(1)+x_(2))`