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) =(rho x_(1))/A, R_(2) =(rhox_(2))/A_(2), R=(rhox)/A`
where `x=x_(1) +x_(2), R = R_(1) + R_(2)`
`(rho x)/A=(rho_(1)x_(1))/A + (rho_(2)x_(2))/A, rho x = rho_(1)x_(1) + rho_(2)x_(2)`
`rho(x_(1)+x_(2)) =rho_(1)(x_(1) + rho_(2)x_(2)) [therefore x=x_(1)=x_(2)]`
`therefore rho=(rho_(1)x_(1) + rho_(2)x_(2))/(x_(1)+x_(2))` also `1=(x_(1)/1 + x_(2)/2)/(x_(1) +x_(2))`