Two wires of equal length and equal diameter and having resisticities `rho_(1) and rho_(2)` are connected in series. The equivalent resistivity of the combination is .... .

Options

A. `(rho_(1) + rho_(2)) `
B. `(rho_(1) rho_(2))/(rho_(1) + rho_(2))`
C. `(rho_(1) + rho_(2))/(2) `
D. `sqrt(rho_(1) rho_(2))`

Answer

C

Explanation

`(rho_(1) + rho_(2))/(2) `
R = `R_(1) + R_(2) `
`(rho (l + l))/(A) = (rho_(1)l)/(A) = (rho_(2)l)/(A)`
`2 rho = rho_(1) + rho_(2)`
`therefore rho = (rho_(1) + rho_(2))/(2) `