Two wires of equal diameters, of resistivities `rho_(1)` and `rho_(2)` and length `l_(1)` and `l_(2)`, respectively of the combination is

Options

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

Answer

A

Explanation

(a) `R_(1) = (rho_(1) l_(1))/(A)` and `R_(2) = (rho_(2) l_(2))/(A)`.In series `R_(eq) = R_(1) + R_(2)`
`(rho_(eq).(l_(2) + l_(2)))/(A) = (rho_(1) l_(1))/(A) + (rho_(2) l_(2))/(A) implies rho_(eq) = (rho_(1) l_(1) + P_(2) l_(2))/(l_(1) + l_(2))`\