Two wires of equal diameters of resistivies `rho_(1)` and `rho_(2)` and lengths l and l respectively are joined in series. The equivalent resistivity 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))/(I_(1) - I_(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

`R_(eq) = R_(1) + R_(2) = (rho_(1)l_(1))/A + (rho_(2)l_(2))/A`
`rho = (rho_(1)l_(1)+rho_(2)l_(2))/(l_(1) + l_(2))`