Two wires of the same dimensions but resistivities `rho_(1) and rho_(2)` are connected in series. The equivalent resistivity of the combination is

Options

A. `(rho_(1)+rho_(2))/(2)`
B. `rho_(1)+rho_(2)`
C. `2(rho_(1)+rho_2)`
D. `sqrt(rho_(1)rho_(2))`

Answer

A

Explanation

`R_(1)=rho_(1)(l)/(A),R_(2)=rho_(2)(l)/(A)`
`R=R_(1)+R_(2)`
`=rho_(1)(l)/(A)+rho_(2)(l)/(A)=(rho_(1)+rho_(2))(l)/(A)` . . (i)
For equivalent resistance
`R=rho*(2l)/(A)` . .. (ii)
From equations (i) and (ii), `2rho=rho_(1)+rho_(2)`
`rho=(1)/(2)(rho_(1)+rho_2)`