Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the effective conductivity of the combination is

Options

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

Answer

C

Explanation

As both metal wires are of identical dimesnions, so their length and area of cross-section will be same. Let them be l and A respectively. Then
The resistance of the first wire is `R_(1)=(l)/(sigma_(1)A)" "…(i)`
and that of the second wire is `R_(2)=(l)/(sigma_(2)A)" "...(ii)`

As they are connected in series, so their effective resistance is `R_(s)=R_(1)+R_(2)`
`=(l)/(sigma_(1)A)+(l)/(sigma_(2)A)=(l)/(A)((1)/(sigma_(1))+(1)/(sigma_(2)))" "...(iii)` (using (i) and (ii))
If `sigma_(eff)` is the effective conductivity of the combination, then `R_(s)=(2l)/(sigma_(eff)A)" "...(iv)`
Equating eqns. (iii) and (iv), we get
`(2l)/(sigma_(eff)A)=(l)/(A)((1)/(sigma_(1))+(1)/(sigma_(2)))`
`implies (2)/(sigma_(eff))=(sigma_(2)+sigma_(1))/(sigma_(1)sigma_(2))impliessigma_(eff)=(2sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))`