Question

Two metal wires of identical dimensions are connected in series. If $\sigma_1$ and $\sigma_2$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is

• A. $\frac{\sigma_1+\sigma_2}{\sigma_1\sigma_2}$
• B. $\frac{\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
• C. $\frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
• D. $\frac{\sigma_1+\sigma_2}{2\sigma_1\sigma_2}$

Answer 1

Answer: (C)

$\frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$

Answer 2

As both metal wires are of identical dimensions, 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
$R_1=\frac{l}{\sigma_1 A}$ ...(i)
and that of the second wire is
$R_2=\frac{l}{\sigma_2 A}$ ...(ii)

As they are connected in series, so their effective
resistance is
$R_s = R_1+R_2$
$=\frac{l}{\sigma_2 A} +\frac{l}{\sigma_2 A} \, \, \,$ (using (i) and (ii))
$=\frac{1}{A}\bigg(\frac{1}{\sigma_1} +\frac{l}{\sigma_2}\bigg)$ ...(iii)
If a $\sigma _{eff}$ is the effective conductivity of the combination, then
$R_s=\frac{2l}{\sigma_{eff}A}$ ...(iv)
Equating cqns. (iii) and (iv), we get
$\frac{2l}{\sigma_{eff}A}=\frac{l}{A}\bigg(\frac{l}{\sigma_1} +\frac{l}{\sigma_2}\bigg)$
$\frac{2l}{\sigma_{eff}}=\frac{\sigma_2+\sigma_1}{\sigma_1\sigma_2}\, \, \, \sigma_{eff}=\frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$