Two metal wires of identical dimensions are connected in series. If and are the conductivities of the metal wires respectively, the effective conductivity of the combination is [2015]

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Solution

Q.
Two metal wires of identical dimensions are connected in series. If $σ_{1}$ and $σ_{2}$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is [2015]

1 $\frac{σ_{1} + σ_{2}}{σ_{1} σ_{2}}$

2 $\frac{σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

3 $\frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

4 $\frac{σ_{1} + σ_{2}}{2 σ_{1} σ_{2}}$

Ans.
(3)

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

$R_{1} = \frac{l}{σ_{1} A}$ ...(i)

and that of the second wire is $R_{2} = \frac{l}{σ_{2} A}$ ...(ii)

As they are connected in series, their effective resistance is

   $R_{s} = R_{1} + R_{2} = \frac{l}{σ_{1} A} + \frac{l}{σ_{2} A}$ (using (i) and (ii))

   $= \frac{l}{A} (\frac{1}{σ_{1}} + \frac{1}{σ_{2}})$ ...(iii)

If $σ_{eff}$ is the effective conductivity of the combination, then

   $R_{s} = \frac{2 l}{σ_{eff} A}$ ...(iv)

Equating eqns. (iii) and (iv), we get

$\frac{2 l}{σ_{eff} A} = \frac{l}{A} (\frac{1}{σ_{1}} + \frac{1}{σ_{2}});$ $\frac{2}{σ_{eff}} = \frac{σ_{2} + σ_{1}}{σ_{1} σ_{2}} or σ_{eff} = \frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

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