Question 93 of 126

Two metallic wires of identical dimensions are connected is series. If ${\sigma }_{1}$ and ${\sigma }_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is :

A$\dfrac{{\sigma }_{1}{\sigma }_{2}}{{\sigma }_{1}+{\sigma }_{2}}$B$\dfrac{2{\sigma }_{1}{\sigma }_{2}}{{\sigma }_{1}+{\sigma }_{2}}$C$\dfrac{{\sigma }_{1}+{\sigma }_{2}}{2{\sigma }_{1}{\sigma }_{2}}$D$\dfrac{{\sigma }_{1}+{\sigma }_{2}}{{\sigma }_{1}{\sigma }_{2}}$

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Answer: (B) $\dfrac{2{\sigma }_{1}{\sigma }_{2}}{{\sigma }_{1}+{\sigma }_{2}}$

Let length of wire be $l$,
Area of wire as $A$
For equivalent wire length $=2l$ & area will be $A$
As the wires are connected in series combination, the thermal resistance
${R}_{eq}={R}_{1}+{R}_{2}$
$\Rightarrow \dfrac{2l}{{\sigma }_{eq}A}=\dfrac{l}{{\sigma }_{1}A}+\dfrac{l}{{\sigma }_{1}A}$
$\Rightarrow \dfrac{2}{{\sigma }_{eq}}=\dfrac{1}{{\sigma }_{1}}+\dfrac{1}{{\sigma }_{2}} \Rightarrow {\sigma }_{eq}=\dfrac{2{\sigma }_{1}{\sigma }_{2}}{{\sigma }_{1}+{\sigma }_{2}}$