question_answer

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)  \[\frac{2{{\sigma }_{1}}{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]           

B)  \[\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{2{{\sigma }_{1}}{{\sigma }_{2}}}\]           

C)  \[\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{{{\sigma }_{1}}{{\sigma }_{2}}}\]             

D)  \[\frac{{{\sigma }_{1}}{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]

Correct Answer: A

Solution :

Net resistance of a metal wire having resistivity \[\rho \], we have \[{{R}_{1}}={{\rho }_{1}}\frac{L}{A}\] Similarly               \[{{R}_{2}}={{\rho }_{2}}\frac{L}{A}\] Then, net effective resistance of two metal wires, \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\]                 Þ          \[\rho \frac{2L}{A}={{\rho }_{1}}\frac{L}{A}+{{\rho }_{2}}\frac{L}{A}\]                 Þ           \[2\rho ={{\rho }_{1}}+{{\rho }_{2}}\] As, conductivity \[\sigma =\frac{1}{\rho },\], we have                                 \[\frac{2}{\sigma }=\frac{1}{{{\sigma }_{1}}}+\frac{1}{{{\sigma }_{2}}}\]                 Þ           \[\frac{2}{\sigma }=\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{{{\sigma }_{1}}.{{\sigma }_{2}}}\] Þ Net effective conductivity of combined wires, \[\sigma =\frac{2{{\sigma }_{1}}.{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]