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, \[\sigma =\frac{2{{\sigma }_{1}}.{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]You need to login to perform this action.
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