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A copper wire and an iron wire, each having an area of corss-section A and lengths L_(1)andL_(2) are joined end to end. The copper end is maintained at a potential V_(1) and the iron end at a lower potential V_(2). If sigma_(1)andsigma_(2) are the conductivities of copper and iron respectively, the potential of the junction will be |
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Answer» `(sigma_(1)V_(1)+sigma_(2)V_(2))/((sigma_(1))/(L_(1))+(sigma_(2))/(L_(2)))` Resistance of iron wire, `R_(2)=(L_(2))/(sigma_(2)A)` Let V is the POTENTIAL at the JUNCTION. As the wires are connected in SERIES same CURRENT flows in the wires. `therefore (V_(1)-V)/(R_(1))=(V-V_(2))/(R_(2))` or `V_(1)R_(2)-VR_(2)=VR_(1)-V_(2)R_(1)ORV(R_(1)+R_(2))=V_(1)R_(2)+V_(2)R_(1)` or `V=(V_(1)R_(2)+V_(2)R_(1))/(R_(1)+R_(2))` Substituting the values of `R_(1)andR_(2)`, we get `V=(V_(1)((L_(2))/(sigma_(2)A))+V_(2)((L_(1))/(sigma_(1)A)))/((L_(1))/(sigma_(1)A)+(L_(2))/(sigma_(2)A))=(V_(1)sigma_(1)L_(2)+V_(2)sigma_(2)L_(1))/(sigma_(2)L_(1)+sigma_(1)L_(2))` `V=((V_(1)sigma_(1))/(L_(1))+(V_(2)sigma_(2))/(L_(2)))/((sigma_(2))/(L_(2))+(sigma_(1))/(L_(1)))` 
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