Question

A copper wire and an iron wire, each having an area of cross-section $A$ and lengths $L_{1}$ and $L_{2}$ 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 $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of copper and iron respectively, the potential of the junction will be

• A. $\frac{\sigma_{1} V_{1}+\sigma_{2} V_{2}}{\frac{\sigma_{1}}{L_{1}}+\frac{\sigma_{2}}{L_{2}}}$
• B. $\frac{\frac{\sigma_{1} V_{1}}{L_{1}}+\frac{\sigma_{2} V_{2}}{L_{2}}}{\frac{\sigma_{1}}{L_{1}}+\frac{\sigma_{2}}{L_{2}}}$
• C. $\frac{\frac{\sigma_{1}}{L_{1}}+\frac{\sigma_{2}}{L_{2}}}{\sigma_{1} V_{1}+\sigma_{2} V_{2}}$
• D. $\frac{\sigma_{1} V_{1}-\sigma_{2} V_{2}}{\frac{\sigma_{1}}{L_{1}}-\frac{\sigma_{2}}{L_{2}}}$

Answer 1

Answer: (B)

Resistance of copper wire, $R_{1}=\frac{L_{1}}{\sigma_{1} A}$
Resistance of iron wire, $R_{2}=\frac{L_{2}}{\sigma_{2} A}$
Let $V$ is the potential at the junction.
As the wires are connected in series same current flows in the wires.

$\therefore \frac{V_{1}-V}{R_{1}}=\frac{V-V_{2}}{R_{2}}$
or $V_{1} R_{2}-V R_{2}=V R_{1}-V_{2} R_{1}$ or $V\left(R_{1}+R_{2}\right)=V_{1} R_{2}+V_{2} R_{1}$
or $V=\frac{V_{1} R_{2}+V_{2} R_{1}}{R_{1}+R_{2}}$
Substituting the values of $R_{1}$ and $R_{2},$ we get
$V=\frac{V_{1}\left(\frac{L_{2}}{\sigma_{2} A}\right)+V_{2}\left(\frac{L_{1}}{\sigma_{1} A}\right)}{\frac{L_{1}}{\sigma_{1} A}+\frac{L_{2}}{\sigma_{2} A}}=\frac{V_{1} \sigma_{1} L_{2}+V_{2} \sigma_{2} L_{1}}{\sigma_{2} L_{1}+\sigma_{1} L_{2}}$
$V=\frac{\frac{V_{1} \sigma_{1}}{L_{1}}+\frac{V_{2} \sigma_{2}}{L_{2}}}{\frac{\sigma_{2}}{L_{2}}+\frac{\sigma_{1}}{L_{1}}}$