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
$V=\frac{V_1\left(\frac{L_2}{{\sigma }_{2}A}\right)+V_2\left(\frac{L_1}{{\sigma }_{1}A}\right)}{\frac{L_1}{{\sigma }_{1}A}+\frac{L_2}{{\sigma }_{2}A}}=\frac{V_1{\sigma }_1{L}_2+V_2{\sigma }_2{L}_1}{{\sigma }_2{L}_1+{\sigma }_1{L}_2}$
$V=\frac{\frac{V_1{\sigma }_1}{L_1}+\frac{V_2{\sigma }_2}{L_2}}{\frac{{\sigma }_2}{L_2}+\frac{{\sigma }_1}{L_1}}$