A copper wire and an iron wire, each having an area of cross-section A and lengths L₁ and L₂ are joined end to end. The copper end is maintained at a potential V1 and the iron end at a lower potential V2. If $σ_{2}$ and $σ_{1}$ are the conductivities of copper and iron respectively, then the potential of the junction will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{σ_{1} V_{1} + σ_{2} V_{2}}{(σ_{1} / L_{1}) + (σ_{2} / L_{2})}$

$\frac{\frac{σ_{1} V_{1}}{L_{1}} + \frac{σ_{2} V_{2}}{L_{2}}}{(σ_{1} / L_{1}) + (σ_{2} / L_{2})}$

$\frac{(σ_{1} / V_{1}) + (σ_{2} / V_{2})}{σ_{1} V_{1} + σ_{2} V_{2}}$

$\frac{σ_{1} V_{1} - σ_{2} V_{2}}{(σ_{1} / L_{1}) - (σ_{2} / L_{2})}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{V_{1} - V}{R (Copper)} = \frac{V - V_{2}}{R (Iron)} \Rightarrow \frac{V_{1} - V}{\frac{L_{1}}{σ_{1} A}} = \frac{V - V_{2}}{\frac{L_{2}}{σ_{2} A}} \\ \Rightarrow \frac{σ_{1} (V_{1} - V)}{L_{1}} = \frac{σ_{2} (V - V_{2})}{L_{2}} \end{array}$
$V = \frac{\frac{σ_{1} V_{1}}{L_{1}} + \frac{σ_{2} V_{2}}{L_{2}}}{(\frac{σ_{1}}{L_{1}} + \frac{σ_{2}}{L_{2}})}$