Courses for Kids
Free study material
Offline Centres

Two metal wires of identical dimensions are connected in series. If ${\sigma _1}$and ${\sigma _2}$​ are the conductivities of the metal wires respectively, the effective conductivity of the combination is:
(A) $\dfrac{{{\sigma _1}{\sigma _2}}}{{{\sigma _1} + {\sigma _2}}}$
(B) $\dfrac{{2{\sigma _1}{\sigma _2}}}{{{\sigma _1} + {\sigma _2}}}$
(C) $\dfrac{{{\sigma _1} + {\sigma _2}}}{{2{\sigma _1}{\sigma _2}}}$
(D) $\dfrac{{{\sigma _1} + {\sigma _2}}}{{{\sigma _1}{\sigma _2}}}$

Last updated date: 29th Feb 2024
Total views: 21k
Views today: 0.21k
Rank Predictor
21k+ views
Hint When two conductors are connected in series then their resistance gets added and becomes effective resistance of arrangement i.e. if two resistors of resistance ${R_1}$and ${R_2}$are connected in series then the effective resistance of combination will become \[{R_1} + {R_2}\]. Using this relation for effective resistance, we can find the effective conductivity.
Formula used
1. $R = \dfrac{L}{{\sigma A}}$ (R is the resistance, L is the length of the conductor, A is the cross-sectional area of the conductor,$\sigma $ is the conductivity of the conductor)
2. \[{R_{eff}} = {R_1} + {R_2}\] (${R_1}$and ${R_2}$ are the resistance of the wire in series combination and ${R_{eff}}$ is the effective resistance of the system.)

Complete Step-by-step solution
As the two metal wires are of identical dimension hence their length and the cross-sectional area are the same let their length be L and cross-sectional area be A. They are connected in series hence the effective resistance is the sum of their resistance.
Let, their resistance be ${R_1}$of wire with conductivity ${\sigma _1}$and ${R_2}$ wire with conductivity${\sigma _2}$ , and the effective resistance be ${R_{eff}}$and it’s effective conductivity be${\sigma _{eff}}$.
\[ \Rightarrow {R_{eff}} = {R_1} + {R_2}\]
$ \Rightarrow \dfrac{{L + L}}{{{\sigma _{eff}}A}} = \dfrac{L}{{{\sigma _1}A}} + \dfrac{L}{{{\sigma _2}A}}$
Length is taken as $L+L$ or $2L$ because the two wires are now connected and the length of both the wires together will contribute to the effective resistance.
\[ \Rightarrow \dfrac{2}{{{\sigma _{eff}}}} = \dfrac{1}{{{\sigma _1}}} + \dfrac{1}{{{\sigma _2}}}\]
$ \Rightarrow {\sigma _{eff}} = \dfrac{{2{\sigma _1}{\sigma _2}}}{{{\sigma _1} + {\sigma _2}}}$
$\therefore $The answer is option(B) ${\sigma _{eff}} = \dfrac{{2{\sigma _1}{\sigma _2}}}{{{\sigma _1} + {\sigma _2}}}$

Note If in question it had been mentioned that the wires are connected in parallel then we had to proceed towards the answer differently in that case reciprocal of effective resistance will be equal to the sum of reciprocal of individual resistance of wires i.e. $\dfrac{1}{{{R_{eff}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$. So one must carefully read the question before answering it and take care of whether the wire is connected in series or parallel as such small mistakes could make the solution incorrect and students might end up losing marks.