Question

A composite slab is prepared by pasting two plates of thicknesses and \[{L_2}\]and thermal conductivities \[{K_1}\] and \[{K_2}\]. The slabs have equal cross sectional area. Find the equivalent conductivity \[{L_1}\] of the slab:
A) \[{K_{eq}} = \dfrac{{{L_1} + {L_2}}}{{\dfrac{{{L_1}}}{{{K_1}}} + \dfrac{{{L_2}}}{{{K_2}}}}}\]
B) \[{K_{eq}} = \dfrac{{{L_1} + {L_2}}}{{{K_1}{K_2}}}\]
C) \[{K_{eq}} = \dfrac{{{L_1}}}{{{K_1} + {K_2}}} + \dfrac{{{L_2}}}{{{K_1} + {K_2}}}\]
D) \[{K_{eq}} = \dfrac{{{L_1}{L_2}}}{{{L_1} + {L_2}}}\]

Answer 1

Hint:The degree to which any material conducts electricity is called conductivity. It is defined as the rate at which heat passes through any material per unit time through unit area at a specific temperature gradient of 1 degree. It is basically the ability of electrons to move freely. There are several factors on which the equivalent conductivity of a conductor depends like: Temperature of the material, number of free electrons, dimensions of that atom such as length, area and volume of the atom

Step-by-Step Explanation:
Step I:
Since the two resistors are connected in series and it is a series combination so\[{R_s} = {R_1} + {R_2}\]……(i)
Let \[{R_1}\]Be the resistance of the first slab and \[{R_2}\]be the resistance of the second slab.
\[R = \dfrac{L}{{KA}}\]…….(ii)
For first slab
\[{R_1} = \dfrac{{{L_1}}}{{{K_1}A}}\]…….(iii)
For second slab\[{R_2} = \dfrac{{{L_2}}}{{{K_2}A}}\]……(iv)
Step II: Substituting values in equation (i),
\[\dfrac{{{L_1} + {L_2}}}{{{K_s}A}} = \dfrac{{{L_1}}}{{{K_1}A}} + \dfrac{{{L_2}}}{{{K_2}A}}\]
Taking ‘A’ common from both sides, the above equation reduces to
\[\dfrac{{{L_1} + {L_2}}}{{{K_s}}} = \dfrac{{{L_1}}}{{{K_1}}} + \dfrac{{{L_2}}}{{{K_2}}}\]
\[\dfrac{{{L_1} + {L_2}}}{{{K_s}}} = \dfrac{{{L_1}{K_2} + {L_2}{K_1}}}{{{K_1}{K_2}}}\]
Step III: Evaluating value of \[{K_s},{K_s} = \dfrac{{({L_1} + {L_2})({K_1}{K_2})}}{{{L_1}{K_2} + {L_2}{K_1}}}\]
\[{K_s} = \dfrac{{{L_1} + {L_2}}}{{\dfrac{{{L_1}}}{{{K_1}}} + \dfrac{{{L_2}}}{{{K_2}}}}}\]
Hence option A is the correct answer.

Note:In case when the resistors are connected in series, there is only one path through which the electrons can pass or current can flow. Therefore the value of the current is the same at all points in the circuit when the resistors are connected in series. The equivalent resistance is the sum of all the resistors connected in series. But the voltage drop across the resistors in the circuit is not the same. This is because their individual resistances will result in varying voltage drop across each resistor. This voltage drop can be determined using Ohm’s Law.