Two conductors having same width and length, thickness d_1 and d_2...

Two conductors having same width and length, thickness

d_1

and

d_2

, thermal conductivity

K_1

and

k_2

are placed one above the another. Find the equivalent thermal conductivity. Two conductors having same width and length, thickness d_1 and d_2 , thermal conductivity K_1 and k_2 are placed one

Two conductors having same width and length, thickness d_1 and d_2 , thermal conductivity K_1 and k_2 are placed one

Options

A $\frac{\left(d_1+d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}$
B $\frac{\left(d_1-d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}$
C $\frac{K_1 d_1+K_2 d_2}{d_1+d_2}$
D $\frac{K_1+K_2}{d_1+d_2}$

Correct answer

C. $\frac{K_1 d_1+K_2 d_2}{d_1+d_2}$

Step-by-step solution

Let

t

be the width and

L

be the length of each conductor. Effective thermal resistance,

R_p=\frac{R_1 R_2}{R_1+R_2}

\because \quad R=\frac{L}{K A}

where

k=

thermal conductivity

\frac{L}{A K_{e q}}=\frac{\frac{L}{K_1 A_1} \frac{L}{K_2 A_2}}{\frac{L}{K_1 A_1}+\frac{L}{K_2 A_2}}

\Rightarrow \frac{1}{K_{e q}\left(A_1+A_2\right)}=\frac{1}{K_2 A_2+K_1 A_1}

\Rightarrow K_{e q}=\frac{K_1 A_1+K_2 A_2}{A_1 A_2}

=\frac{K_1 d_1 t+K_2 d_2 t}{d_1 t+d_2 t}=\frac{K_1 d_1+K_2 d_2}{d_1+d_2}

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