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Conductivity and Resistivity

Two conductor...

Question

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 parallel to the contact surface such that the temperature difference across both the faces of both the conductors are same.

A

\frac{(d_{1} + d_{2}) (K_{1} d_{2} + K_{2} d_{2})}{2 (K_{1} + K_{2})}

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B

\frac{(d_{1} - d_{2}) (K_{1} d_{2} + K_{2} d_{2})}{2 (K_{1} + K_{2})}

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C

\frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}

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D

\frac{K_{1} + K_{2}}{d_{1} + d_{2}}

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Solution

The correct option is C $\frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}$
Let $t$ be the width and $L$ be the length of each conductor.

Effective thermal resistance (parallel combination) is,
$\frac{1}{R_{e q .}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$R_{e q .}$ = $\frac{R_{1} R_{2}}{R_{1} + R_{2}} . . . . . . . (i)$

where $R = \frac{L}{K A}$ , $K =$ thermal conductivity

substituting in Eq. (i),

$\frac{L}{K_{e q .} (A_{1} + A_{2})} = \frac{⎛ ⎝ \frac{L}{K_{1} A_{1}} ⎞ ⎠ ⎛ ⎝ \frac{L}{K_{2} A_{2}} ⎞ ⎠}{\frac{L}{K_{1} A_{1}} + \frac{L}{K_{2} A_{2}}}$

$\Rightarrow \frac{L}{K_{e q} (A_{1} + A_{2})} = (\frac{L^{2}}{K_{1} A_{1} \times K_{2} A_{2}}) \times (\frac{(K_{1} A_{1}) (K_{2} A_{2})}{L (K_{1} A_{1} + K_{2} A_{2})})$

$\Rightarrow \frac{1}{K_{e q} (A_{1} + A_{2})} = \frac{1}{K_{2} A_{2} + K_{1} A_{1}}$

$\Rightarrow K_{e q} = \frac{K_{1} A_{1} + K_{2} A_{2}}{A_{1} + A_{2}}$

$\Rightarrow K_{e q} = \frac{K_{1} (d_{1} t) + K_{2} (d_{2} t)}{(d_{1} t) + (d_{2} t)}$

$∴ K_{e q} = \frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}$

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