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Six wire each of cross-sectional area $A$ and length $l$ are combined as shown in the figure. The thermal conductivities of copper and iron are $K_1$ and $K_2$ respectively. The equivalent thermal resistance between points $A$ and $C$ is :-
$\frac{l(K_1+K_2)}{K_1K_2A}$
$\frac{2l(K_1+K_2)}{K_1K_2A}$
$\frac{l}{(K_1+K_2)A}$
$\frac{2l}{(K_1+K_2)A}$
Solution
Let $R_{1}$ and $R_{2}$ be the thermal resistance of copper and iron wires. Then
$\mathrm{R}_{1}=\frac{l}{\mathrm{K}_{1} \mathrm{A}}$ and $\mathrm{R}_{2}=\frac{l}{\mathrm{K}_{2} \mathrm{A}}$
According to the principle of Wheat stone's bridge, the point $\mathrm{B}$ and $\mathrm{D}$ must be at same temperature when the bridge is balanced. Therefore, thermal resistance of arm $BD$ becomes ineffective. Now the equivalent circuit at balance is
The effective resistance between $A$ and $C$ is
$\mathrm{R}=\frac{\left(2 \mathrm{R}_{1}\right)\left(2 \mathrm{R}_{2}\right)}{2 \mathrm{R}_{1}+2 \mathrm{R}_{2}}$
$=\frac{2 \mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
$\mathrm{R}=\frac{2 \frac{l}{\mathrm{K}_{1} \mathrm{A}} \cdot \frac{l}{\mathrm{K}_{2} \mathrm{A}}}{\frac{l}{\mathrm{K}_{1} \mathrm{A}}+\frac{l}{\mathrm{K}_{2} \mathrm{A}}}=\frac{2 l}{\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) \mathrm{A}}$
