Six wires,each of cross-sectional area A and | vedclass

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(D) Let $R_1$ and $R_2$ be the thermal resistances of the copper and iron wires respectively. Then,$R_1 = \frac{l}{K_1 A}$ and $R_2 = \frac{l}{K_2 A}$

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$\frac{2l}{(K_1+K_2)A}$

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(D) Let $R_1$ and $R_2$ be the thermal resistances of the copper and iron wires respectively. Then,$R_1 = \frac{l}{K_1 A}$ and $R_2 = \frac{l}{K_2 A}$.From the figure,the arrangement is a Wheatstone bridge network. The arms $AB$ and $BC$ are copper (resistance $R_1$ each),and $AD$ and $DC$ are iron (resistance $R_2$ each). The central arm $BD$ consists of two wires (one iron and one copper) in parallel,but due to the symmetry of the bridge,the potential (temperature) at $B$ and $D$ is the same. Thus,no heat flows through the central arm $BD$,and it can be removed.The circuit simplifies to two parallel branches: one branch with two copper wires in series $(R_1 + R_1 = 2R_1)$ and one branch with two iron wires in series $(R_2 + R_2 = 2R_2)$.The equivalent thermal resistance $R_{eq}$ between $A$ and $C$ is given by:$R_{eq} = \frac{(2R_1)(2R_2)}{2R_1 + 2R_2} = \frac{4R_1 R_2}{2(R_1 + R_2)} = \frac{2R_1 R_2}{R_1 + R_2}$.Substituting the values of $R_1$ and $R_2$:$R_{eq} = \frac{2 \left( \frac{l}{K_1 A} \right) \left( \frac{l}{K_2 A} \right)}{\frac{l}{K_1 A} + \frac{l}{K_2 A}} = \frac{2 \frac{l^2}{K_1 K_2 A^2}}{\frac{l(K_1 + K_2)}{K_1 K_2 A}} = \frac{2l}{A(K_1 + K_2)}$.

Six wires,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:

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