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(B) The given circuit is a Wheatstone bridge. Since the bridge is balanced (the ratio of resistances in the arms is equal),no heat flows through the c

Vedclass · Accepted Answer

$\frac{2l}{(k_1 + k_2)A}$

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(B) The given circuit is a Wheatstone bridge. Since the bridge is balanced (the ratio of resistances in the arms is equal),no heat flows through the central wire. The circuit simplifies to two parallel branches: $1$. The upper branch consists of two copper wires in series,each of length $l$. The total resistance is $R_1 = \frac{l}{k_1 A} + \frac{l}{k_1 A} = \frac{2l}{k_1 A}$. $2$. The lower branch consists of two steel wires in series,each of length $l$. The total resistance is $R_2 = \frac{l}{k_2 A} + \frac{l}{k_2 A} = \frac{2l}{k_2 A}$. Since these two branches are in parallel,the equivalent thermal resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{k_1 A}{2l} + \frac{k_2 A}{2l} = \frac{(k_1 + k_2)A}{2l}$. Therefore,$R_{eq} = \frac{2l}{(k_1 + k_2)A}$.

Five wires each of cross-sectional area $A$ and length $l$ are combined as shown. The thermal conductivity of copper and steel are $k_1$ and $k_2$ respectively. The equivalent thermal resistance between $A$ and $C$ is

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