Three rods AB, BC and AC having thermal resis | vedclass

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(A) The thermal circuit can be analyzed by considering the thermal resistances. The rods $AB$ and $BC$ are in series with each other,and this combinat

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The thermal circuit can be analyzed by considering the thermal resistances. The rods $AB$ and $BC$ are in series with each other,and this combination is in parallel with rod $AC$.Thermal resistance of the series branch ($AB$ and $BC$) is $R_1 = 10 + 10 = 20 	ext{ units}$.Thermal resistance of the parallel branch $(AC)$ is $R_2 = 20 	ext{ units}$.The equivalent thermal resistance $R_{eq}$ between points $A$ and $C$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.Thus,$R_{eq} = 10 	ext{ units}$.The total rate of heat flow from $A$ to $C$ is $\frac{dH}{dt} = \frac{\Delta T}{R_{eq}} = \frac{100 - 0}{10} = 10 	ext{ units}$.Since the branch $AB+BC$ has a resistance of $20 	ext{ units}$ and the branch $AC$ has a resistance of $20 	ext{ units}$,the total heat flow splits equally between the two branches.Therefore,the rate of heat flow through the junction $B$ (which is part of the $AB+BC$ branch) is $\frac{10}{2} = 5 	ext{ units}$.

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