For the shown figure,calculate the equivalent | vedclass

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(B) The thermal resistance of a material is given by $R = \frac{l}{KA}$,where $l$ is the length,$A$ is the area,and $K$ is the thermal conductivity.$1

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$\frac{3l}{{KA}}$

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(B) The thermal resistance of a material is given by $R = \frac{l}{KA}$,where $l$ is the length,$A$ is the area,and $K$ is the thermal conductivity.$1$. For the first section (left): There are three parallel bricks each of length $l$ and area $A/3$. The resistance of each brick is $R_1 = \frac{l}{K(A/3)} = \frac{3l}{KA}$. Since they are in parallel,the equivalent resistance $R_{eq1}$ is $\frac{1}{R_{eq1}} = \frac{1}{R_1} + \frac{1}{R_1} + \frac{1}{R_1} = \frac{3}{R_1} = \frac{3}{3l/KA} = \frac{KA}{l}$. Thus,$R_{eq1} = \frac{l}{KA}$.$2$. For the middle section: There is a single brick of length $l$ and area $A$. The resistance is $R_{eq2} = \frac{l}{KA}$.$3$. For the third section (right): There are two parallel bricks each of length $l$ and area $A/2$. The resistance of each brick is $R_3 = \frac{l}{K(A/2)} = \frac{2l}{KA}$. Since they are in parallel,the equivalent resistance $R_{eq3}$ is $\frac{1}{R_{eq3}} = \frac{1}{R_3} + \frac{1}{R_3} = \frac{2}{R_3} = \frac{2}{2l/KA} = \frac{KA}{l}$. Thus,$R_{eq3} = \frac{l}{KA}$.$4$. Since these three sections are in series,the total equivalent thermal resistance is $R_{total} = R_{eq1} + R_{eq2} + R_{eq3} = \frac{l}{KA} + \frac{l}{KA} + \frac{l}{KA} = \frac{3l}{KA}$.

For the shown figure,calculate the equivalent thermal resistance if the bricks are made of the same material of thermal conductivity $K$.

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