In a composite rod,two rods of different leng | vedclass

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(C) For rods connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.The for

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$\frac{\ell_1}{K_1} + \frac{\ell_2}{K_2}$

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(C) For rods connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.The formula for thermal resistance is $R = \frac{\ell}{KA}$,where $\ell$ is length,$K$ is thermal conductivity,and $A$ is the cross-sectional area.Substituting this into the series equation:$\frac{\ell_1 + \ell_2}{K A} = \frac{\ell_1}{K_1 A} + \frac{\ell_2}{K_2 A}$Since the cross-sectional area $A$ is the same for both rods,we can cancel $A$ from both sides:$\frac{\ell_1 + \ell_2}{K} = \frac{\ell_1}{K_1} + \frac{\ell_2}{K_2}$

In a composite rod,two rods of different lengths and the same cross-sectional area are joined end to end. If $K$ is the equivalent coefficient of thermal conductivity of the composite rod,then $\left( \frac{\ell_1 + \ell_2}{K} \right)$ is equal to:

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