Two conducting rods A and B of same length an | vedclass

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Question

(A) The heat current $H$ is given by $H = \frac{\Delta 	heta}{R}$,where $R$ is the thermal resistance.For a rod,thermal resistance $R = \frac{l}{KA}$

Vedclass · Accepted Answer

$\frac{16}{3}$

Vedclass · Answer

(A) The heat current $H$ is given by $H = \frac{\Delta 	heta}{R}$,where $R$ is the thermal resistance.For a rod,thermal resistance $R = \frac{l}{KA}$.Let $R_A = \frac{l}{(3K)A}$ and $R_B = \frac{l}{KA}$.In series combination,the equivalent resistance is $R_S = R_A + R_B = \frac{l}{3KA} + \frac{l}{KA} = \frac{l + 3l}{3KA} = \frac{4l}{3KA}$.The heat current in series is $H_S = \frac{\Delta 	heta}{R_S} = \frac{\Delta 	heta \cdot 3KA}{4l}$.In parallel combination,the equivalent resistance is $R_P = \frac{R_A R_B}{R_A + R_B} = \frac{(\frac{l}{3KA})(\frac{l}{KA})}{\frac{l}{3KA} + \frac{l}{KA}} = \frac{\frac{l^2}{3K^2 A^2}}{\frac{4l}{3KA}} = \frac{l}{4KA}$.The heat current in parallel is $H_P = \frac{\Delta 	heta}{R_P} = \frac{\Delta 	heta \cdot 4KA}{l}$.The ratio of heat current in parallel to series is $\frac{H_P}{H_S} = \frac{\frac{\Delta 	heta \cdot 4KA}{l}}{\frac{\Delta 	heta \cdot 3KA}{4l}} = \frac{4}{1} \cdot \frac{4}{3} = \frac{16}{3}$.

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