A composite slab is prepared by pasting three slab of thickness $L_{1}, L_{2}, L_{3}$ and thermal conductivity $K_{1}, K_{2}, K_{3}$ .The slab have equal cross-sectional area. Find the equivalent thermal conductivity.

\frac{K_{1} + K_{2} + K_{3}}{L_{1} + L_{2} + L_{3}}

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\frac{K_{1} K_{2} K_{3} (L_{1} + L_{2} + L_{3})}{K_{1} K_{2} L_{3} + L_{1} K_{2} K_{3} + K_{1} L_{2} K_{3}}

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\frac{K_{1} K_{2} L_{3} + L_{1} K_{2} K_{3} + K_{1} L_{2} K_{3}}{K_{1} + K_{2} + K_{3}}

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\frac{L_{1} + L_{2} + L_{3}}{K_{1} + K_{2} + K_{3}}

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Solution

The correct option is B $\frac{K_{1} K_{2} K_{3} (L_{1} + L_{2} + L_{3})}{K_{1} K_{2} L_{3} + L_{1} K_{2} K_{3} + K_{1} L_{2} K_{3}}$
Effective Resistance $R = R_{1} + R_{2} + R_{3}$
$\Rightarrow \frac{L_{1} + L_{2} + L_{3}}{K_{e q} A} = \frac{L_{1}}{K_{1} A} + \frac{L_{2}}{K_{2} A} + \frac{L_{3}}{K_{3} A} \Rightarrow \frac{L_{1} + L_{2} + L_{3}}{K_{e q}} = \frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}} + \frac{L_{3}}{K_{3}} \Rightarrow \frac{L_{1} + L_{2} + L_{3}}{K_{e q}} = \frac{K_{3} K_{2} L_{1} + K_{1} K_{3} L_{2} + K_{1} K_{2} L_{3}}{K_{1} K_{2} K_{3}} \Rightarrow K_{e q} = \frac{K_{1} K_{2} K_{3} (L_{1} + L_{2} + L_{3})}{K_{3} K_{2} L_{1} + K_{1} K_{3} L_{2} + K_{1} K_{2} L_{3}}$

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A composite slab is prepared by pasting three slab of thickness L1,L2,L3 and thermal conductivity K1,K2,K3.The slab have equal cross-sectional area. Find the equivalent thermal conductivity.

The correct option is B K1K2K3(L1+L2+L3)K1K2L3+L1K2K3+K1L2K3Effective Resistance R=R1+R2+R3⇒L1+L2+L3KeqA=L1K1A+L2K2A+L3K3A⇒L1+L2+L3Keq=L1K1+L2K2+L3K3⇒L1+L2+L3Keq=K3K2L1+K1K3L2+K1K2L3K1K2K3⇒Keq=K1K2K3(L1+L2+L3)K3K2L1+K1K3L2+K1K2L3

A composite slab is prepared by pasting three slab of thickness $L_{1}, L_{2}, L_{3}$ and thermal conductivity $K_{1}, K_{2}, K_{3}$ .The slab have equal cross-sectional area. Find the equivalent thermal conductivity.