A composite slab is prepared by pasting two plates of thickness L1 and L2 and thermal conductivites K1 and K2- The slabs have equal cross-sectional area- Find the equivalent conductivity of the composite slab-

It is equivalent to the series combination of 2 resistors. ∴ RS = R1 +R2 Resistance of a conducting slab, R=LKA RS = R1 + R2 L1+L2Ks.A=L1K1A+L2K2AL1+L2KS=L1K1+ ...

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Standard XII

Physics

Conductivity and Resistivity

A composite s...

Question

A composite slab is prepared by pasting two plates of thickness L₁ and L₂ and thermal conductivites K₁ and K₂. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

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Solution

It is equivalent to the series combination of 2 resistors.
∴ R_S = R₁ +R₂
Resistance of a conducting slab, $R = \frac{L}{K A}$
R_S = R₁ + R₂
$\frac{L_{1} + L_{2}}{K_{s} . A} = \frac{L_{1}}{K_{1} A} + \frac{L_{2}}{K_{2} A} \frac{L_{1} + L_{2}}{K_{S}} = \frac{L_{1}}{K_{1}} + \frac{L_{2}}{K_{2}} \frac{L_{1} + L_{2}}{K_{S}} = \frac{L_{1} K_{2} + L_{2} K_{1}}{K_{1} \times K_{2}} K_{S} = \frac{(L_{1} + L_{2}) (K_{1} K_{2})}{L_{1} K_{2} + L_{2} K_{1}}$

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A composite slab is prepared by pasting two plates of thickness L1 and L2 and thermal conductivites K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

It is equivalent to the series combination of 2 resistors. ∴ RS = R1 +R2 Resistance of a conducting slab, R=LKA RS = R1 + R2 L1+L2Ks.A=L1K1A+L2K2AL1+L2KS=L1K1+L2K2L1+L2KS=L1K2+L2K1K1×K2KS=L1+L2 K1 K2L1 K2+L2K1

A composite slab is prepared by pasting two plates of thickness L₁ and L₂ and thermal conductivites K₁ and K₂. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.