Two metal plates of same area and thickness $l_{1}$ and $l_{2}$ are arranged in series. If the thermal conductivities of the materials of the two plates are $k_{1}$ and $k_{2}$ . The thermal conductivity of the combination is :

\frac{2 k_{1} k_{2}}{k_{1} + k_{2}}

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\frac{k_{1} + k_{2}}{2}

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\frac{k_{1} k_{2} (l_{1} + l_{2})}{k_{1} l_{2} + k_{2} l_{1}}

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k_{1} + k_{2}

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Solution

The correct option is C $\frac{k_{1} k_{2} (l_{1} + l_{2})}{k_{1} l_{2} + k_{2} l_{1}}$
As we know that for a plate $Q = \frac{K A (T_{1} - T_{2})}{L}$
where $R_{t h} = \frac{L}{K A}$
So we can solve it same as we solve the resistance in circuit.
$\frac{L_{1} + L_{2}}{K} = {\frac{L_{1}}{K}}_{1} + {\frac{L_{2}}{K}}_{2}$
So, equivalent thermal conductivity will be given as: $K = \frac{K_{1} K_{2} (L_{1} + L_{2})}{L_{1} K_{2} + L_{2} K_{1}}$

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