Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively are connected in series. The effective thermal conductivity of the combination is:

$\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

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$\frac{K_{1} + K_{2}}{K_{1} K_{2}}$

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$\frac{K_{1} + K_{2}}{2 K_{1} K_{2}}$

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$\frac{K_{1} K_{2}}{K_{1} + K_{2}}$

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Solution

The correct option is A
$\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

Step 1: Given
Thermal conductivity of first metal wire is $K_{1}$ .
Thermal conductivity of second metal wire is $K_{2}$ .
Equivalent thermal conductivity of metal wires is $K_{e q}$ .
Resistance of first metal wire is $R_{1}$ .
Resistance of second metal wire is $R_{2}$ .
Equivalent resistance of metal wires is $R_{e q}$ .
Length of the metal wires is $l$ . (Both wires have same length)
Area of cross-section of the metal wires is $A$ . (Both wires have same length)
Step 2: Formula Used
When two wires are connected in series, their equivalent resistance is $R_{e q} = R_{1} + R_{2}$ .
The resistance of a wire in terms of thermal conductivity $(K)$ is given by $R = \frac{l}{K A}$ , where $l$ is length of wire and $A$ is area of wire.
Step 3: Find the effective thermal conductivity
Substitute $R = \frac{l}{K A}$ , in $R_{e q} = R_{1} + R_{2}$ . The new length will be twice the original length and the area of cross-section will remain same.
$R_{e q} = R_{1} + R_{2} \Rightarrow \frac{2 l}{K_{e q}} = \frac{l}{K_{1} A} + \frac{l}{K_{2} A} \Rightarrow \frac{2}{K_{e q}} (\frac{l}{A}) = (\frac{1}{K_{1}} + \frac{1}{K_{2}}) (\frac{l}{A}) \Rightarrow \frac{2}{K_{e q}} = (\frac{1}{K_{1}} + \frac{1}{K_{2}})$
Solve the above equation further.
$\frac{2}{K_{e q}} = \frac{K_{1} + K_{2}}{K_{1} K_{2}} \Rightarrow \frac{K_{e q}}{2} = \frac{K_{1} K_{2}}{K_{1} + K_{2}} \Rightarrow K_{e q} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$
Hence, option A is correct.

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