Question:

The temperature \(\theta\) at the junction of two insulating sheets, having thermal resistances \(R_1\) and \(R_2\) as well as top and bottom temperatures \(\theta_1\) and \(\theta_2\) (as shown in figure) is given by :

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In steady state the same heat current flows through both sheets: \((\theta_1-\theta)/R_1 = (\theta-\theta_2)/R_2\).
Updated On: Nov 4, 2026
  • \(\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}\)
  • \(\frac{\theta_1 R_1 + \theta_2 R_2}{R_1 + R_2}\)
  • \(\frac{\theta_1 R_2 - \theta_2 R_1}{R_2 - R_1}\)
  • \(\frac{\theta_2 R_2 - \theta_1 R_1}{R_2 - R_1}\)
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The Correct Option is A

Solution and Explanation

In steady state the heat current is the same through both sheets (thermal resistances in series).
The sheet of resistance \(R_1\) lies between \(\theta_1\) and \(\theta\), and the sheet of resistance \(R_2\) lies between \(\theta\) and \(\theta_2\).
\(\frac{\theta_1-\theta}{R_1} = \frac{\theta-\theta_2}{R_2}\)
\(\theta_1 R_2 - \theta R_2 = \theta R_1 - \theta_2 R_1\)
\(\theta (R_1 + R_2) = \theta_1 R_2 + \theta_2 R_1\)
\(\theta = \frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}\)
Answer: (1) \(\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}\)
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