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ExamsJEE AdvancedPhysics

Two insulating slabs are placed in series with thermal resistances R1 (top) and R2 (bottom). The top surface is held at temperature theta1 and the bottom at theta2. In steady state, what is the temperature theta at the junction between the two slabs?

  1. (theta2*R2 - theta1*R1)/(R2 - R1)
  2. (theta1*R2 - theta2*R1)/(R2 - R1)
  3. (theta1*R2 + theta2*R1)/(R1 + R2)
  4. (theta1*R1 + theta2*R2)/(R1 + R2)

Correct answer: (theta1*R2 + theta2*R1)/(R1 + R2)

Solution

Equating the heat current through both slabs, (theta1 - theta)/R1 = (theta - theta2)/R2, and solving gives theta = (theta1*R2 + theta2*R1)/(R1 + R2).

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