There slabs of thickness `d_(1)`, `d_(2)`, `d_(3)` and thermal conductivites `lambda_(1)`, `lambda_(2)` and `lambda_(3)` are placed in contact. Show that in the steady the combination behaves as a single slab of material of conductivity `lambda` given by `(d_(1)+d_(2)+d_(3))/(lambda)=(d_(1))/(lambda_(1))+(d_(2))/(lambda_(2))+(d_(3))/(lambda_(3))`

Answer

NA