Two rods of the same area of cross-section, but of length l_(1) and l_(2) and conductivites K_(1) and K_(2) are joined in series. Show that the combination is equivalent of a material of conductivity K = (l_(1)+l_(2))/((l_(1)/(K_(2)))+((l_(2))/(K_(2))))

Two rods of the same area of cross-section, but of length l_(1) and l_

1.	Two rods of the same area of cross-section, but of length l_(1) and l_(2) and conductivites K_(1) and K_(2) are joined in series. Show that the combination is equivalent of a material of conductivity K = (l_(1)+l_(2))/((l_(1)/(K_(2)))+((l_(2))/(K_(2))))
Answer» Answer :Since they are in series, the rate of flow of HEAT energy is the same. But the SUM of the DIFFERENCE in TEMPERATURE is the difference across their FREE ends. `(##RES_HAT_PHY_XI_C03_E01_006_A01##)` `:. (theta_(1) -theta) +(theta -theta_(1)) = (theta_(1)- theta_(2))` i.e `(Q)/(t) (l_(1))/(K_(1)A) +(Q)/(t).(l_(2))/(K_(2)A) = (Q)/(t) ((l_(1)+l_(2)))/(K_(2)A)` `rArr (l_(1))/(K_(1)) +(l_(2))/(K_(2)) = (l_(1)+l_(2))/(K_(eq)) :. K_(eq)= (l_(1)+l_(2))/((l_(1)/(K_(1))+l_(2)/(K_(2))))`