A composite slab is prepared by pasting two plates of thickness `L_(1)` and `L_(2)` and thermal conductivites `K_(1)` and `K_(2)` . The slab have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Options

A. `(d_(1))/(K_(1))`
B. `d_(1)K_(1) + d_(1) K_(2) = K`
C. `(d_(1))/(K_(1)) + (d_(2))/(K_(2)) = (d_(1) + d_(2))/(K)`
D. `K (d_(1) + d_(2)) = K_(1) K_(2) d_(1) d_(2)`

Answer

C

Explanation

Thermal resistances behave analogous to electrical resistances
In series. `R_(eq) = R_(1) +R_(2)` Now, `R_("thermal")= (Delta x)/(KA)`
Here, `((d_(1) + d_(2)))/(KA) = (d_(1))/(K_(1)A) + (d_(2))/(K_(2)A) rArr (d_(1))/(K_(1)) + (d_(2))/(K_(2)) = (d_(1) + d_(2))/(K)`
`rArr` c is correct