A compound slab is made up two slabs. If thermal conductivity of their material are k_(1) and k_(2) respectively of their cross-sectional areas are the same, the equivalent thermal conductivity of the slab is. . . . . (consider series connection)

A compound slab is made up two slabs. If thermal conductivity of their

1.	A compound slab is made up two slabs. If thermal conductivity of their material are k_(1) and k_(2) respectively of their cross-sectional areas are the same, the equivalent thermal conductivity of the slab is. . . . . (consider series connection)
Answer» `(2k_(1)k_(2))/(k_(1)+k_(2))` `sqrt(k_(1)k_(2))` `(k_(1)k_(2))/(k_(1)+k_(2))` `k_(1)+k_(2)` Solution :In SERIES connection, `(t)/(k_(1))+(t)/(k_(2))=(2t)/(k)` `(1)/(k_(1))+(1)/(k_(2))=2/k` `rArrk=(2k_(1)k_(2))/(k_(1)+k_(2))` Thermal resistance of compound slab, `R_(H)=(L)/(kA)` As the thickness of both SLABS is same, `L_(1)=L_(2)=L` As the cross SECTION of both slabs is same, `A_(1)=A_(2)=A` For series resistance connection, `R_(H)=(R_(H))_(a)+(R_(H))_(2)` Equivalent resistance `R_(H)=(2L)/(kA)`, [total length of slab = 2L] `(R_(H))_(1)=(L)/(k_(1)A)` and `(R_(H))_(2)=(L)/(k_(2)A)` `:.(2L)/(kA)=(L)/(k_(1)A)+(L)/(k_(2)A)` `:.2/k=(k_(2)+k_(1))/(k_(1)k_(2))` `:.k=(2k_(1)k_(2))/(k_(1)+k_(2))`