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)

Options

A. `(2k_(1)k_(2))/(k_(1)+k_(2))`
B. `sqrt(k_(1)k_(2))`
C. `(k_(1)k_(2))/(k_(1)+k_(2))`
D. `k_(1)+k_(2)`

Answer

A

Explanation

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))`