Three rods of material `x` and three of material `y` are connected as shown in figure. All the rods are identical in length and cross sectional area. If the end `A` is maintained at `60^(@)C` and the junction `E` at `10^(@)C` , calculate the temperature of the junction `B`. The thermal conductivity of `x` is `800Wm^(-1).^(@)C^(-1)` and that of `y` is `400Wm^(-1).^(@)C^(-1)` .

Explanation

It is clear from the symmetry of the figure that the points `C` and `D` are equivalent in all respect and hence, they are at the same temperature, say `T`. No heat will flow through the rod `CD`. We can, therefore neglect this rod in further analysis. (Treated as balance wheat stone bridge)
Let `L` and `A` be the length and the area of cross-section of each rod. The thermal resistance of `AB`, `BC` and `BD` are equal. Each has a value
`R_(1)=(1)/(K_(x))(L)/(A)`.......`(i)`
Similarly, thermal resistances of `CE` and `DE` are equal, each having a value
`R_(2)=(1)/(K_(y))(L)/(A)`........`(ii)`
As the rod `CD` has no effect, we can say that the rods `BC` and `CE` are joined in series. Their equivalent thermal resistance is
`R_(3)=R_(BC)+R_(CE)=R_(1)+R_(2)`.
Also, the rods `BD` ad `DE` together have an equivalent thermal resistance
`R_(4)=R_(BD)+R_(DE)=R_(1)+R_(2)`.

The resistances `R_(3)` and `R_(4)` are joined in parallel and hence their equivalent thermal resistance is given by
`(1)/(R_(5))=(1)/(R_(3))+(1)/(R_(4))=(2)/(R_(3))` or `R_(5)=(R_(3))/(2)=(R_(1)+R_(2))/(2)`

The resistance `R_(5)` is connected in series with `AB`. Thus, the total arrangement is equivalent to a thermal resistance.
`R=R_(AB)+R_(5)=R_(1)+(R_(1)+R_(2))/(2)=(3R_(1)+R_(2))/(2)`

The heat current through `A` is
`i=(T_(A)-T_(B))/(R )=(2(T_(A)-T_(B)))/(3R_(1)+R_(2))`......`(i)`
This current passes through the rod `AB`. We have
`i=(T_(A)-T_(B))/(R_(AB))`.......`(ii)`
by using `(i)` and `(ii)` we get
`T_(A)-T_(B)=(2K_(y)(T_(A)-T_(B)))/(K_(x)+3K_(y))=(2xx400)/(800+3xx400)=20^(@)C`
`T_(B)-T_(A)-20^(@)C=40^(@)C`