Question

Six identical conducting rods are joined as shown in the figure. Points A and D are
maintained at temperatures of $200^\circ \,C$ and $20^\circ \,C$ respectively. The temperature of junction B will be:

A. $120^\circ \,C$
B. $100^\circ \,C$
C. $140^\circ \,C$
D. $80^\circ \,C$

Answer 1

Hint:Here, we will use the concept of Kirchhoff’s law. According to Kirchhoff’s law, the current between points A and B will be equal to the current between the points B and D. Therefore, the ratio of change in temperature to the thermal resistance of points A and B will be equal to the ratio of change in temperature to the thermal resistance between points B and D.

Complete Step by Step Answer:
Consider six identical rods which are as shown in the figure between the points A and B. As all the rods are identical, therefore, the thermal resistance of each rod is $$R$$ which is shown below

Now, the resistances between points B and C are in series and parallel to each other. As we know, the resistance in parallel becomes half of the resistance in series. Therefore, the above diagram becomes

Now, the resistance between the points B and D will be the sum of the resistance between the points B and C and the resistance between the points C and D.

Also, the electrical concept and the electrical thermal concept both are same, therefore, the
resistance in both the cases will also be same, which are shown below
${R_{eq}} = R$

Now, the thermal resistance between the points A and B $ = R$

Also, the thermal resistance between the points B and D $ = R + R = 2R$

Now, the heat flowing through A and B will be equal to the heat flowing through B and D.

Therefore, the ratio of change in temperature between A and B and resistance between A and B will be equal to the ratio of the change in temperature between B and D and the resistance between B and D, which is shown below

$\dfrac{{200 - T}}{R} = \dfrac{{T - 20}}{{2R}}$
$ \Rightarrow \,\dfrac{{200 - T}}{1} = \dfrac{{T - 20}}{2}$
$ \Rightarrow \,2\left( {200 - T} \right) = T - 20$
$ \Rightarrow \,400 - 2T = T - 20$
$ \Rightarrow \,3T = 420$
$ \Rightarrow \,T = \dfrac{{420}}{3}$
$ \Rightarrow \,T = 140^\circ \,C$

Hence, the temperature of the junction B will be $140^\circ \,C$ .

Hence, option (C) is the correct option.

Note:Here, there are four resistances between the points B and C. The two resistances are in series forming two pairs and these pairs are parallel to each other. When the resistances in series will be parallel to each other than the resistance in series will become.