Question

Three rods of material $x$ and three rods of material $y$ are connected as shown in the figure. All the rods are of identical length and cross section. If the end $A$ is maintained at 60${}^o$C and the junction E at 10${}^o$C, find the effective thermal resistance. Given length of each rod$=l$, area of cross-section$=A$, conductivity of $x=K$ and conductivity of $y=2K$.

• A. $\frac{4l}{3KA}$
• B. $\frac{7l}{6KA}$
• C. $\frac{4KA}{3l}$
• D. $\frac{7KA}{3l}$

Answer 1

Answer: (B)

$\frac{7l}{6KA}$

Thermal Resistivity is defined as $R=\frac{L}{KA}$

$R_x=\frac{L}{KA}$

$R_y=\frac{L}{2KA}$

If we take $R_x=2R$ then, $R_y=R$

Since the bridge is balanced i.e $\frac{R_{BC}}{R_{CE}}=\frac{R_{BD}}{R_{DE}}$

The temp of junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.

The thermal resistance of the bridge between junction B and E is

$\frac{1}{R_{BE}}=\frac{1}{2R+2R}+\frac{1}{R+R}$

$\therefore R_{BE}=\frac{4}{3}R$

The total resistance between A and E will be,

$R_{eq}=R_{AB}+R_{BE}=R+\frac{4}{3}R=\frac{7}{3}R$

Substituting $R=\frac{L}{2KA}$ we get $R_{eq}=\frac{7}{3}\times \frac{L}{2KA}=\frac{7L}{6KA}$