Properties of Matter

For a steady flow, we have $\frac{\Delta Q}{\Delta t}$ = $\frac{\Delta T}{R_{Th}}$ where $R_{Th}$ is the thermal resistance, which is expressed as $R_{Th}$ = $\frac{l}{KA}$ If the width of each rod is taken as w, then we have $R_1$ = $\frac{L}{2k(4Lw)}$ = $\frac{1}{8kw}$ $R_2$ = $\frac{4L}{3k(Lw)}$ = $\frac{4}{3kw}$ $R_3$ = $\frac{4L}{4k(2Lw)}$ = $\frac{1}{2kw}$ $R_4$ = $\frac{4L}{5k(Lw)}$ = $\frac{4}{5kw}$ $R_5$ = $\frac{L}{6k(4Lw)}$ = $\frac{1}{24kw}$ The equivalent electrical circuit for the above set-up can be depicted as shown in the following figure: The heat flow through each rod is equivalent to current through each thermal resistance: $I_1$ = $I_5$ = I Therefore, $V_{PQ}$ = $IR_1$ ; $V_{QR}$ = $\frac{I}{\frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}}$; $V_{RS}$ = $I_{RS}$ $V_{PQ}$ = $\frac{1}{8kw}$; $V_{QR}$ = $\frac{1}{4kw}$ Therefore, the value of $V_{RS}$ is the least. That is, $V_{RS}$ = $\frac{I}{24kw}$ Now, $I_3$ = $\frac{V_{QR}}{R_3}$ = $\frac{I}{4kw} \times 2kw$ = $\frac{I}{2}$ $I_2$ = $\frac{V_{QR}}{R_2}$ = $\frac{I}{4kw} \times \frac{3kw}{4}$ = $\frac{3I}{16}$ $I_4$ = $\frac{V_{QR}}{R_4}$ = $\frac{I}{4kw} \times \frac{5kw}{4}$ = $\frac{5I}{16}$ Therefore, $I_2 + I_4$ = $I_2$ = $I_3$