Q.
Consider two insulating sheets with thermal resistances $R_{1}\ and\ R_{2}\ $as shown in figure. The temperature $\theta\ is$
A) $\frac{\theta_{1}R_{2} + \theta_{2}R_{1}}{R_{1} + R_{2}}$
B) $\frac{\left( \theta_{1} + \theta_{2} \right)R_{1}R_{2}}{R_{1}^{2} + R_{2}^{2}}$
C) $\frac{\theta_{1}R_{1} + \theta_{2}R_{2}}{R_{1} + R_{2}}$
D) $\frac{\theta_{1}\theta_{2}\ R_{1}R_{2}}{\left( \theta_{1} + \theta_{2} \right)\left( R_{1}R_{2} \right)}$
Explanation
(a) For the two sheets, shown in figure, rate of heat transfer is same, $ie,$
$\frac{dQ_{1}}{dt} = \frac{dQ_{2}}{dt}$
$\therefore\ \frac{dT_{1}}{R_{1}} = \frac{dT_{2}}{R_{2}}$
$\frac{\theta_{1} - \theta}{R_{1}}\ = \ \frac{\theta - \theta_{2}}{R_{2}}$
$\theta_{1}R_{2} - \theta R_{2} = \theta R_{1} - \theta_{2}R_{1},\ \ \ \ \theta\frac{\theta_{1}R_{2} + \theta_{2}R_{1}}{R_{1} + R_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.