Consider the circuit shown. Resistance R, initially at ${50}^{o}C$  at and having resistance  $15\Omega$, is kept inside a chamber maintained at ${20}^{0}C$ . Rate of change of resistance of R with temperature is  $1\Omega {/}^{o}C$. Assuming that resistance cools following Newton’s law with proportionality constant equal to $\frac{\mathcal{l}n\left(\frac{9}{4}\right)}{100}$  per s, time after which the resistance R consumes maximum power is 

The given circuit can be simplified first as shown in Fig.1 and finally as in Fig.2   
                                           
As per maximum power theorem, for maximum power consumption , R should have value equal to resistance of remaining circuit which is  $5\Omega$. As rate of change of resistance is  $1\Omega {/}^{\text{O}}\text{C}$, therefore  $\frac{\Delta R}{\Delta T}=1$
 $\Rightarrow \frac{5-15}{\Delta T}=1\Rightarrow -{10}^{0}C$
$\Rightarrow$ Final temperature of R
  $=50-10={40}^0\text{C}$
As per Newton’s law of cooling,
 ${\text{T}}_{\text{f}}-T_{\text{surrounding}}=(T-T_{\text{surrounding}})e^{-\text{kt}}$
 $\Rightarrow 40-20=(50-20)e^{-\text{kt}}$
 $\Rightarrow \text{ t =}\frac{\ln \frac{3}{2}}{k}=\frac{\ln \frac{3}{2}}{\frac{\ln \frac{9}{4}}{100}}=\frac{\ln \frac{3}{2}}{\frac{2.\ln \frac{3}{2}}{100}}=50\text{ s}$