In the circuit shown in figure ${\mathrm{R}}_1={\mathrm{R}}_2=10\mathrm{\Omega },{\mathrm{E}}_1=20\mathrm{V},{\mathrm{E}}_2=40\mathrm{V}$ For what value of R, the thermal power developed in it will be maximum (in$\mathrm{\Omega }$)?

Assuming I is the current through the cell E1 and I1 is the current through resistor in the centre: 

Using KVL for two loops I and II

 $-40+10\mathrm{I}+{\mathrm{I}}_1\mathrm{R}=0$

$10(\mathrm{I}-{\mathrm{I}}_1)-20-{\mathrm{I}}_1\mathrm{R}=0$

Solving above equation we get $I_1=\frac{60}{100+20R}$

Power developing in R is given by, $\mathrm{P}={\mathrm{I}}_1^2\mathrm{R}$

For P to be maximum  $\frac{\mathrm{dP}}{\mathrm{dR}}=0$

which gives $\mathrm{R}=5\mathrm{\Omega }$