For the circuit shown in figure the value of R so that 25W of electrical energy is delivered to a combination of 11V battery and 0.04Ω resistor is

Let I is the current in circuit for given
requirement 25 = 0.04 x I² + 11 x I
Solving above equation we get I = 2.25 A
Using KVL, $\mathrm{I}=\frac{14-11}{0.05+\mathrm{R}+0.04}$
Equating above two values of I, we get $\mathrm{R}=1.24\mathrm{\Omega }$