The inductors of two 𝐿𝑅 circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced 𝐸𝑀𝐹 in the inductors by the time the currents reach their steady state values is________ mJ.

When current in an $$LR$$ circuit is growing, the cell has to do work against the back-emf of the inductor. The total work done by all the cells up to the instant the currents attain their steady values is equal to the total magnetic energy finally stored in the inductors.

For two magnetically coupled coils the magnetic energy in the steady state is
$$ U=\frac12\,L_1I_1^{2}+\frac12\,L_2I_2^{2}+MI_1I_2 $$ (the mutual-energy term is positive because the dots marked on the coils in the given figure indicate that the fluxes aid each other).

The steady currents are obtained from Ohm’s law for each branch:
$$I_1=\dfrac{V_1}{R_1},\qquad I_2=\dfrac{V_2}{R_2}.$$ Substituting the numerical data given in the figure,
$$I_1=\dfrac{10\;\text{V}}{10\;\Omega}=1.0\;\text{A}, \qquad I_2=\dfrac{20\;\text{V}}{5.0\;\Omega}=4.0\;\text{A}. $$

Now the final magnetic energy is $$ \begin{aligned} U&=\frac12(20\times10^{-3})\,(1.0)^{2} +\frac12(10\times10^{-3})\,(4.0)^{2} +(5.0\times10^{-3})(1.0)(4.0)\\ &=\frac12(0.020)(1)+\frac12(0.010)(16)+0.005(4)\\ &=0.010+0.080+0.020\\ &=0.110\;\text{J}. \end{aligned} $$

Finally convert the energy into milli-joule: $$ W=U=0.110\;\text{J}=110\;\text{mJ}. $$

Hence the total work done by the batteries against the induced emfs is 110 mJ.