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.

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answer is 55.

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Detailed Solution

$l_{1} = \frac{5}{5} = 1 A I_{2} = \frac{20}{10} = 2 A$

$∴$ Total work done by battery against induced emf = Increase in mag. energy
$= U_{1} + U_{2} + U_{12}$

$∴ W = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2} = \frac{1}{2} \times 10 \times 10^{- 3} \times 1^{2} + \frac{1}{2} \times 20 \times (2)^{2} \times 10^{- 3} + 5 \times 10^{- 3} \times 1 \times 2 = \frac{1}{2} \times 10 \times 10^{- 3} \times 1^{2} + \frac{1}{2} \times 20 \times (2)^{2} \times 10^{- 3} + 5 \times 10^{- 3} \times 1 \times 2 = 55 mJ$