The inductors of two 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. [2020]

Question

The inductors of two $L R$ 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 $E M F$ in the inductors by the time the currents reach their steady-state values is ________ mJ. [2020]

Smart Achievers · Accepted Answer

(55)

$Given: Mutual inductance, M = 5 mH$

$L_{1} = 10 mH, V_{1} = 5 V, L_{2} = 20 mH, V_{2} = 20 V$

$I_{1} = \frac{V_{1}}{R_{1}} = \frac{5}{5} = 1 A, I_{2} = \frac{V_{2}}{R_{2}} = \frac{20}{10} = 2 A$

After both the switches are closed simultaneously, the total work done by the batteries against the induced emf equals the increase in magnetic energy.

$∴$ $W = Δ U = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2}$

$= \frac{1}{2} \times (10 \times 10^{- 3}) \times 1^{2} + \frac{1}{2} \times (20 \times 10^{- 3}) \times 2^{2} + (5 \times 10^{- 3}) \times 1 \times 2$

$= (5 + 40 + 10) \times 10^{- 3} J$

$∴$ $W = 55 mJ$

Solution

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