JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Electromagnetic Induction

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 EMF in the inductors by the time the currents reach their steady state values is ____ mJ.

Answer: 55

Solution

At steady state:

I_{1} = \frac{V _{1}}{R _{1}} = \frac{10}{10} = 1

I_{2} = \frac{V _{2}}{R _{2}} = \frac{20}{10} = 2

A

Total energy stored (mutual inductance produces flux in same direction as self inductance):

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} (10 \times 1 0^{- 3}) (1)^{2} + \frac{1}{2} (20 \times 1 0^{- 3}) (2)^{2} + (5 \times 1 0^{- 3}) (1) (2)

= 5 \times 1 0^{- 3} + 40 \times 1 0^{- 3} + 10 \times 1 0^{- 3}

= 55 \times 1 0^{- 3}

= 55

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Electromagnetic Induction

Solution

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