Skip to content
dronacharya.ai
← JEE Advanced 2020 · Paper 2
PhysicsNumerical+4Question 17 of 54

JEE Advanced 2020 · Paper 2, Physics question 17

The inductors of two LRLR 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 EMFEMF in the inductors by the time the currents reach their steady state values is ______ mJ.

Two separate LR circuits side by side. Left: V₁ = 5 V, R₁ = 5 Ω, L₁ = 10 mH, switch, current I₁. Right: V₂ = 20 V, R₂ = 10 Ω, L₂ = 20 mH, switch, current I₂. The two inductors are adjacent, with mutual inductance M = 5 mH.
Answer and worked solution

Official answer: 55

The work done against the induced emfs equals the magnetic energy stored once the currents are steady.

Steady currents.

  • I1=5/5=1I_1=5/5=1 A.
  • I2=20/10=2I_2=20/10=2 A.

Energy, with the mutual term (fluxes aiding):
U=12L1I12+12L2I22+MI1I2=12(10)(1)+12(20)(4)+5(1)(2)=5+40+10=55 mJ.U=\tfrac12L_1I_1^2+\tfrac12L_2I_2^2+MI_1I_2=\tfrac12(10)(1)+\tfrac12(20)(4)+5(1)(2)=5+40+10=55\text{ mJ}.
Answer: 55

Topics:Energy in an inductorMutual inductanceLR circuits
← Previous questionNext question →