MHT CET20255 May 2025Evening ShiftPhysicsElectromagnetic InductionActual
Two inductors of 80 mH each are joined in parallel. The current passing through the combination is 2.1 A . The energy stored in this combination of inductors is
Options
- A4.84 10⁻² ~J
- B7.26 10⁻² ~J
- C8.82 10⁻² ~J
- D10.85 10⁻² ~J
Correct answer
C. 8.82 10⁻² ~J
Step-by-step solution
Energy stored in parallel inductor combination For two identical inductors L₁ = L₂ = L = 80 10⁻³ H connected in parallel, the equivalent inductance is L_ eq = L 2 = 40 10⁻³ H. The energy stored in an inductor carrying current I = 2.1 A is given by U = 1 2 L_ eq I^2 . Substituting values: U = 1 2 (40 10⁻³) (2.1)^2 = 1 2 (40 10⁻³) (4.41) = 88.2 10⁻³ = 8.82 10⁻² J This matches option C .