NTA Abhyas JEE Main2020PhysicsAlternating CurrentPractice
The steady-state current through L 1 after the switch has been closed for a very long time is
Options
- AV 0 R
- BV 0 L 1 R L 1 + L 2
- CV 0 L 2 R L 1 + L 2
- DNone of these
Correct answer
C. V 0 L 2 R L 1 + L 2
Step-by-step solution
After a very long time t → ∞ , I 0 = V R ....(i) Since the inductors are in parallel, L 1 d I 1 d t = L 2 d I 2 d t After a long time, L 1 I 1 = L 2 I 2 .... (ii) The total current I 0 through the battery is divided into the inductors, hence I 0 = I 1 + I 2 ..... (iii) From equations (ii) & (iii) I 1 = ( L 2 L 1 + L 2 ) I 0 & I 2 = ( L 1 L 1 + L 2 ) I 0 The current through L 1 is   I 1 = ( L 2 L 1 + L 2 ) I 0 = ( L 2 L 1 + L 2 ) V 0 R