NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
In the circuit shown below, i 1 and i 2 are the steady-state values of the current through L 1 and L 2 respectively, then i 1 is
Options
- Ai 1 = E L 2 R L 1 + L 2
- Bi 1 = E L 1 R L 1 + L 2
- Ci 1 = E L 2 R L 1 L 2
- Di 1 = E L 1 R L 1 L 2
Correct answer
A. i 1 = E L 2 R L 1 + L 2
Step-by-step solution
Since the potential difference across the inductors is same, we get L 1 d i 1 d t = L 2 d i 2 d t L 1 d i 1 = L 2 d i 2 L 1 ∆ i 1 = L 2 ∆ i 2 L 1 i 1 = L 2 i 2 (Since the initial value of current through both the inductors was zero) i 1 i 2 = L 2 L 1 The steady current i passing through R is i = E R i 1 = L 2 L 1 + L 2 i = E L 2 R L 1 + L 2