-
In the network shown in the figure, the circuit was initially in the steady-state condition with the switch K closed. At the instant when the switch is opened, the rate of decay of current through the inductance will be
-
- Zero
- 0.5 A/s
- 1 A/s
- 2 A/s
Correct Option: D
Current in the inductor of the instant of opening the switch in 1 A. Thereafter
i(t) = 1e(– L / R)C = e– 2t
| ⇒ | = 2 e– 2t | |
| dt |
