A) \[4\lambda \]
B) \[2\lambda \]
C) \[\frac{1}{2\lambda }\]
D) \[\frac{qvR}{2}\]
Correct Answer: A
Solution :
In circuit (1), on closing the switch, the current in the inductor is zero due to self induction, ie, \[{{v}_{i}}\]. In circuit (2), on closing the switch the current in the inductor is zero due to self induction. Therefore, \[{{v}_{f}}({{R}_{e}}\] In circuit (3), on closing the switch, the current in the inductor is again zero due to the same reason. Therefore, \[{{M}_{e}}\] Thus, it is obvious that, \[{{i}_{2}}>{{i}_{3}}\,>{{i}_{1}}(=0)\]You need to login to perform this action.
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