A) \[\frac{E}{2R}amp,\frac{E}{2}volt\]
B) \[\frac{E}{9R}amp,E/2volt\]
C) \[\frac{E}{3R}amp,\frac{E}{4}volt\]
D) \[\frac{E}{9R}amp,E/4volt\]
Correct Answer: A
Solution :
The circuit in steady state is shown in figure. Capacitor would be open circuit while inductor would be short circuited. Two 2R resistors have no role, whatever current is coming that goes the inductor branch. \[l=\frac{E}{2R}\]same current goes to inductor branch \[{{V}_{capacitor}}=IR=E/2\]You need to login to perform this action.
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