Alternating Current Question 2
Question 2 - 24 January - Shift 2
Three identical resistors with resistance $R=12 \Omega$ and two identical inductors with sell inductance $L=5 mH$ are connected to an ideal battery with emf of $12 V$ as shown in figure. The current through the battery long after the switch has been closed will be A.
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Answer: (3)
Solution:
Formula: Growth Of Current in Series R-L Circuit
After long time an inductor behaves as a resistance-less path.
So current through cell
$ I=\frac{12}{R / 3}=3 A{\because R=12 \Omega} $
