A) I = 2 mA at all time t
B) I oscillates between 1 mA & 2mA
C) 1= 1mA at all time t
D) At t = 0,l = 2mA & with time it goes to 1 mA
Correct Answer: D
Solution :
At t = 0, when the .switch is closed the capacitor is uncharged & offers zero resistance for the charge to flow i.e. it behaves as short circuit and hence total current flow through capacitor branch i.e. at \[t=0,l=\frac{2}{1000}A=2mA\] As time increases charging of capacitor is taking place and when capacitor is fully charged it behaves as open circuit and current in the circuit is\[l=\frac{2}{1000+1000}A=1mA\] i.e. current in resistor AB decreases from 2 mA to 1 mA
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