Figure shows a circuit that contains three identical resistors with resistance R = 9.0 $\Omega$ each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf $\epsilon$ = 18 V. The current i through the battery just after the switch closed is

 

 

 At time,  t = 0  i.e., when switch is closed, inductor in the circuit provides very high resistance (open circuit) while capacitor starts charging with maximum current (low resistance). Equivalent circuit of the given circuit Current drawn from battery,

$i=\frac{\epsilon }{(R/2)}=\frac{2\epsilon }{R}=\frac{2\times 18}{9}=4 \mathrm{A}$

NOTE: 

The original options given in NEET 2017 were:

1. 0.2 A

2. 2 A

3. 0 A

4. 2 mA

 None of the options was correct and hence marks were awarded for all the students who attempted the question.

We have modified option (2) as 4 A in this test to make it correct the correct answer.