b
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,
\(i=\frac{\varepsilon}{(R / 2)}=\frac{2 \varepsilon}{R}=\frac{2 \times 18}{9}=4 \,\mathrm{A}\)
