1. Use conservation of energy in an ideal LC circuit

When the charged capacitor is connected to the inductor, energy oscillates between the capacitor and the inductor.

• Initial energy stored in capacitor: $U_C = \frac{1}{2}CV^2$
• Maximum energy stored in inductor: $U_L = \frac{1}{2}LI_{\max}^2$

At maximum current, all the energy of the capacitor is transferred to the inductor, so $\frac{1}{2}CV^2 = \frac{1}{2}LI_{\max}^2$

Thus, $I_{\max} = V\sqrt{\frac{C}{L}}$

2. Substitute the given values

Given: $C = 100\,\mu F = 100 \times 10^{-6} F = 10^{-4} F$ $L = 6.4\,mH = 6.4 \times 10^{-3} H$ $V = 12\,V$

So, $I_{\max} = 12\sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}}$

3. Simplify the expression

$\frac{10^{-4}}{6.4 \times 10^{-3}} = \frac{1}{6.4}\times 10^{-1} = 0.015625$

Therefore, $I_{\max} = 12\sqrt{0.015625}$

Now, $\sqrt{0.015625} = 0.125$

Hence, $I_{\max} = 12 \times 0.125 = 1.5\,A$

4. Match with the options

The maximum current is $\boxed{1.5\,A}$

So the correct option is C.