A capacitor of capacitance $$100 \mu F$$ is charged to a potential of $$12$$ V and connected to a $$6.4$$ mH inductor to produce oscillations. The maximum current in the circuit would be :

We need to find the maximum current in an LC oscillation circuit where a capacitor of $$100 \, \mu F$$ charged to $$12$$ V is connected to a $$6.4$$ mH inductor.

In an LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. The total energy is conserved:

$$\frac{1}{2}CV^2 = \frac{1}{2}LI_{max}^2$$

The left side is the maximum energy stored in the capacitor (when current is zero), and the right side is the maximum energy stored in the inductor (when the capacitor is fully discharged).

From the energy conservation equation:

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

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

$$I_{max} = 12\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$$

$$= 12\sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}} = 12\sqrt{\frac{1}{64}}$$

$$= 12 \times \frac{1}{8} = 1.5 \text{ A}$$

The correct answer is Option (2): 1.5 A.