A capacitor having capacitance of 100 μF is charged with a potential difference of 12 V is connected to an inductor of inductance 10 mH. Find the maximum current through the inductor.

To find the maximum current through the inductor in an LC circuit, we can use the principle of conservation of energy.

Initially, the capacitor is fully charged with a capacitance of $C=100\mu F=100\times {10}^{-6}F$ and a potential difference of $V=12V$. The initial energy stored in the electric field of the capacitor ($E_C$) is given by the formula:
$E_C=\frac{1}{2}C{V}^2$

When the capacitor is connected to the inductor of inductance $L=10mH=10\times {10}^{-3}H$, the electrostatic energy of the capacitor starts converting into electromagnetic energy in the inductor.

The maximum current ($I_{\text{max}}$) flows through the inductor when all the electrostatic energy initially stored in the capacitor is completely converted into magnetic energy stored in the inductor ($E_L$). The maximum energy stored in the inductor is:
$E_L=\frac{1}{2}L{I}_{\text{max}}^2$

By conservation of energy:
$\frac{1}{2}L{I}_{\text{max}}^2=\frac{1}{2}C{V}^2$

We can simplify this to solve for $I_{\text{max}}$:
$L{I}_{\text{max}}^2=C{V}^2$
$I_{\text{max}}^2=\frac{C}{L}{V}^2$
$I_{\text{max}}=V\sqrt{\frac{C}{L}}$

Now, substitute the given values into the equation:
$I_{\text{max}}=12\times \sqrt{\frac{100\times {10}^{-6}}{10\times {10}^{-3}}}$
$I_{\text{max}}=12\times \sqrt{\frac{{10}^{-4}}{{10}^{-2}}}$
$I_{\text{max}}=12\times \sqrt{{10}^{-2}}$
$I_{\text{max}}=12\times {10}^{-1}$
$I_{\text{max}}=1.2A$

Therefore, the maximum current through the inductor is 1.2 A.