An oscillating LC circuit consists of a $75 ~\mathrm{mH}$ inductor and a $1.2 ~\mu \mathrm{F}$ capacitor. If the maximum charge to the capacitor is $2.7 ~\mu \mathrm{C}$. The maximum current in the circuit will be ___________ $\mathrm{mA}$

<p>The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:</p> <p>$I_{\text{max}} = \omega Q_{\text{max}}$,</p> <p>where:</p> <ul> <li>$I_{\text{max}}$ is the maximum current,</li> <li>$\omega$ is the angular frequency, and</li> <li>$Q_{\text{max}}$ is the maximum charge on the capacitor.</li> </ul> <p>The angular frequency $\omega$ for an LC circuit is given by:</p> <p>$\omega = \frac{1}{\sqrt{LC}}$,</p> <p>where:</p> <ul> <li>$L$ is the inductance, and</li> <li>$C$ is the capacitance.</li> </ul> <p>Given that $L = 75 \, \text{mH} = 75 \times 10^{-3} \, \text{H}$, $C = 1.2 \, \mu \text{F} = 1.2 \times 10^{-6} \, \text{F}$, <br/><br/>and $Q_{\text{max}} = 2.7 \, \mu \text{C} = 2.7 \times 10^{-6} \, \text{C}$, <br/><br/>we can substitute these values into the formulas to find $I_{\text{max}}$:</p> <p>$\omega = \frac{1}{\sqrt{(75 \times 10^{-3})(1.2 \times 10^{-6})}} = 3333.33 \, \text{rad/s}$,</p> <p>$I_{\text{max}} = \omega Q_{\text{max}} = 3333.33 \times 2.7 \times 10^{-6} = 0.009 \, \text{A}$.</p> <p>Therefore, the maximum current in the circuit is $0.009 \, \text{A}$, or equivalently, $9 \, \text{mA}$.</p>