LC Circuit (Oscillations)
By An LC circuit, also known as a resonant circuit or tank circuit, consists of an inductor (L) and a capacitor (C). It is a resonant circuit with a resonance frequency \begin{align} \omega=\frac{1}{2\pi}\sqrt{\frac{1}{LC}}. \end{align} The energy oscillates between the inductor and the capacitor at the resonant frequency. At the resonant frequency, the reactance of the inductor and the capacitor cancel each other out, allowing a maximum transfer of energy between the two components.
The charge $q$ on the capacitor follows the differential equation (SHM) \begin{align} \frac{\mathrm{d}^2q}{\mathrm{d}t^2}+\frac{1}{LC}q=0 \end{align} If initial charge on the capacitor is $q_\text{max}$ then its variation with time is given by \begin{align} q=q_\text{max}\cos(\omega t). \end{align} The current in the circuit is \begin{align} i=\frac{\mathrm{d}q}{\mathrm{d}t}=-q_\text{max}\omega\sin(\omega t) \end{align}
The electric field energy in the capacitor varies with time as \begin{align} U_E=\frac{q^2}{2C}=\frac{q_\text{max}^2}{2C}\cos^2(\omega t). \end{align} The magnetic field energy in the inductor varies with time as \begin{align} U_B=\frac{1}{2}Li^2=\frac{1}{2}q_\text{max}^2 \omega^2 L^2\sin^2(\omega t). \end{align}
Problems from IIT JEE
Problem (IIT JEE 1998): An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 $\mu$F and the resulting LC circuit is set oscillating at its natural frequency. Let $Q$ denotes the instantaneous charge on the capacitor and $I$ the current in the circuit. It is found that the maximum value of $Q$ is 200 $\mu$C.
- When $Q=100\,\mu$C, what is the value of $|\frac{\text{d}I}{\text{d}t}|$?
- When $Q=200\,\mu$C, what is the value of $I$?
- Find the maximum value of $I$.
- When $I$ is equal to one-half of its maximum value, what is the value of $|Q|$?
Solution: The charge in LC circuit oscillates with an angular frequency $\omega$ given by \begin{align} \omega&=\frac{1}{\sqrt{LC}} \\ &=\frac{1}{\sqrt{2.0\times{10}^{-3}\times 5.0\times{10}^{-6}}} \\ &={10}^{4} \,\mathrm{s^{-1}}. \nonumber \end{align} The charge $Q(t)$ on the capacitor, the current $I(t)$ in the circuit, and time derivative of the current, $\mathrm{d}I/\mathrm{d}t$, are given by \begin{align} Q(t)&=Q_0\sin(\omega t+\phi), \\ I(t)&=\frac{\mathrm{d}Q}{\mathrm{d}t}\\ &=Q_0\omega\cos(\omega t+\phi), \\ \frac{\mathrm{d}I(t)}{\mathrm{d}t}&=-Q_0\omega^2\sin(\omega t+\phi), \end{align} where $Q_0=200\,\mu$C is the maximum value of the charge and $\phi$ is the phase. Substitute $Q(t)=100\,\mu$C in the above equation to get \begin{align} \sin(\omega t+\phi)={1}/{2}. \end{align} Substitute this in the other equation to get \begin{align} \left|\frac{\mathrm{d}I(t)}{\mathrm{d}t}\right| &=(200\times{10}^{-6})({10}^{8})({1}/{2}) \\ &={10}^{4}\,\mathrm{A/s}. \nonumber \end{align}
Substitute $Q=200\,\mu$C in the above equation to get \begin{align} \sin(\omega t+\phi)=1 \end{align} and \begin{align} \cos(\omega t+\phi)=0. \end{align} Substitute this in the other equation to get $I(t)=0$.
Maximum value of $I(t)$ occurs at $\cos(\omega t+\phi)=1$. Substitute this in the above equation to get \begin{align} I_\text{max}=Q_0\omega=2\,\mathrm{A}. \end{align}
The energy of the system is conserved i.e., \begin{align} &\frac{1}{2}LI_\text{max}^2=\frac{1}{2}LI^2+\frac{1}{2}\frac{Q^2}{C}. \nonumber \end{align} Simplify to get \begin{align} Q=\sqrt{LC(I_\text{max}^2-I^2)}. \end{align} Substitute $I=\frac{I_\text{max}}{2}=1$ A to get \begin{align} Q=\sqrt{3}\times{10}^{-4}\,\mathrm{C}. \end{align}
