A capacitor$\mathbf{C}$ is charged up to a voltage $\mathbf{V}$ and connected to an inductor $\mathbf{L}$, as shown schematically in Fig. 7.39. At time $t=\mathbf{0}$, the switch $\mathbf{S}$ is closed. Find the current in the circuit as a function of time. How does your answer change if a resistor R is included in series with $\mathbf{C}$ and$\mathbf{L}$ ?

Consider acapacitor $\text{C}$ is charged up to a voltage $\text{V}$ and connected to an inductor $\text{L}$, as shown schematically in Fig. 7.39.

Write the formula of current in the circuit as a function of time.

$I\left(t\right)=-\mathbf{A}\omega \mathbf{sin}\omega \mathbf{t}+\mathbf{B}\omega \mathbf{cos}\omega \mathbf{t}$…… (1)

Here, role="math" localid="1658232291576" $\mathbf{A}$is constant, $\omega$is angular frequency of the circuit, $\mathbf{B}$is constant.

Write the formula of current if a resistor $\text{R}$ is included in series with $\text{C}$and$\text{L}$.

$I=-\mathbf{C}\frac{\mathrm{dV}}{\mathrm{dt}}$ …… (2)

Here, $\mathbf{C}$ is capacitance and $\mathbf{V}$ is voltage.

When the switch$\text{S}$ is closed, the circuit conducts electricity as seen in figure 7.39. Assume that the inductor's emf is as follows and that current $I$ is flowing in a clockwise direction:

$\epsilon =-L\frac{dI}{dt}$

Here, $L$ is the inductance of the inductor and $\frac{dI}{dt}$is the rate of the change of the current.

Determine the emf across the capacitor is,

$\epsilon =\frac{Q}{C}$

Here, $Q$is the charge on the capacitor and $C$ is the capacitance of the capacitor.

Determine the current passing through the circuit is equal to the rate of change of the charge.

$I=\frac{dQ}{dt}$

Substitute $\frac{dQ}{dt}$for $I$ into equation$\epsilon =-L\frac{dI}{dt}$ and solve for .localid="1658400158306" $\epsilon$

$\begin{array}{c}\epsilon =-L\frac{d}{dt}\left(\frac{dQ}{dt}\right)\\ =-L\frac{d^{2}Q}{d{t}^2}\end{array}$

Substitute $\frac{Q}{C}$ for $\epsilon$into above equation.

$\begin{array}{c}\frac{Q}{C}=-L\frac{d^{2}Q}{d{t}^2}\\ \frac{d^{2}Q}{d{t}^2}=-\frac{Q}{LC}\\ \frac{d^{2}Q}{d{t}^2}+\left(\frac{1}{LC}\right)Q=0\\ \frac{d^{2}Q}{d{t}^2}+{\omega }^{2}Q=0\end{array}$

Here,$\omega =\frac{1}{\sqrt{LC}}$is the angular frequency of the circuit.

Then general solution for the second order differential equation is as follows:

$Q\left(t\right)=A\cos \omega t+B\sin \omega t$

Here, $A$and $B$are constants.

At the time$t=0s$, the charge across the capacitor is as follows:

$Q=CV$

Substitute $CV$for $Q$and $0$ for $t$ in the equation $Q\left(t\right)=A\cos \omega t+B\sin \omega t$ and simplify.

$\begin{array}{c}CV=A\cos \omega \left(0\right)+B\sin \omega \left(0\right)\\ CV=A\left(1\right)+0\\ A=CV\end{array}$

Differentiate the equation $Q\left(t\right)=A\cos \omega t+B\sin \omega t$ with respect to time.

$\begin{array}{l}\frac{dQ}{dt}=-A\omega \sin \omega t+B\omega \cos \omega t\\ I\left(t\right)=-A\omega \sin \omega t+B\omega \cos \omega t\end{array}$

At, $t=0s$the current passing through the circuit is equal to zero.

$I(t=0)=0$

Substitute $0$for $I\left(t\right)$ and $0$ for $I$ into above equation$I\left(t\right)$.

$\begin{array}{l}0=-A\omega \sin \omega \left(0\right)+B\omega \cos \omega \left(0\right)\\ 0=0+B\omega \left(1\right)\end{array}$

The angular frequency is not equal to zero according to the equation above. As a result, $B$value must be zero.

$B=0$

Determine thecurrent in the circuit as a function of time.

Substitute $CV$ for $A$, $\frac{1}{\sqrt{LC}}$for $\omega$ and $0$ for $B$ in the equation (1).

$\begin{array}{c}I\left(t\right)=-CV\sqrt{\frac{1}{LC}}\sin \left(\frac{1}{\sqrt{LC}}t\right)\\ =-V\sqrt{\frac{C}{V}}\sin \left(\frac{t}{\sqrt{LC}}\right)\end{array}$

Therefore, thevalue of current in the circuit as a function of time is $-V\sqrt{\frac{C}{L}}\sin \left(\frac{t}{\sqrt{LC}}\right)$The resistor's addition dampens the circuit's oscillation. Hence, the voltage across the resistor is IR.

$-L\frac{dI}{dt}=\frac{Q}{C}+IR$

Substitute $\frac{dQ}{dt}$ for $I$into above equation.

$\begin{array}{c}-L\frac{d\left(\frac{dQ}{dt}\right)}{dt}=\frac{Q}{C}+IR\\ \frac{d^{2}Q}{d{t}^2}+\left(\frac{R}{L}\right)\frac{dQ}{dt}+\frac{Q}{LC}=0\end{array}$

Substitute $CV$for $Q$into above equation.

$C(\frac{d^{2}V}{d{t}^2}+\left(\frac{R}{L}\right)\frac{dV}{dt}+\frac{V}{LC})=0$

The Capacitance will never equal 0 according to the equation above. Hence,

$\frac{d^{2}V}{d{t}^2}+\left(\frac{R}{L}\right)\frac{dV}{dt}+\frac{V}{LC}=0$

The second order differential equation with constant coefficients is represented by the aforementioned equation.

Hence, the solution for the above equation is as follows:

$V=A{e}^{-at}\cos \omega t$

Here, $A$, $\alpha$and $\omega$are constants.

Differential the equation $V=A{e}^{-at}\cos \omega t$ with respect to time on both sides.

$\frac{dV}{dt}=A{e}^{-at}(-\alpha \cos \omega t-\omega \sin \omega t)$

Again differential the above equation with respect to the time.

$\frac{d^{2}V}{d{t}^2}=A{e}^{-at}[({\alpha }^2-{\omega }^2)\cos \omega t+2\alpha \omega \sin \omega t]$

Substitute the values of $V$,$\frac{dV}{dt}$ for $\frac{d^{2}V}{d{t}^2}$ in the equation $\frac{d^{2}V}{d{t}^2}+\left(\frac{R}{L}\right)\frac{dV}{dt}+\frac{V}{LC}=0$.

$\left[\begin{array}{l}A{e}^{-at}[(a^2-{\omega }^2)\cos \omega t+2\alpha \omega \sin \omega t]\\ +\left(\frac{R}{L}\right)\left(A{e}^{-at}(-\alpha \cos \omega t-\omega \sin \omega t)\right)+\frac{1}{LC}\left(A{e}^{-at}\cos \omega t\right)\end{array}\right]=0$

As both $\sin \omega t$and $\cos \omega t$coefficients are equal to zero, the aforementioned equation is fulfilled.

$2\alpha \omega -\frac{R\omega }{L}=0$

Solve the equation for $\alpha$.

$\alpha =\frac{R}{2L}$

Equate the $\cos \omega t$ coefficients equal to zero.

${\alpha }^2-{\omega }^2-\alpha \frac{R}{L}+\frac{1}{LC}=0$

Solve the equation for${\omega }^2$.

$\begin{array}{c}{\omega }^2=\frac{1}{LC}-\left(\frac{R}{2L}\right)\frac{R}{L}+{\left(\frac{R}{2L}\right)}^2\\ =\frac{1}{LC}-\frac{R^2}{4L^2}\end{array}$

There are more options than using the formula $A{e}^{-at}\cos \omega t$. Consequently, the following is the general solution to the second order differential equation:

$V\left(t\right)=e^{-at}(A\cos \omega t+B\sin \omega t)$

The following is the current flowing through the circuit as a function of time:

Determine thecurrent if a resistor $\text{R}$ is included in serieswith $\text{C}$and $\text{L}$.

Substitute $e^{-at}(A\cos \omega t+B\sin \omega t)$for $V\left(t\right)$into equation (2).

$\begin{array}{c}I=C\frac{d}{dt}\left(e^{-at}(A\cos \omega t+B\sin \omega t)\right)\\ =AC\omega (\sin \omega t+\frac{\alpha }{\omega }\cos \omega t)e^{-at}\end{array}$

Substitute the values of $R$ and $\omega$ in the above equation and simplify.

$I=AC{(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}(\sin {(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}t+\frac{\frac{R}{2L}}{{(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}}\cos {(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}t)e^{\frac{R}{2L}I}$

The current flowing through the circuit when resistance is introduced is represented by the equation above.