Q7.27P - Introduction to Electrodynamics

Question

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 $\mathbf{L}$ ?

Step-by-Step Solution

Answer

The value of current in the circuit as a function of time is  $-V\sqrt{\frac{C}{L}}\sin \left(\frac{t}{\sqrt{LC}}\right)$.

The value of current if a resistor $\text{R}$  is included in series with  $\text{C}$ and $\text{L}$ is  $I=AC{(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}(\sin {(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}t+\frac{\frac{R}{2L}}{{(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}}\cos {(\frac{1}{LC}-\frac{R^2}{4L^2})}^{1/2}t)e^{\frac{R}{2L}I}$.

Step 1: Write the given data from the question.

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.

Step 2: Determine the formulaof current in the circuit as a function of time and current if a resistor R is included in series with C and L .

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,   $\mathbf{A}$ 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.

Step 3:Determine the current in the circuit as a function of time and current if a resistor R is included in series with C and L .

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 .$\epsilon$

 

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

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 the current 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, the value 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 the current if a resistor $\text{R}$  is included in series  with   $\text{C}$ 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.