Question

Find the charge on the capacitor and the current in an L C-series circuit when$\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\gamma }\mathit{t}\mathbf{\text{V}}\mathbf{,}\mathit{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{q}}_{\mathbf{0}}\mathbf{\text{C}}$, and $\mathit{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{i}}_{\mathbf{0}}\mathbf{\text{A}}$.

Short answer

$q\left(t\right)=q_{0}C\cos \frac{t}{\sqrt{LC}}+i_0\sqrt{LC}\sin \frac{t}{\sqrt{LC}}+\frac{E_{0}C}{1-CL{\gamma }^2}\left(\cos \gamma t-\cos \frac{t}{\sqrt{LC}}\right)$

Step-by-step solution

Definition Of LRC Series circutes

LRC-SERIES CIRCUITS As was mentioned in the introduction to this chapter, many different physical systems can be described by a linear second-order differential equation similar to the differential equation of forced motion with damping:

$m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right)$

Using LRC series and find characteristic equation

In the -series electric circuits, we have

$L\frac{d^{2}q}{d{t}^2}+\frac{1}{C}q=E\left(t\right).$

Simply,

$\frac{d^{2}q}{d{t}^2}+\frac{1}{LC}q=\frac{E_0}{L}\cos \gamma t.$

The characteristic equation is given

$m^2+\frac{R}{L}=0$

whose roots are $\pm \frac{1}{\sqrt{LC}}i$. Thus the complementary solution is

$q_c\left(t\right)=c_1\cos \frac{t}{\sqrt{LC}}+c_2\sin \frac{t}{\sqrt{LC}}.$

If$\gamma \ne \frac{1}{\sqrt{LC}}$, assume that the particular solution is$q_p\left(t\right)=A\cos \gamma t+B\sin \gamma t$and substitutingthe differential equation to get$A=\frac{E_{0}C}{1CL{\gamma }^2}$and$B=0$. Therefore

$q_p\left(t\right)=\frac{E_{0}C}{1-CL{\gamma }^2}\cos \gamma t$

and so

$q\left(t\right)=c_1\cos \frac{t}{\sqrt{LC}}+c_2\sin \frac{t}{\sqrt{LC}}+\frac{E_{0}C}{1-CL{\gamma }^2}\cos \gamma t$

 Step 3: Substitute the initial conditions

Using the initial conditions,$q\left(0\right)=q_{0}C$and$q^{\text{'}}\left(0\right)=i\left(0\right)=i_0$, we get

$q_{0}C=c_1+\frac{E_{0}C}{1-LC{\gamma }^2}$

implies

$c_1=q_{0}C-\frac{E_{0}C}{1-LC{\gamma }^2},$

and

$q^{\text{'}}\left(t\right)=-c_1\frac{1}{\sqrt{LC}}\sin \frac{t}{\sqrt{LC}}+c_2\frac{1}{\sqrt{LC}}\cos \frac{t}{\sqrt{LC}}-\frac{E_{0}C\gamma }{1-LC{\gamma }^2}\sin \gamma t$

which makes

$i_0=c_2\frac{1}{\sqrt{LC}}$

implies

$c_2=i_0\sqrt{LC}$

Thus

$q\left(t\right)=\left(q_{0}C-\frac{E_{0}C}{1-LC{\gamma }^2}\right)\cos \frac{t}{\sqrt{LC}}+i_0\sqrt{LC}\sin \frac{t}{\sqrt{LC}}+\frac{E_{0}C}{1-CL{\gamma }^2}\cos \gamma t$

$q\left(t\right)=q_{0}C\cos \frac{t}{\sqrt{LC}}+i_0\sqrt{LC}\sin \frac{t}{\sqrt{LC}}+\frac{E_{0}C}{1-CL{\gamma }^2}\left(\cos \gamma t-\cos \frac{t}{\sqrt{LC}}\right)$