Question

Find the charge on the capacitor and the current in an LRC circuit when L = 1 h, R= 100 Ω , C= 0.0004f and E(t)= 30 V, q(0)=0C, i(0)=2A.

Short answer

The charge on the capacitor is ${\mathrm{q}}_{\max }=0.0187\mathrm{C}$and the current in an LRC circuit isi(t) = e^-50t (2-70t) A.

Step-by-step solution

Definition:

As in the case of forced oscillations of a spring-mass system with damping, we call Qp the steady state charge on the capacitor of the LRC circuit. Since $\mathbf{I}\mathbf{=}\mathbf{Q}\mathbf{\prime }\mathbf{=}\mathbf{Q}\mathbf{\prime }\mathbf{c}\mathbf{+}\mathbf{Q}\mathbf{\prime }\mathbf{p}$and Q′c also tends to zero exponentially as $\mathrm{t}\to \infty$, you can say that Ic=Q′c is the transient current and Ip=Q′p is the steady state current.

In LCR circuit:

In the LRC series electric circuits, you have

$\begin{array}{l}\mathrm{L}\frac{{\mathrm{d}}^2\mathrm{q}}{{\mathrm{dt}}^2}+\mathrm{R}\frac{\mathrm{dq}}{\mathrm{dt}}+\frac{1}{\mathrm{C}}\mathrm{q}=\mathrm{E}\left(\mathrm{t}\right)\\ \\ \frac{{\mathrm{d}}^2\mathrm{q}}{{\mathrm{dt}}^2}+100\frac{\mathrm{dq}}{\mathrm{dt}}+2500\mathrm{q}=30\end{array}$

The characteristics equation is given by:

$m^2+100m+2500=0$

Whose a repeated root -50. Thus the complementary solution is:

${\mathrm{q}}_{\mathrm{c}}\left(\mathrm{t}\right)={\mathrm{e}}^{-50\mathrm{t}}({\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2\mathrm{t})$

Solve further:

Assume that the particular solution is q_p(t)= A and substituting A= 3/250 ,

${\mathrm{q}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{3}{250}$

role="math" localid="1668508899073" ${\mathrm{q}}_{\mathrm{c}}\left(\mathrm{t}\right)={\mathrm{e}}^{-50\mathrm{t}}\left(\frac{7}{5}\mathrm{t}-\frac{3}{250}\right)+\frac{3}{250}\mathrm{c}$

Using the initial condition,

You get c₁= -3/250 and c₂= 7/5 . Therefore,

$\begin{array}{l}q\left(t\right)=e^{-50t}\left(\frac{7}{5}t-\frac{3}{250}\right)+\frac{3}{250}c\\ \\ i\left(t\right)=q\text{'}\left(t\right)=e^{-50t}(2-70t)A\\ \\ e^{-50t}(2-70t)=0\\ \\ \therefore q(1/35)=\frac{1}{250}(7e^{-10/7}+3)\\ \end{array}$

Thus, the final answer is :

$q_{\max }\approx 0.0187c$