Question

Find the charge on the capacitor in an L R C-series circuit when L=1/2 h, R=100 Ω , C=0.01 f, E(r)=150 V, q(0)=1C 56 and i(0)= 0 A. What is the charge on the capacitor after a long time?

Short answer

After a along time, the charge tends to be $1.5C$.

Step-by-step solution

Definition Of LRC Series circuits:

LRC-SERIES CIRCUITS As 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:

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

Using LRC series and find characteristic equation:

In the LRC series electric circuits, you have

$L\left(\frac{d^{2}q}{d{t}^2}\right)+R\left(\frac{dq}{dt}\right)+\frac{q}{C}=E\left(t\right)$

So that

$\frac{1}{2}\left(\frac{d^{2}q}{d{t}^2}\right)+10\left(\frac{dq}{dt}\right)+\frac{q}{0.01}=150$

Simply,

role="math" localid="1668585654887" $\left(\frac{d^{2}q}{d{t}^2}\right)+20\left(\frac{dq}{dt}\right)+200q=300$

The characteristic equation is given by

role="math" localid="1668585705176" $m^2+20m+200=0$

Whose roots are $-10+10i$. Thus the complementary solution is

$q_c\left(t\right)=e(c_1\cos 10t+c_2\sin 10t)$

Assume that the particular solution is $q_p\left(t\right)=A$and substituting n the differential equation to get $A=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Therefore,

$q_p\left(t\right)=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

And so,

$q_p\left(t\right)=e^{-10t}\left(c_1\cos 10t+c_2\sin 10t\right)+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Hence. after a long time the charge tends to be role="math" localid="1668585954728" $1.5C$.