Question 3.8.10: Find the steady-state solution qp(t) and the steady-state cu......
Find the steady-state solution q_{p}(t) and the steady-state current in an L R C-series circuit when the impressed voltage is E(t)=E_{0} \sin \gamma t.
The steady-state solution q_{p}(t) is a particular solution of the differential equation
L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{1}{C} q=E_{0} \sin \gamma t
Using the method of undetermined coefficients, we assume a particular solution of the form q_{p}(t)=A \sin \gamma t+B \cos \gamma t. Substituting this expression into the differential equation, simplifying, and equating coefficients gives
A=\frac{E_{0}\left(L \gamma-\frac{1}{C \gamma}\right)}{-\gamma\left(L^{2} \gamma^{2}-\frac{2 L}{C}+\frac{1}{C^{2} \gamma^{2}}+R^{2}\right)}, \quad B=\frac{E_{0} R}{-\gamma\left(L^{2} \gamma^{2}-\frac{2 L}{C}+\frac{1}{C^{2} \gamma^{2}}+R^{2}\right)}
It is convenient to express A and B in terms of some new symbols.
\begin{aligned} & \text { If } X=L \gamma-\frac{1}{C \gamma}, \quad \text { then } \quad X^{2}=L^{2} \gamma^{2}-\frac{2 L}{C}+\frac{1}{C^{2} \gamma^{2}} \\ & \text { If } Z=\sqrt{X^{2}+R^{2}}, \quad \text { then } \quad Z^{2}=L^{2} \gamma^{2}-\frac{2 L}{C}+\frac{1}{C^{2} \gamma^{2}}+R^{2}. \end{aligned}
Therefore A=E_{0} X /\left(-\gamma Z^{2}\right) and B=E_{0} R /\left(-\gamma Z^{2}\right), so the steady-state charge is
q_{p}(t)=-\frac{E_{0} X}{\gamma Z^{2}} \sin \gamma t-\frac{E_{0} R}{\gamma Z} \cos \gamma t.
Now the steady-state current is given by i_{p}(t)=q_{p}^{\prime}(t) :
i_{p}(t)=\frac{E_{0}}{Z}\left(\frac{R}{Z} \sin \gamma t-\frac{X}{Z} \cos \gamma t\right). (35)