6 Engineering Example 2

6.1 An LC circuit with sinusoidal input

The differential equation governing the flow of current in a series LC circuit when subject to an applied voltage $v\left(t\right)=V_{0}sin\omega t$ is $L\frac{d^{2}i}{d{t}^2}+\frac{1}{C}i=\omega V_{0}cos\omega t$

Figure 3

Obtain its general solution.

Solution

The homogeneous equation is $L\frac{d^2{i}_{\text{cf}}}{d{t}^2}+\frac{i_{\text{cf}}}{C}=0.$

Letting $i_{\text{cf}}={\text{e}}^{kt}$ we find the auxiliary equation is $L{k}^2+\frac{1}{C}=0$ so that $k=\pm \text{i}∕\sqrt{LC}$ . Therefore, the complementary function is:

$i_{\text{cf}}=Acos\frac{t}{\sqrt{LC}}+Bsin\frac{t}{\sqrt{LC}}\text{ where}A\text{and}B\text{arbitrary constants.}$

To find a particular integral try $i_{\text{p}}=Ecos\omega t+Fsin\omega t$ , where $E$ , $F$ are constants. We find:

$\frac{d{i}_{\text{p}}}{dt}=-\omega Esin\omega t+\omega Fcos\omega t\frac{d^2{i}_{\text{p}}}{d{t}^2}=-{\omega }^{2}Ecos\omega t-{\omega }^{2}Fsin\omega t$

Substitution into the inhomogeneous equation yields:

$L\left(-{\omega }^{2}Ecos\omega t-{\omega }^{2}Fsin\omega t\right)+\frac{1}{C}\left(Ecos\omega t+Fsin\omega t\right)=\omega V_{0}cos\omega t$

Equating coefficients of $sin\omega t$ gives: $-{\omega }^{2}LF+\left(F∕C\right)=0$ .

Equating coefficients of $cos\omega t$ gives: $-{\omega }^{2}LE+\left(E∕C\right)=\omega V_0$ .

Therefore $F=0$ and $E=C{V}_0\omega ∕\left(1-{\omega }^{2}LC\right)$ . Hence the particular integral is

$i_{\text{p}}=\frac{C{V}_0\omega }{1-{\omega }^{2}LC}cos\omega t.$

Finally, the general solution is:

$i=i_{\text{cf}}+i_{\text{p}}=Acos\frac{t}{\sqrt{LC}}+Bsin\frac{t}{\sqrt{LC}}+\frac{C{V}_0\omega }{1-{\omega }^{2}LC}cos\omega t$