Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

$\mathbf{RLC}$Series Circuit. In the study of an electrical circuit consisting of a resistor, capacitor, inductor, and an electromotive force (see Figure), we are led to an initial value problem of the form

$\begin{array}{r} \mathbf{\left(}\mathbf{20}\mathbf{\right)}\mathbf{L}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{+}\mathbf{RI}\mathbf{+}\frac{q}{C}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{;}\\ \mathbf{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{q}}_0\\ \mathbf{I}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{I}}_0\mathbf{,}\end{array}$

where $\mathbf{L}$ is the inductance in henrys, $\mathbf{R}$ is the resistance in ohms, $\mathbf{C}$ is the capacitance in farads,  $\mathbf{E}\left(t\right)$ is the electromotive force in volts,$\mathbf{q}\left(t\right)$  is the charge in coulombs on the capacitor at the time $\mathbf{t}$, and   $\mathbf{I}\mathbf{=}\mathbf{dq}\mathbf{/}\mathbf{dt}$ is the current in amperes. Find the current at time t if the charge on the capacitor is initially zero, the initial current is zero,  $\mathbf{L}\mathbf{=}\mathbf{10}\mathbf{H}\mathbf{,}\mathbf{R}\mathbf{=}\mathbf{20}\mathbf{\Omega }\mathbf{,}\mathbf{C}\mathbf{=}\mathbf{(}\mathbf{6260}{\mathbf{)}}^{-1}\mathbf{ }\mathbf{F}$ , and  $\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{ }\mathbf{V}$ . 

Step-by-Step Solution

Answer

The current at the time $\mathrm{t}$  is $\mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{I}}_0\cos 25\mathrm{t}+\frac{10-{\mathrm{I}}_0-6260{\mathrm{q}}_0}{}\sin 25\mathrm{t}\right)$.

Step 1: Differentiate by time

Given equation is $\mathbf{L}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{+}\mathbf{RI}\mathbf{+}\frac{q}{C}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}.$

 

We will differentiate both sides of the given equation with respect to time: 

$\mathbf{L}\frac{{\mathbf{d}}^2\mathbf{I}}{{\mathbf{dt}}^2}\mathbf{+}\mathbf{R}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{+}\frac{1}{C}\frac{\mathrm{dq}}{\mathrm{dt}}\mathbf{=}\frac{\mathrm{dE}}{\mathrm{dt}}$

We know that  $\mathrm{E}\left(\mathrm{t}\right)=100 \mathrm{V}$, so the electromotive force is constant with time and therefore  $\frac{\mathrm{dE}}{\mathrm{dt}}=0$.

Substituting this and $\frac{\mathrm{dq}}{\mathrm{dt}}\mathbf{=}\mathbf{I}$ into the previous equation, we get that;

$\mathbf{L}\frac{{\mathbf{d}}^2\mathbf{I}}{{\mathbf{dt}}^2}\mathbf{+}\mathbf{R}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{+}\frac{I}{C}\mathbf{=}\mathbf{0}$ .

 

Since $\mathrm{L}=10\mathrm{H},\mathrm{R}=20\mathrm{\Omega }$ and  $\mathrm{C}=(6260{)}^{-1} \mathrm{F}$ this is a homogeneous linear second-order equation and its auxiliary equation is;

${\mathrm{Lr}}^2+\mathrm{Rr}+\frac{1}{\mathrm{C}}=0$ .

Step 2: Finding roots.

Now we will find its roots ${\mathbf{Lr}}^2\mathbf{+}\mathbf{Rr}\mathbf{+}\frac{1}{C}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{r}}_{1,2}\mathbf{=}\frac{\mathbf{-}\mathbf{R}\mathbf{\pm }\sqrt{{\mathbf{R}}^2\mathbf{-}\mathbf{4}\mathbf{L}\mathbf{/}\mathbf{C}}}{2L}$

 

All quantities are given in their basic units. From now on we will drop the units, so our result will be valid only if we measure the current in Amperes $\left(\mathrm{A}\right)$  and time in seconds $\left(\mathrm{s}\right)$.

$\begin{array}{c}{\mathrm{r}}_{1,2}=\frac{-20\pm \sqrt{400-4\times 10\times 6260}}{2\times 10}\\ =\frac{-20\pm 500\mathrm{i}}{20}\\ =-1\pm 25\end{array}$

Since the roots of the auxiliary equation are complex, the general solution has the form of  $\mathrm{I}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$, for  $\mathrm{r}=\mathrm{\alpha }\pm \mathrm{\beta i}$ 

 

Since $\mathrm{\alpha }=-1$  and  $\mathrm{\beta }=25$, the general solution is  $\mathrm{I}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}\cos 25\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}\sin 25\mathrm{t}$. 

Step 3: Finding c 1 and c 2

Now we will find the constants ${\mathrm{c}}_1$  and ${\mathrm{c}}_2$  from the initial conditions.

$\begin{array}{l}\mathrm{I}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^0\cos 0+{\mathrm{c}}_2{\mathrm{e}}^0\sin 0\\ {\mathrm{c}}_1={\mathrm{I}}_0\\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{I}}_0{\mathrm{e}}^{-\mathrm{t}}\cos 25\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}\sin 25\mathrm{t}\end{array}$

The second initial condition is $\mathbf{q}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{q}}_0$. From the initial differential equation, we have that  $\mathrm{q}=\mathrm{C}\times \mathrm{E}-\mathrm{C}\times \mathrm{L}\frac{\mathrm{dl}}{\mathrm{dt}}-\mathrm{C}\times \mathrm{RI}$ . So, we need to find the derivative of $\mathbf{l}$ with respect to $\mathbf{t}$ :

 

 $\begin{array}{c}\frac{\mathrm{dI}}{\mathrm{dt}}\mathbf{=}\frac{d}{\mathrm{dt}}({\mathbf{I}}_0{\mathbf{e}}^{-t}\mathbf{cos}\mathbf{25}\mathbf{t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{sin}\mathbf{25}\mathbf{t})\\ \mathbf{=}\mathbf{-}{\mathbf{I}}_0{\mathbf{e}}^{-t}\mathbf{cos}\mathbf{25}\mathbf{t}\mathbf{-}\mathbf{25}{\mathbf{I}}_0{\mathbf{e}}^{-t}\mathbf{sin}\mathbf{25}\mathbf{t}\mathbf{-}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{sin}\mathbf{25}\mathbf{t}\mathbf{+}\mathbf{25}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{sin}\mathbf{25}\mathbf{t}\\ \mathbf{=}(\mathbf{25}{\mathbf{c}}_2\mathbf{-}{\mathbf{I}}_0){\mathbf{e}}^{-t}\mathbf{cos}\mathbf{25}\mathbf{t}\mathbf{-}(\mathbf{25}{\mathbf{I}}_0\mathbf{+}{\mathbf{c}}_2){\mathbf{e}}^{-t}\mathbf{sin}\mathbf{25}\mathbf{t}\\ \frac{\mathrm{dI}\left(0\right)}{\mathrm{dt}}\mathbf{=}(\mathbf{25}{\mathbf{c}}_2\mathbf{-}{\mathbf{I}}_0){\mathbf{e}}^0\mathbf{cos}\mathbf{0}\mathbf{-}(\mathbf{25}{\mathbf{I}}_0\mathbf{+}{\mathbf{c}}_2){\mathbf{e}}^0\mathbf{sin}\mathbf{0}\\ \mathbf{=}\mathbf{25}{\mathbf{c}}_2\mathbf{-}{\mathbf{I}}_0\end{array}$

Now we have;

 $\begin{array}{l}\mathbf{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{C}\mathbf{\times }\mathbf{E}\mathbf{-}\mathbf{C}\mathbf{\times }\mathbf{L}\frac{\mathrm{dI}\left(0\right)}{\mathrm{dt}}\mathbf{-}\mathbf{C}\mathbf{\times }\mathbf{RI}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\\ \mathbf{=}\mathbf{(}\mathbf{6260}{\mathbf{)}}^{-1}\mathbf{\times }\mathbf{100}\mathbf{-}\mathbf{(}\mathbf{6260}{\mathbf{)}}^{-1}\mathbf{\times }\mathbf{10}\mathbf{\times }(\mathbf{25}{\mathbf{c}}_2\mathbf{-}{\mathbf{I}}_0)\mathbf{-}\mathbf{(}\mathbf{6260}{\mathbf{)}}^{-1}\mathbf{\times }\mathbf{20}\mathbf{\times }{\mathbf{I}}_0\mathbf{=}{\mathbf{q}}_0\\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{100}\mathbf{-}\mathbf{250}{\mathbf{c}}_2\mathbf{+}\mathbf{10}{\mathbf{I}}_0\mathbf{-}\mathbf{20}{\mathbf{I}}_0\mathbf{=}\mathbf{6260}{\mathbf{q}}_0\\ {\mathbf{c}}_2\mathbf{=}\frac{\mathbf{10}\mathbf{-}{\mathbf{I}}_0\mathbf{-}\mathbf{6260}{\mathbf{q}}_0}{25}\end{array}$

Finally, the current at the time  $\mathrm{t}$ is  $\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{-t}\left(I_0\cos 25t+\frac{10-I_0-6260q_0}{}\sin 25t\right)$.