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Chris L T521 · #10 December 27th, 2008, 10:56 AM

Something told me now is the best time to start updating this tutorial, especially since I'm currently on vacation from school. This post will be on electrical circuits.

Electrical Circuits

First, let me start off with a diagram.

The type of circuit we will analyze will be a RLC circuit.

In a RLC circuit, there is a(n):

- resistor with resistance $R$ ohms ( $\Omega$ )
- inductor with inductance $L$ henries
- capacitor with capacitance $C$ farads.
- Source with voltage $E$ volts.

There is a nice relationship between current, $I$ , and charge, $Q$ . Current is the rate of change of the flow of charges. Thus, we can say $I=\frac{\,dQ}{\,dt}$ .

According to the fundamental elementary principles of electricity, we see that the voltage drop across the three elements are as follows:

- Across a resistor, the voltage drop is resistance times current, or $RI$ .

- Across an inductor, the voltage drop is the inductance times the rate of change in the current, or $L\frac{\,dI}{\,dt}$ .

- Across a capacitor, the voltage drop is the charge divided by capacitance, or $\frac{1}{C}Q$ .

Now, we can analyze the behavior of the circuit by using one of Kirchoff's Laws:

The (Algebraic) sum of the voltage drops across the elements in a simple loop of an electrical circuit is equal to the applied voltage.

Thus, we see that if $E\left(t\right)$ is the applied voltage from the source, we get the equation $L\frac{\,dI}{\,dt}+RI+\frac{1}{C}Q=E\left(t\right)$ .

Using the relationship between current and charge, we can rewrite this as the second order non-homogeneous DE $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$ .

The solution to this differential equation, of course, is the amount of charge, $Q$ , at any given time, $t$ .

However, we usually want to solve for current. You can solve for current in one of two ways:

1) If you differentiate both sides of the differential equation, we get $L\frac{\,d^3Q}{\,dt^3}+R\frac{\,d^2Q}{\,dt^2}+\frac{1}{C}\frac{\,dQ}{\,dt}=E^{\prime}\left(t\right)\implies L\frac{\,d^2I}{\,dt^2}+R\frac{\,dI}{\,dt}+\frac{1}{C}I=E^{\prime}\left(t\right)$ . Now the solution to the DE is current, $I$ .

2) Once you solve $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$ , differentiate the solution to get current, $I$ .

Now, what happens if we are not dealing with a RLC Circuit?!?! We make slight modifications.

- If we have a RL Circuit, we solve the differential equation $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}=E\left(t\right)$

- If we have a RC Circuit, we solve the differential equation $R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$

- If we have a LC Circuit, we solve the differential equation $L\frac{\,d^2Q}{\,dt^2}+\frac{1}{C}Q=E\left(t\right)$

Let us go through a couple of examples:

Example 22

Suppose that in an RLC Circuit, we have a resistance of 60 $\mathit{\Omega}$ , an inductance of 2 henries, and a capacitance of 0.0025 farads. Now, let the circuit have an emf of $\mathit{E\left(t\right)=100e^{-10t}}$ volts. Find the current in the circuit, given that the inital current in the circuit is zero, and the charge on the capacitor is one coloumb.

This is another initial value problem.

Here, we are to solve the DE $2\frac{\,d^2Q}{\,dt^2}+60\frac{\,dQ}{\,dt}+\frac{1}{0.0025}Q=100e^{-10t}$ , where $I\left(0\right)=0,~Q\left(0\right)=1$ .

The DE can be rewritten as $\frac{\,d^2Q}{\,dt^2}+30\frac{\,dQ}{\,dt}+200Q=50e^{-10t}$

In solving the homogeneous equation, we get the characteristic equation $r^2+30r+200=0\implies \left(r+20\right)\left(r+10\right)=0$ . We now see that this gives us $r_1=-20$ and $r_2=-10$ .

Thus, our complimentary solution is $Q_c=c_1e^{-20t}+c_2e^{-10t}$

Now, to find the particular solution, I will apply the method of the annihilator (in a sense, the annihilator method leads to the method of undetermined coefficients).

Rewriting the DE in differential operator notation, we get $\left(D^2+30D+200\right)\left(y\right)=50e^{-10t}$

The term that annihilates $50e^{-10t}$ is $\left(D+10\right)$

Thus, applying the annihilator to both sides of the DE, we can then convert the DE to the characteristic equation $(r+10)\left(r^2+30r+200\right)=0$

The particular solution is $r_p=-10$

Thus, it will take on the form $Q_p=Ate^{-10t}$

Now, substituting this into the original DE, we get $10Ae^{-10t}=50e^{-10t}$ (Verify)

You now get that $A=5$

Thus, the general solution to the DE is $Q\left(t\right)=c_1e^{-20t}+c_2e^{-10t}+5te^{-10t}$

Let us now apply the initial conditions:

$Q\left(0\right)=1\implies 1=c_1+c_2$

To apply the second condition, find $I\left(t\right)$

$I\left(t\right)=\frac{\,dQ}{\,dt}=-20c_1e^{-20t}-10c_2e^{-10t}-50te^{-10t}+5e^{-10t}$

Thus, $I\left(0\right)=0\implies 0=-20c_1-10c_2+5$ .

Solving these two equations for $c_1$ and $c_2$ , we get $c_1=-\tfrac{1}{2}$ and $c_2=-\tfrac{3}{2}$ (Verify)

Thus, the current is $I\left(t\right)=-\tfrac{1}{2}\left(-20\right)e^{-20t}-\left[\tfrac{3}{2}\left(-10\right)+5\right]e^{-10t}-50te^{-10t}=\color{red}\boxed{10e^{-20t}+10e^{-10t}-50te^{-10t}}$

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I will post another example when I find the time later today

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