Differential Equations RLC Circuit

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Understanding RLC Differential Equation. (Krantz)

* .... Differential Equations for solving Engineering problems.

* .... have method that we know to solve these equations.

Formulation: Mathematics in physics can be used to make predictions even in the case of electric circuits. This can be shown in the case of periodic motion in particular as in the case of springs or wave motion. Applying the same math to circuits that oscillate can be done as well.

When we solved previous differential equations we used the fact that they were Separable. We shall use the same problem solving style in this set of solutions when we get to the ezamples.(Krantz)

Let us begin below by looking at the different circuit elements: the resistor (R), the inductor (L), and the capacitor (C). Also included in the diagram below is the symbol for a power supply or source E. At the bottom of the page is depicted a simple single loop RLC circuit. The resistor acts as a simple current inducer drawing current from the voltage source. The inductor and capacitor, however, act differently with characteristics that actually oscillate or display periodic motion. For this reason differential equations are ideal for finding the equation for the circuit. Of course the current in the circuit can be characterized by a flow of charge carriers, or Q (representing a single charge). A stream of charges over time, moveover, can be described as I = dQ/dt or the number of charge carriers per unit time. Ohm's law helps to describe the voltage over the resistor, V = I R or E = I R in using the variables shown.

Next, below we use Kirchhof's voltage law describing the fact that the sum of the voltages around any closed loop to be zero in order to achieve our first equation. Looking back we see that the voltage for the resistor is E = I R. The voltage for the inductor is E = L dI/dt. Finally, the voltage for the capacitor is E = Q / C.

So, now our equation has the form E = RI + L dI/dt + Q/C. We want an expression that is described purely in terms of I, thus we take the derivative of E with respect to t, or dE/dt getting dE/dt = R dI/dt + L d^2I/dt^2 + (1/C)(dQ/dt). But, remember that I = dQ/dt so that dE/dt = R dI/dt + L d^2I/dt^2 + (1/C)I

In other cases we may want to keep the Q and eliminate I. To do this we need I = dQ/dt AND dI = d^2Q/dt^2. Solving for E we get E = R dQ/dt + L d^2Q/dt^2 + Q/C.

That being said we now have what we need to get started with some examples....

Example of an R-L circuit ... a simple R-L circuit

Back to the Hanging Chain ... Hanging Chain

Go back to Gauss' Law Problem from here!

1st Gauss' Law Example problem ... Gauss Law Example 1.

Navigate below to go back to the start!

You can go to DE Ex2 here...D.E. Ex2.

You can go to DE Ex1 here ...D.E. Ex1.