0:03

The topic of this problem is The Complete Response of RLC Circuits.

Â The problem is to find the 2nd order differential equation expression for

Â the voltage, Vc(t) in the circuit shown below.

Â So, we have a circuit that has a series combination of R, Ls and Cs.

Â It also has a voltage source, VS sub t.

Â The voltage Vc(t) is the voltage that's across

Â the compositor in the top of our circuit.

Â So we're going to use Kirchhoff's voltage law, and

Â sum up the voltages around our single loop circuit.

Â To find an expression which will ultimately give us our route

Â to finding our second order differential equation for Vc(t).

Â So if we use Kirchhoff's voltage law, it's a single loop circuit,

Â so only one loop to choose from, and we're going to travel

Â clockwise around our circuit starting in the lower left corner.

Â The first thing we encounter is a negative polarity of the voltage source Vs(t).

Â As we continue around this, we run into the inductor and

Â the voltage across the inductor.

Â We know that the voltage across

Â the inductor is L di, dt.

Â And it's an I of t it's a function of time,

Â and this is our single loop current I of t.

Â We also have our voltage Vc(t), it's part of this expression

Â as well because we capture the capacitance next and then we have the resistor.

Â And the voltage throughout across the resistor is R times i(t),

Â and so the sum of those is equal to 0.

Â So if we use our well known expression for the capacitor that is,

Â that the current through the capacitor is equal to the capacitance times,

Â 2:19

The time derivative of the voltage across a capacitor

Â then we can rewrite our expression that we have above.

Â And so we have our Vc (t) here.

Â We have a Vc(t) in our expression, our Kirchhoff's voltage law expression,

Â and ultimately we're going to rewrite this, so that we end up with L times C.

Â And then the second time derivative of the voltage across the capacitor.

Â 2:54

And what we've done is we've plugged our expression for

Â the current into our second term,

Â take in the derivative of it and we end up with this.

Â And then we have still our Vc (t) for

Â the voltage or up across the capacitor a third term.

Â We have our fourth term that we can plug in for using our expression that we R for

Â the current and voltage associated with the capacitor.

Â And so, that would be Rc(dVc(t) / dt) and

Â all those are going to be equal to the source

Â voltage Vs(t) pointing to the other side.

Â So if we continue to put this in a form that is standard for

Â second-order differential equations,

Â just moving terms around, we'll get this expression for

Â our series combination of R, Ls and Cs.

Â 4:22

Again, it's equal to our forcing function on the right hand side of this circuit.

Â And the forcing function for us is Vs(t) / LC.

Â So, here's our standard format for our second order

Â differential equation expression for the voltage Vc(t).

Â