Homework Help: Second Order RLC Circuit

1. Oct 4, 2015

Roy Fokker

1. The problem statement, all variables and given/known data
Find DEQ, circuit for time t>0.

2. Relevant equations
Ic= CdVc/dt
Vl=LdiL/dt

3. The attempt at a solution
At T=inf
L Short , C Open
IL(inf)=4mA, VL(inf)=0, Ic(inf)=0, Vc(inf)=0

T=0-

All at 0

T=0+
L=Open C=Short
IL(0+)=0, VL(0+)=0, Ic(0+)=4mA, Vc(0+)=0

20v=L diL/dt + Vc + iL5k

I need to make a sub for iL but I feel that I am wrong in my subbing in.
iL= Vc/2k + C dVc/dt

I keep wanting to just say iL=Vs/5k but I need to account for the transient effect of the cap charging and discharging... it just seems odd not to include R1 in iL since thats how I came up with it in the first place.

2. Oct 4, 2015

axmls

Regarding your initial conditions, the values for the current and voltage of inductors and capacitors cannot change instantaneously (discontinuously), so the values at 0- are the same as the values at 0+ (but I haven't had a chance to look closely at your work yet).

Here's a hint to avoid substitutions: you can write one nodal equation and solve for that, and you can derive all of the currents using that.

3. Oct 4, 2015

Roy Fokker

Yes Vc(0-)=Vc(0+) and iL(0+)=iL(0-)

But cant current can change instantaneously in a capacitor and voltage can instantaneously change in an inductor.
I say this because the equations for Ic and Vl where the derivatives are time based so it would.. at T=0 be zero.
Ic(0-)=0ma because the circuit is not complete and no current flows.
Ic(0+)=4ma because the capacitor is seen as a short so all current flows here?
I got 4ma from 20v/2k = 4ma

4. Oct 4, 2015

axmls

The main focus really is on the initial conditions (voltage for capacitor, current for inductor).

What exactly are you trying to describe with the circuit? You say you want the differential equation, but the differential equation for what? I suggest the nodal equation method using the capacitor voltage, because it would allow you to find any other value of the circuit easily (and it only requires one equation).

You equations from what I can tell are incorrect. For instance, your expression for iL is a little off. Write Kirchoff's current law at that junction to see that you're missing a term (and a sign).

5. Oct 4, 2015

Roy Fokker

I suppose I have some freedom to solve the deq for what I want. I guess i can leave it in the form of Vs= d^2/dt(x) + d/dt(y) +c
Ok so I am going over my equations and looking at the node equations as you suggested.

KCL

I1=I2+I3+I4

(20-VL)/5k = LdiL/dt + CdVc/dt + Vc/ 2k

**Kinda confused here** since LdiL/dt = VL
Can I just integrate all this? I am sure I am making this more difficult that it has to be..
But I know IL... will be ... (20-VL)/5k

Then

iL=Vc/2k + CdVc/dt + (20-VL)/5k

Man I really need to get working with latex :)

6. Oct 4, 2015

axmls

Why?

Also, notice in your KCL equation, the first term on the right side is the voltage across the inductor (not the current through it). Also notice that VL = VC.

7. Oct 4, 2015

Roy Fokker

I was just staring at the same thing... Totally not right.
I am looking at it from a steady state point of view.

I see that since this is all in parallel I could say VL=VC.
The inductor current is confusing me since it has no resistance. All I can see to do is integrate?

Integral VL/L and that would be the current I2?

Or I could just leave it as IL and make substitutions for VC = VL

Last edited: Oct 4, 2015
8. Oct 4, 2015

Roy Fokker

Ok so I did some fixin.

(20-VL)/5k = IL + CdVc/dt + Vc/ 2k

since VC = VL

(20-VL)/5k = IL + CdVL/dt + VL/ 2k

in the end...

IL= -d^2IL/dt (LC) - dIL/dt (L 700m) + 4m

9. Oct 5, 2015

Staff: Mentor

My instinct would be to go after the node voltage as was previously suggested. With that, all the other currents can be found in a straightforward manner.

Use the fundamental relationships for voltage and current for the reactive components to construct your node equation, solve the DE, then determine the individual currents:

$v_C = \frac{1}{C} \int{i_C dt}$

$i_C = C \frac{dv_C}{dt}$

$v_L = L \frac{di_L}{dt}$

$i_L = \frac{1}{L} \int{v_L dt}$

(An aside: You'll be really happy to know that when you've covered Laplace Transforms, solving this sort of problem becomes almost trivial )