# Homework Help: Two commutation processes for RL circuit

1. Nov 25, 2013

### evol_w10lv

1. The problem statement, all variables and given/known data
I have to calculate first commutation process (t=0) and second commutation process (t1 = 2*τ)

Both swithes are closing in the given time moment.

2. Relevant equations
Form of second commutation process:

3. The attempt at a solution
I have already calculated first commutation process:

And also some important figures for second commutation process:

But how to get final result of second commutation process?

I have to use mesh current method and write 2 equations (loops with current IL1 and IL2 in the picture) and after that I have to calculate complicated system of differential equations or what?

2. Nov 25, 2013

### Staff: Mentor

What you have to do depends upon what it is you want to find. What do you mean when you say "calculate a commutation process"? Is there a particular voltage or current function that you need to find? What was the exact problem statement?

3. Nov 26, 2013

### evol_w10lv

I guess, I wasn't formulated my problem.
So.. I have to calculate current through the inductor L2 and voltage across the L2, when switch 1 is closed in the time moment t=0, and switch 2 is closed in the time moment t = 1.6364 us (as I calculated).
I tried to simulate my task using PSpice:

And here is results of voltages:

We know that:

Am I wright, that I have to use mesh current method and write equations, including dIL1/dt and dIL2/dt?

Here is my attempt:

Is this correct? And to get IL2(t) I have to calculate this complicated system of differential equations?

4. Nov 26, 2013

### Staff: Mentor

Oh, that makes things much easier! Just the voltage at those moments in time?

Except I'm not so sure that you are right in thinking that's what you are to find. http://imageshack.us/a/img27/6946/sulkoff.gif [Broken] It seems too easy, no differential equations are needed. For example, when s1 is closed, s2 remains open, so the voltage across L2 remains at zero. Is that too easy??

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon2.gif [Broken] Please confirm that all you are required to determine is the instantaneous voltage and current applying to L2 at two moments in time.

Last edited by a moderator: May 6, 2017
5. Nov 26, 2013

### evol_w10lv

Not exactly. I have to find expression that describes all process of voltage and current, how they are changing in the time. And we know, that second switch is closed in the time moment 1.6364 us.
I know, that answer should be in the form like this:

Where t-t1 is shifting time between first and second commutation process, I guess.
I see, that form is simular to answer of differential equation.
Any suggestions?

Last edited by a moderator: May 6, 2017
6. Nov 26, 2013

### Staff: Mentor

Yes, mesh or nodal or just KVL/KCL equations.

Those are KVL/KCL equations (you've labeled individual currents for each branch). The second equation doesn't look right; iL2 doesn't flow through R1.

That leading (t - t1) looks suspicious. It would be more credible if it was U(t - t1), where U is the unit step function. And I doubt the final iL2(t) should be there, since you're looking for iL2(t). Maybe iL2(0) would be appropriate. But you will find this out when you solve your differential equations.

You might consider the Laplace Domain for solving the equations, if you've taken Laplace Transforms that is.

7. Nov 27, 2013

### evol_w10lv

I know about LT, but I have to use standart method.
So, it means, that I have to find initial conditions for I_L2 before I have started to calculate differential equation.

And here is my problem, I don't know, how exactly I can find it.
I know, that in my case I_L2(t1)=0, but I have to find dIL_2/dt in the time moment.
I have got an example. There is standart cicuit and after that - equivalent cicuit. It looks, that this example is very simular to my circuit. But still I don't understand, how exactly I can find this value. I circuled an expression in the example. I have to do the same with my circuit.

So.. can you explain, how to get this value?

To be clear.. it's initial conditon for my second order differential equation. I converted my system of differential equations to second order differential equation and using it, I will find I_L2(t). As I said, I know IL_2(0) = 0, but I have to find dIL_2/dt(0) like in the example. To say easier: x(0)=0, x'(0) -?

I hope that you understand my problem.

8. Nov 27, 2013

### Staff: Mentor

Since inductor currents cannot change instantaneously at commutation, for purposes of analyzing the situation at that instant they replaced L1 with a current source equal to the current through L1 at that instant. They also "removed" L2 since it will look like an open circuit at that instant. Then they used superposition to determine the potential ##U_{L2}(t_{1+}) across L2:

1) Current source suppressed: R1,R2, and R3 form a voltage divider for E.
2) Current source suppressed: Current divider to find current through R3+R4 branch, then multiply by R4 to find the potential across L2.
3) Sum the contributions.

9. Nov 27, 2013

### evol_w10lv

Unfortunately, this theme is not of my favourite.
My circuit:

I guess, then equivalent circuit should look like that (without given/real values):

In my case:
Voltage divider:

Current divider:

I have to sum these equations? But it seems to me, that I am not on the right track..

10. Nov 27, 2013

### Staff: Mentor

No, you're doing very well

Here's a hint that may help. Any components in series with a current source are irrelevant. A current source will produce its designated current no matter what; It will generate any amount of voltage necessary to get that job done. So R1 and V1 in your circuit make no difference and you can remove them, leaving just the current source driving current through R2, the only available complete path. That should make determining the potential across the "L2 open" very easy indeed.

11. Nov 27, 2013

### evol_w10lv

Then voltage across R2 is V = J * R2 and it's the same in the branch, where is R3 and "L2 open". To find voltage drop accross "L2 open" I should from J*R2 subtract current divider? Or actually current divider is output of branch where is L2?
Ouch, sorry.. it becomes awkward.. not good day for my thinking.

12. Nov 27, 2013

### Staff: Mentor

There's no current divider in this case; all the current is flowing through R2. So the potential across the "open" is the same as the potential across R2.

13. Nov 29, 2013

### evol_w10lv

Thanks. Alles klar. :)