1. The problem statement, all variables and given/known data Find instantaneous value UL after commutation. 2. Relevant equations type of commutation - switch key(I hope I've translated it correct) Use operator method and classic method 3. The attempt at a solution K1=1.2 K2=0.7 K3=1 J=8*K1 L=10*K1 R1=K2*8 R2=K2*8 R3=4*K3 R4=12*K3 I need to find UL in the moment of commutation?
Re: Transients(L) Does current flow in R1 before commutation ? After ? J is J... it's a current generator and it does its job.
Re: Transients(L) I don't know. I think before commutation all currents and voltage was 0. But what is the main algorithm at all for this task for classic method?
Re: Transients(L) Is it only the voltage across the inductor immediately after commutation of the switch that is required? Not the voltage as a function of time? I don't know exactly what is meant by "classical method" versus "operator method", unless the first wants a 'down and dirty' differential equation approach and the latter a Laplace transform approach or something along those lines. Personally, I'd just recognize that the circuit is going to transition between two states, the first being the steady state before the switch is thrown, and the second the steady state that will eventually hold a long time after the switch is thrown. Throwing the switch will be like hitting the circuit with a step function in terms of change of current. The circuit has only L and R components, so it'll have a time constant. It might be instructive to derive the Norton equivalents for the before and after states and see what the difference really is. Things to keep in mind: Inductors "don't like" sudden changes in current, they'll happily react with any voltage they have to in order to try to maintain the status quo (at least very briefly!); So, constant current through the inductor over the instant of commutation. Also, when drawing up the post-switching event equivalent circuit, it's fair game to put a current source in series with the inductor that drives the same current as the inductor carried prior to the switch being thrown. This has the effect of carrying the previous state of the circuit along as the initial condition for the new one. That current has to go somewhere... and it won't be through the main current supply!
Re: Transients(L) I only have translated the task. I think the result need to be like this u(t)=U0*(1*e^(t/T)) Yes
Re: Transients(L) The main problem is how to calculate circuit in every case(after,before,in the moment of commutation)
Re: Transients(L) You should be able to use what you know about the properties of inductors in DC circuits to determine the steady state conditions that will hold before switching and a "long time" after. Have you done that?
Re: Transients(L) Before comutation - L1 does not exist in the circuit 'cause there is not U? After commutation UL exists right? After commutation U(t)=(Ur3+Ul(t)+Ur4) because Ur2=Ur1=Ur3+UL(t)+Ur4?
Re: Transients(L) When you say, "L1 does not exist in the circuit", you need to be a bit careful. At steady state in a DC powered circuit the voltage across an inductor will be zero, but it will be carrying current and storing energy in its magnetic field, so that when some change occurs to alter the circuit conditions that energy is available to generate counteracting voltages. So just before the switch is closed UL will be zero and the current through the inductor will be, say, IL. When the switch is closed, the circuit will be driven towards a new steady state where IL will be different. The inductor wants to keep its current flowing at the IL rate, and "resists" the change of state by producing a voltage to try to keep it so. Have you drawn the Norton equivalent circuits of the networks feeding the inductor for both steady states?
Re: Transients(L) But before commutation I have only resistors and current source. I can only find the resitance of circuit.
Re: Transients(L) Exactly. Replace the whole resistor network and its current source with a single resistor and current source. Do it for the before- and after-switching steady states. You want to reduce it to the form shown in the attached figure.
Re: Transients(L) Yes, but of course with the inductor paralleling it, at steady state it will draw no current since all the current will be passing through the inductor. Don't worry too much about the details of the end states yet. Get the Norton equivalents drawn and then we'll examine the results.
Re: Transients(L) Perhaps you're not familiar with Norton equivalents? Both before and after circuits should look just as I showed in post #13. Of course, the values for the resistor and the current source will differ for the two cases. To find the Norton equivalent, temporarily remove the inductor from the circuit and also remove the current source. That will leave you with just the resistor network. Find the equivalent resistance of the resistor network looking into where the inductor was connected. Next, replace the current source and place a short circuit across where the inductor was connected. Find the current passing through that shorting wire. Assign the resistance you found to RN in the equivalent circuit, and the current value you found to the new current supply of that equivalent.
Re: Transients(L) Yes.I've never heard about it.Maybe in Russian it has another name.(Like Thevenin equivalent for example) (R1*R3+R4*R1+R2*R1+R2*R3+R2*R4)/(R1*R2*(R3+R4)? I don't understand. When did I find it?I only have resistance of all circuit.