Finding Instantaneous UL and iL After Commutation for Transients (LC) Physics

[gneill]

Before proceeding, let's make an educated guess at what the voltage across the inductor is going to look like when the switch closes. That way we'll know where we're headed.

When the switch closes you'll have the current supply J = 12A in parallel with R2 and C1 and the branch R3&&L1 (using your && notation). The capacitor has an initial voltage of UC = UR2 = 9.6V (correct?). The inductor is carrying initial current iL = 9.6A (correct?).

The instant the switch closes, the capacitor will want to maintain its voltage and the inductor will want to maintain its current. So the top of R3 will be at potential UC = 9.6V. The current through R3, however, will still be iL = 9.6A, so the voltage drop on R3 is going to be 19.2V. That means the voltage across L1 must shoot up to 9.6V (- on top, + on bottom of L1) in order to satisfy Kircchoff, because UR3 + UL1 = UC at that instant.

Eventually the voltage across L1 must return to zero in a new steady state. So there will be an exponential decay (with time constant to be determined) from 9.6V down to zero. Superimposed on that will be whatever oscillations, if any, that result from the interaction of L and C. But the "big picture" is that UL1 will shoot up to 9.6V and decay.

Does all that make sense?

 

[builder_user]

Before proceeding, let's make an educated guess at what the voltage across the inductor is going to look like when the switch closes. That way we'll know where we're headed.

When the switch closes you'll have the current supply J = 12A in parallel with R2 and C1 and the branch R3&&L1 (using your && notation). The capacitor has an initial voltage of UC = UR2 = 9.6V (correct?). The inductor is carrying initial current iL = 9.6A (correct?).

The instant the switch closes, the capacitor will want to maintain its voltage and the inductor will want to maintain its current. So the top of R3 will be at potential UC = 9.6V. The current through R3, however, will still be iL = 9.6A, so the voltage drop on R3 is going to be 19.2V. That means the voltage across L1 must shoot up to 9.6V (- on top, + on bottom of L1) in order to satisfy Kircchoff, because UR3 + UL1 = UC at that instant.

Eventually the voltage across L1 must return to zero in a new steady state. So there will be an exponential decay (with time constant to be determined) from 9.6V down to zero. Superimposed on that will be whatever oscillations, if any, that result from the interaction of L and C. But the "big picture" is that UL1 will shoot up to 9.6V and decay.

Does all that make sense?

Why 12?J=4*K1=4*1.2=4.8

 

[gneill]

Isn't J = 4*K1 = 4*3 = 12? Or are the units something other than amps?

 

[builder_user]

Isn't J = 4*K1 = 4*3 = 12? Or are the units something other than amps?

But K1=1.2

 

[gneill]

Ah! My mistake. Very sorry. These K's will be the end of me...

So, make that:

iL = 3.84A
UC = 3.84V
UR3 = 7.68V
UL = 3.84V (+ on bottom end)

The general picture still stands, though, with these modified values. Do your calculated values agree, and does the argument I presented make sense?

 

[builder_user]

All values are the same

Ur3 will be iL*R3=7.68 after commutation?

 

[gneill]

All values are the same

Ur3 will be iL*R3 after commutation?

Right. The inductor maintains the same current as before the switch closes (at least for a brief instant).

So after the switch closes the circuit looks as in the attached figure, keeping in mind that C1 and L1 have their initial conditions. The idea now will be to develop the differential equations for the circuit. The current supply J and resistor R2 could be converted to a Thevenin equivalent, leaving two loops for KVL. Or you might try using a KCL approach.

A good approach might be to write the equations in differential form, convert to Laplace (allowing you to insert initial conditions), then solve for the desired variable.

 

[builder_user]

i2=i1+i3
i2*R2+Uc=E
i3*R3+Ldi/dt=0

and Uc=3.84?

But UL=-3.84?

 

[gneill]

That's fine. Keep in mind that UL and UC are initial conditions that will go into the s-domain (Laplace transformed) equations.

 

[builder_user]

I forgot Uc in my third equatation.

Why I need two previous equatations at all?
I think one will be enough
this one
i3*R3+Ldi3/dt+Uc=0

 

[gneill]

The voltage on the capacitor is not going to stay at Uc indefinitely; it's only an initial condition. If you want to find UL(t), then you'll have to solve the equations.

You should be able to notice that there are R L and C components in the circuit, so you should expect a second order system. There will be a decaying oscillation if the damping is low, or a smooth decay if the system is overdamped.

 

[builder_user]

My result

i(t)=-Uc/L+e^((-r3/L)*t)*C

Uc=3.84

I think it's wrong.

 

[builder_user]

How to do it correct?

 

[gneill]

Using whatever is your choice of circuit analysis technique you need to write out the equations for the loops (or nodes). Convert to Laplace form (s-domain) including the initial conditions, then solve.

I have not taken the time to write them out and solve them at this point, and what equations are written will depend upon your choice of analysis method. Perhaps you should post your equation attempt so that we can look at it.

 

[builder_user]

I have already posted them

I use this equatation to find i(t)
i3*R3+Ldi3/dt+Uc=0

 

[builder_user]

I've solved it correct, but It seems strange to me that equatation...System must have and i(t)=Cdu/dt and Ldi/dt

The problem is to make correct system of equatation

If I find correct equatation to solve i'll can transform it in Laplas form

At the mail problem is that I need to have my work done today

 

[gneill]

I've solved it correct, but It seems strange to me that equatation...System must have and i(t)=Cdu/dt and Ldi/dt

The problem is to make correct system of equatation

If I find correct equatation to solve i'll can transform it in Laplas form

At the mail problem is that I need to have my work done today

Well, I'm glad that you've solved it. Congratulations.

I don't find it surprising that the equations will involve changing values on both the capacitor and inductor; it's a second order system, and the configuration is capable of producing oscillations (ringing) or other "excursions" when its perturbed (like the throwing of the switch does). The precise response depends upon the component values (damping).

 

[builder_user]

I make new equatation and solve for transformed scheme
J2=(J*1/sC+L*i0-U/s)/(R3+s*L+1/s*C)
Here my result.Is it correct now?
i(t)=(1.9-34464*j)*e^((-1428+1421j)*t)+(34464-1.9j)*e^((-1428-14211*j)*t)*i
u(t)=L*di/dt=(342835-34469*j)*e^((-1428+14211)*jt)+(342835+34469*j)*e^(-1428+14211j)t

 

[gneill]

I'm not understanding how you arrived at your equation for J2 (is that the current for the inductor?). There are two loops so there should be two loop equations. In your figure, is J/p meant to be a current source in series with resistor R2? How did you get from the original circuit with a current source in parallel with Rs to one with it in series? I see that you've drawn in the s-domain sources associated with the initial conditions. That's good.

If I plug t = 0 into your u(t) solution the answer I see looks a bit large. Is it scaled somehow?

 

[builder_user]

I'm not understanding how you arrived at your equation for J2 (is that the current for the inductor?). There are two loops so there should be two loop equations.

But there is only one variable.---J2(on resistor R3).Besides if I make 2 equatations Mathcad can't find J2

 

[gneill]

But there are two loops that interact. You only need to solve for the current in the second loop, but the first loop still plays a part!

 

[builder_user]

But there are two loops that interact. You only need to solve for the current in the second loop, but the first loop still plays a part!

So,I do not need first loop as equatation to solve?

 

[gneill]

You need the first loop equation in order to eliminate its current (say, i1) from the second loop's equation. But you don't have to solve for i1.

You have solved circuits with two loops before. You know that the loop currents in both loops affect each other in the solution. So you have to write the equations for both loop currents (i1, i2), and eliminate i1 from the the loop 2 equation. Solve for i2.

 

[builder_user]

You need the first loop equation in order to eliminate its current (say, i1) from the second loop's equation. But you don't have to solve for i1.

You have solved circuits with two loops before. You know that the loop currents in both loops affect each other in the solution. So you have to write the equations for both loop currents (i1, i2), and eliminate i1 from the the loop 2 equation. Solve for i2.

But I know J - so it's i1?

 

[gneill]

But I know J - so it's i1?

I don't believe that your circuit with J in series with R2 is correct. The original circuit (with the switch closed) has J in parallel with R2 and in parallel with C1. I don't see how you could go from there to having J in series with R2.

If J were a voltage source, then yes, I would agree. Then you would have the Thevenin equivalent of J and R2.

 

[builder_user]

Correct result is sth like this.(I've only changed direction of Uc. It was mistake but it wasn't a big problem)
i(t)=7.3*e^(-2863t)-3.89*e^2.73t*cos⁡(133t)-0.4175*e^2.73t*sin⁡(133t)