What is the instantaneous value of UL after commutation?

[gneill]

A Thevenin equivalent circuit consists of a voltage source (the Thevenin Voltage) and a resistance (the Thevenin Resistance). What are they for the circuit in question?

 

[builder_user]

A Thevenin equivalent circuit consists of a voltage source (the Thevenin Voltage) and a resistance (the Thevenin Resistance). What are they for the circuit in question?

Rth=(R1*R3+R4*R1+R2*R1+R2*R3+R2*R4)/(R1*R2*(R3+R4)) or Rth=R2.I'm not sure.

 

[gneill]

To find the Thevenin resistance you suppress the sources (Remove them by shorting voltage sources and removing current sources) and then find the equivalent resistance of what's left. What's the equivalent resistance in the attached figure?

 

[builder_user]

Oh.R=R2+R3+R4
I found for all curcit...But not for this.

This results for state "before commutation",right?

 

[gneill]

Okay! So you have the Thevenin equivalent for the network that's driving the inductor. To put numbers to them, R_TH = 21.6 and V_TH = 53.76. Correct?

Now, the Norton equivalent circuit has the same resistor value as the Thevenin equivalent, and the voltage source is replaced with a current source of V_TH/R_TH. The resistor is in parallel with the current source.

Do the same work for the case after the switch is closed. Note that all the switch does is put R1 in parallel with R2. What numbers do you get for the Norton equivalent in this case?

 

[builder_user]

Do the same work for the case after the switch is closed. Note that all the switch does is put R1 in parallel with R2. What numbers do you get for the Norton equivalent in this case?

But when switch is closed there is the inductive element.It resistance add too?




This circuit after commutation?

 

[gneill]

But when switch is closed there is the inductive element.His resistance add too?

We're treating the inductor as the load being driven by a Norton equivalent of the source network. It plays no part in finding the equivalent circuit for the source network.

So no, there is no inductor added to the source network by the switch. It only adds R1 in parallel with R2 to the circuit we just analyzed. We will deal with the inductor after we have the equivalent circuits for the sources.

What we want to end up with is two Norton equivalent circuits, one for each state of the switch. These two equivalent circuits replace the current source and resistor networks of the original circuit. These equivalent circuits will be "driving" the inductor. They will make it easy to analyze what happens when the switch is closed.

 

[builder_user]

We're treating the inductor as the load being driven by a Norton equivalent of the source network. It plays no part in finding the equivalent circuit for the source network.

So no, there is no inductor added to the source network by the switch. It only adds R1 in parallel with R2 to the circuit we just analyzed. We will deal with the inductor after we have the equivalent circuits for the sources.

What we want to end up with is two Norton equivalent circuits, one for each state of the switch. These two equivalent circuits replace the current source and resistor networks of the original circuit. These equivalent circuits will be "driving" the inductor. They will make it easy to analyze what happens when the switch is closed.

After replacing there will be diff. equatations?
So this scheme?
and
R=R1*R2/(R1+R2)
U=J*R

 

[gneill]

I'll show you how to analyze the circuit without differential equations to begin with. That way you'll know what result you're looking for!

Yes, the figure you provided is the circuit of the source network after commutation of the switch. You want to find its Norton equivalent, just like you did for when the switch was open. What's its Norton resistance and current?

 

[builder_user]

I'll show you how to analyze the circuit without differential equations to begin with. That way you'll know what result you're looking for!

But I must find U as time function...like this
U=900/375*e^375t+C


The problem is...I must use differential equatations.But they only need at the moment of commutation.With Laplace and without it(especially for LC).It's the task.

 

[builder_user]

What's its Norton resistance and current?

Strange but currents are the same.

 

[gneill]

But I must find U as time function...like this
U=900/375*e^375t+C


The problem is...I must use differential equatations.But they only need at the moment of commutation.With Laplace and without it(especially for LC).It's the task.

The equivalent circuits will allow you to write the differential equations very easily. In fact, you will probably already have done so for circuits in this basic form!

I will show you how to write the desired U(t) function by inspection, so that you can check your differential equation result. Will that work for you?

 

[gneill]

Strange but currents are the same.

I think you may have forgotten about R3 and R4...

 

[builder_user]

I think you may have forgotten about R3 and R4...

I found U.
U/R3=J3
U/R4=J4?

 

[builder_user]

The equivalent circuits will allow you to write the differential equations very easily. In fact, you will probably already have done so for circuits in this basic form!

I will show you how to write the desired U(t) function by inspection, so that you can check your differential equation result. Will that work for you?

I need equations like this

from example
moment of commutation
UL=Ldi/dt

i2+i3=i1
i1*R1+i2*R3=E
-i2*R3+i3*R4+Ldi3/Dt=0

-E-i3*R1/(R1+R3)*R3+i3*R4+Ldi3/dt=0
i3=2.4+c*e^-375t

i(0)=1.7(before commut.)
2.4+c*e^0=1.7
c=-0.7

i3(t)=2.4-0.7*e^-375t

 

[gneill]

I found U.
U/R3=J3
U/R4=J4?

What are J3 and J4?

If you are looking at the Thevenin equivalent of your circuit, there is no current through R3 or R4. The Thevenin voltage is produced by the current flowing through the parallel combination of R1 and R2. The Thevenin resistance is the sum of R3 and R4 and the parallel combination of R1 and R2.

The Norton resistance, R_N, is the same as the Thevenin resistance. The Norton current is V_TH/R_TH.

 

[builder_user]

I see.Operator method it seems to similar
find i before and after commutation
replace inductor with operator resistor
p=j*w - operator.

 

[gneill]

I need equations like this

from example
moment of commutation
UL=Ldi/dt

i2+i3=i1
i1*R1+i2*R3=E
-i2*R3+i3*R4+Ldi3/Dt=0

-E-i3*R1/(R1+R3)*R3+i3*R4+Ldi3/dt=0
i3=2.4+c*e^-375t

i(0)=1.7(before commut.)
2.4+c*e^0=1.7
c=-0.7

i3(t)=2.4-0.7*e^-375t

The equations will be easy to write from the equivalent circuits. You will have only to deal with one current source, one resistor, and one inductor. Your textbook probably has this form as an example!

 

[builder_user]

The equations will be easy to write from the equivalent circuits. You will have only to deal with one current source, one resistor, and one inductor. Your textbook probably has this form as an example!

so, for L i'll just need to use U=Ldi/dt?

 

[gneill]

I see.


Operator method it seems to similar
find i before and after commutation
replace inductor with operator resistor
p=j*w - operator.

Yes. Certainly.

What I've been trying to do is reduce the before and after commutation circuits to their very simplest forms so that it will be very easy to apply any method of circuit analysis you wish to use. They will be so basic that your textbook should already provide a solved example.

 

[builder_user]

Ok.So I just need to make system of equatations and only to solve it?

 

[gneill]

so, for L i'll just need to use U=Ldi/dt?

Essentially, yes. The transition that occurs when the switch is closed will be the same as applying a step change in the source current. Or if you wish, it will be equivalent to applying a voltage source to an RL circuit.

 

[gneill]

Ok.So I just need to make system of equatations and only to solve it?

If that is your goal, sure. And it will be one differential equation; The one for a parallel RL circuit driven by a current source.

 

[gneill]

Perhaps I should summarize the work done so far. The equivalent circuits for before the switch is closed and after the switch is closed are as in the attached image. These are simple circuits!

To analyze the instant of switch closure, you can produce a "differential" circuit that reflects the change between the two. The difference is the sudden change in current and change in parallel resistance. Since you are looking for the voltage across the inductor after the switch closes, the circuit to write the equation for will consist of a current supply with a value equal to the change in current, the parallel resistance of the "after switch closed" circuit, and the inductor.

 

[builder_user]

Final part.At the moment of commutation scheme(pic.)

It has only one equatation?

(Req+R3+R4)*i+Ldi/dt=J*Req

and I must find i??

How to do it in Mathcad?

 

[builder_user]

I've found i(t)=1.43+1.06*e^-1567t

To find U i only need to
U=i(t)'*L?

U(t)=-19.93*e^(-1567t)

Is it correct?About Laplace.
In what state I need to replace inductor with pL.before commutation and after commutation I do not have inductor.Does it mean that I only need to replace inductor at the moment of commutation?

 

[builder_user]

After operator method results are different

i(t)=1.43+1.06*e^-1567t vs i(t)=1.43-0.0000915*e^-1563t

 

[gneill]

I've found i(t)=1.43+1.06*e^-1567t

To find U i only need to
U=i(t)'*L?

U(t)=-19.93*e^(-1567t)

Is it correct?


About Laplace.
In what state I need to replace inductor with pL.before commutation and after commutation I do not have inductor.Does it mean that I only need to replace inductor at the moment of commutation?

Hi builder-user, sorry to be away so long...

Your equations look fine except the time constant parameter seems to be missing a decimal place; I see R/L as 1.567 given your component values.

My own method when I solve these sorts of problems where there's a sudden transition involved (like the switch being closed), is to find the equivalent circuits for before and after steady state conditions, then produce a "difference" circuit to analyze. The difference circuit incorporates the final steady state circuit components, and where the source is the change in source between the old and new states. So in this case, looking at the Norton equivalents, the "old" current source was 2.49, the "new" source was 1.43, so the change in source is -1.06. See the attached figure. I then analyze this circuit assuming no prior state ("power-on" is at t=0).

The results of this analysis will be the changes that will occur in the original circuit after the switch closes. It will give the voltage across the inductor, and the change in current through the inductor. So if the pre-switching steady state current through the inductor was I₀, and the analysis provided current I₁(t), then we have I(t) = I₀ + I₁(t) for when the switch is closed.

This method may or may not suit you.

To incorporate the initial inductor current into the Laplace transform method, it appears as an initial condition for the current -- the i_o term in L*(s*I - i_o).

So if you have the Thevenin equivalent of the circuit at time t = 0 (when the switch has just been closed), with voltage source V and series resistance R and series inductor L, the equation you write for the Laplace transform of the current looks like:

V/s = I*R + L*(s*I - i₀)

Solve for I, take the inverse Laplace transform, and you're done (for the current, at least).

 

[builder_user]

Your equations look fine except the time constant parameter seems to be missing a decimal place; I see R/L as 1.567 given your component values.

You mean 1.43+1.06*e^-1.567t?

So if you have the Thevenin equivalent of the circuit at time t = 0

At time t=0 I need to add inductor to all resistance?And where is the potensial difference?

 

[gneill]

You mean 1.43+1.06*e^-1.567t?

Yes. I'm taking the component values as:

R1 = 5.6
R2 = 5.6
R3 = 4
R4 = 12
L = 12

Those are what I get after applying your 'K' constants.

These yield a Thevenin resistance of 18.8 when the switch is closed. So the time constant for the circuit should be τ = 12/18.8 = 0.6383, and 1/τ is 1.567

At time t=0 I need to add inductor to all resistance?And where is the potensial difference?

I not sure that I understand what you're asking. If you're referring to the Laplace transform equation, then I started with the Thevenin equivalent circuit and wrote KVL around the loop. The V is the Thevenin source voltage. i₀ is the initial current flowing in the inductor at t=0 when the switch is closed.

V = i*R + L*di/dt

Laplace transform:

V/s = R*I + L*(s*I - i₀)

I = (V + L*i₀*s)/(L*s² + s*R)

inverse Laplace transform:

i(t) = (i₀ - V/R)e(-R*t/L) + V/R