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Transients(L) physics

by builder_user
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gneill
#37
May24-11, 07:08 PM
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Quote Quote by builder_user View Post
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
#38
May24-11, 07:11 PM
P: 196
Quote Quote by gneill View Post
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
Attached Thumbnails
f.JPG  
gneill
#39
May24-11, 07:16 PM
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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
#40
May24-11, 07:24 PM
P: 196
Quote Quote by gneill View Post
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
#41
May24-11, 07:26 PM
P: 196
Quote Quote by gneill View Post
What's its Norton resistance and current?
Strange but currents are the same.
gneill
#42
May24-11, 07:30 PM
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Quote Quote by builder_user View Post
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
#43
May24-11, 07:30 PM
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Quote Quote by builder_user View Post
Strange but currents are the same.
I think you may have forgotten about R3 and R4...
builder_user
#44
May24-11, 07:33 PM
P: 196
Quote Quote by gneill View Post
I think you may have forgotten about R3 and R4...
I found U.
U/R3=J3
U/R4=J4?
builder_user
#45
May24-11, 07:40 PM
P: 196
Quote Quote by gneill View Post
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
#46
May24-11, 07:42 PM
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Quote Quote by builder_user View Post
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, RN, is the same as the Thevenin resistance. The Norton current is VTH/RTH.
builder_user
#47
May24-11, 07:46 PM
P: 196
I see.


Operator method it seems to similar
find i before and after commutation
replace inductor with operator resistor
p=j*w - operator.
gneill
#48
May24-11, 07:48 PM
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Quote Quote by builder_user View Post
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
#49
May24-11, 07:51 PM
P: 196
Quote Quote by gneill View Post
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
#50
May24-11, 07:56 PM
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Quote Quote by builder_user View Post
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
#51
May24-11, 08:01 PM
P: 196
Ok.So I just need to make system of equatations and only to solve it?
gneill
#52
May24-11, 08:08 PM
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Quote Quote by builder_user View Post
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
#53
May24-11, 08:13 PM
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Quote Quote by builder_user View Post
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
#54
May24-11, 08:32 PM
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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.
Attached Thumbnails
Fig20.gif  


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