
#37
May2411, 07:08 PM

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P: 11,415

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. 



#38
May2411, 07:11 PM

P: 196

So this scheme? and R=R1*R2/(R1+R2) U=J*R 



#39
May2411, 07:16 PM

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P: 11,415

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? 



#40
May2411, 07:24 PM

P: 196

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. 



#41
May2411, 07:26 PM

P: 196





#42
May2411, 07:30 PM

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P: 11,415

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? 



#43
May2411, 07:30 PM

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#44
May2411, 07:33 PM

P: 196

U/R3=J3 U/R4=J4? 



#45
May2411, 07:40 PM

P: 196

from example moment of commutation UL=Ldi/dt i2+i3=i1 i1*R1+i2*R3=E i2*R3+i3*R4+Ldi3/Dt=0 Ei3*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)=2.40.7*e^375t 



#46
May2411, 07:42 PM

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P: 11,415

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}. 



#47
May2411, 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. 



#48
May2411, 07:48 PM

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#49
May2411, 07:51 PM

P: 196





#50
May2411, 07:56 PM

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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. 



#51
May2411, 08:01 PM

P: 196

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




#52
May2411, 08:08 PM

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#53
May2411, 08:13 PM

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#54
May2411, 08:32 PM

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P: 11,415

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. 


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