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Transients(L) |
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| May24-11, 06:58 PM | #35 |
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Transients(L)
Okay! So you have the Thevenin equivalent for the network that's driving the inductor. To put numbers to them, RTH = 21.6 and VTH = 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 VTH/RTH. 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? |
| May24-11, 07:01 PM | #36 |
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This circuit after commutation? |
| May24-11, 07:08 PM | #37 |
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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. |
| May24-11, 07:11 PM | #38 |
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So this scheme? and R=R1*R2/(R1+R2) U=J*R |
| May24-11, 07:16 PM | #39 |
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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? |
| May24-11, 07:24 PM | #40 |
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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. |
| May24-11, 07:26 PM | #41 |
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| May24-11, 07:30 PM | #42 |
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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? |
| May24-11, 07:30 PM | #43 |
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| May24-11, 07:33 PM | #44 |
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U/R3=J3 U/R4=J4? |
| May24-11, 07:40 PM | #45 |
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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 |
| May24-11, 07:42 PM | #46 |
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![]() 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. |
| May24-11, 07:46 PM | #47 |
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I see.
Operator method it seems to similar find i before and after commutation replace inductor with operator resistor p=j*w - operator. |
| May24-11, 07:48 PM | #48 |
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| May24-11, 07:51 PM | #49 |
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| May24-11, 07:56 PM | #50 |
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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. |
| May24-11, 08:01 PM | #51 |
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Ok.So I just need to make system of equatations and only to solve it?
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