Question about current round this short circuit

Discussion in 'Homework Help' started by KevinEamon, Apr 21, 2017.

  1. KevinEamon

    Thread Starter Member

    Apr 9, 2017
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    See the current in Isc here. Is that just simply = 10a?

    is it just zipping away round there happily. Unaffected by anything else? It seems to have a complete circuit so why not.

    The math when I consider the 5 ix amp source keeps coming out with zero values.

    Something Wbahm said earlier got me thinking about this. Is it simply the case that View attachment 125158 Isc = 10
    Or Isc = 10 + 4ix ?

    10ampswirl (2).jpg
     
  2. MrAl

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    Jun 17, 2014
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    Hi,

    Yes it looks like 10A because the short has much lower resistance than anything else.
     
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  3. WBahn

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    Mar 31, 2012
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    That's very faulty reasoning.

    Replace the CSCS source with, say, a 9 A constant current source. Now the 1 Ω and 2 Ω resistors form a current divider with 3 A going through the 2 Ω resistor. That current then combines with the current from the 10 A source to make Isc equal to 13 A.

    The reason that Ix is equal to zero is because that is the only value of current that can satisfy KCL and KVL. If you were to replace the 2 Ω resistor with a 0.25 Ω resistor, then Ix would become indeterminate because ANY value of Ix would satisfy KCL and KVL.

    It would be trivial to change the circuit so that some determinate, non-zero current flows through Isc from the right hand side. It would also be possible to make it so that zero current flows through Isc because all 10 A from the left hand source would have to flow through the 2 Ω resistor to satisfy KCL and KVL.
     
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  4. KevinEamon

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    Apr 9, 2017
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    Cool this is good. That's exactly the problem I was running into with this circuit. I didn't know how to define it clearly. I thought I was making mistakes with the zero values for ix.
    so the current would be indeed 10 + 4ix. But its just in the case the kvls for ix = 0. Hmmm... K... Much to think about. Thx once again guys
     
  5. WBahn

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    But you need to be able to show that ix is identically zero. Doing so is quite straightforward.

    With the switch closed, the voltage across the 2 Ω resistor is the same as across the 1 Ω resistor (since the closed switch places them in parallel).

    If the current in the 1 Ω resistor is ix, then the current in the 2 Ω resistor must be 4·ix (by KCL at the junction of those two resistors and the CCCS), flowing away from that junction.

    Setting the voltages equal, we have

    (2 Ω)·(4·ix) = (1 Ω)·(ix)

    8·ix = ix

    The only value of ix for which this equation is true is for ix = 0.

    Some people will divide both sides by ix and get 8 = 1 and then claim that this means that the circuit can't be realized. Nothing of the sort. To divide by ix requires that ix not be zero, since division by zero is not defined. So the result is that IF ix is NOT zero, then 8 = 1. But the equation makes no such claim IF ix IS zero. If ix=0 then we have 0 = 0 which is perfectly true and valid.
     
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  6. MrAl

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    Jun 17, 2014
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    Hi,

    Sorry i dont know what you mean when you quote a "CSCS" source and i dont want to guess.

    But it sounds like you are trying to change the circuit to a different circuit and then state that the reasoning for the original circuit is faulty. It's like you are trying to solve some general universal problem rather than the actual circuit given as is. That's ok too, but it does not necessarily mean the reasoning for the original circuit is wrong. I'll explain a little more.

    First, i see this as a circuit with one direct short in it, and that short is a regular short that has zero resistance. Second, since the resistor connected to it (and the topology itself) is 2 ohms this is not a special case. The short has no competition like it might with say another parallel short where it gets hard to say if both shorts have the same current or what. That's a special case where we have to define the shorts more carefully, but i dont believe this is one of those cases. Third, since there are no independent sources to the right of that 2 ohm resistor there are no additional energy supplies to the circuit meaning that nothing but the 10 amp supply has the ability to provide a sense current through the 1 ohm sense resistor. This means that if some of the current of that 10 amp supply can get through the 1 ohm resistor then we have an additional current source to consider, the 5ix source. Since there is no competition for the current through the short, no current can flow through the 1 ohm resistor, therefore no additional current can come from the 5ix source.

    So you see this reasoning is straightforward and does not seek to solve a more general problem in circuit analysis but simply solves the circuit at hand, the actual unchanged circuit at hand. If you want to analyze the circuit differently that is up to you (by doing a full nodal for example) but we also can use experience about how sub circuits work in order to come up with a solution to problems in circuit analysis.

    So the analysis points are:
    1. Short has no competition in this circuit.
    2. No current sensed equals no additional current generated.

    Note that there are cases when the short has competition, but this isnt one of those cases. We can look at other resistor values too if you would like to do that, but i consider that another case.
     
  7. WBahn

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    Typo on my part -- should have been CCCS (current-controlled current source).

    This is a VERY different analysis than, "Yes it looks like 10A because the short has much lower resistance than anything else."

    That justification is completely faulty as it WOULD apply to a circuit with any of the modifications that I suggested.

    Furthermore, even your new reasoning would be faulty if the 2 Ω resistor would have been stated as being 0.25 Ω (or if, instead, the 1 Ω resistor had been 4 Ω) because then ANY current flowing through the short would be perfectly consistent with KVL and KCL for the entire circuit, making the current indeterminate. None of your reasoning relies on the 2 Ω resistor being any particular value and certainly 0.25 Ω is just as reasonable as 2 Ω. That the current ix HAPPENS to be identically zero is a consequence of factors in THAT circuit that were completely ignored in your reasoning (old or new) and that just, by sheer coincidence, HAPPEN to work out.
     
  8. KevinEamon

    Thread Starter Member

    Apr 9, 2017
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    I'm glad Wbahm stated it the way he did because I was tempted to think for a second, that in all circuits, of that configuration, the current would simply be the 10A. I suppose its difficult advising noobeenoobs like me, whether to give a more detailed answer, or to hold back for fear of over complication. There's nothing worse than getting an answer more complex than the dang question. In this case however I really did want to know about that 5ix source and how it and others like it, affects things. During this discussion you 2 are having, I also discovered that 5ix is completely dependant upon the 10 A for any response. Which you would have thought was obvious... But let's just say it reinforced that idea in my mind
     
  9. WBahn

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    I'm glad the discussion is helping you. It, indeed, if very hard to know at what level an explanation should be targeted and I don't think anyone here has a particularly stellar track record of getting it right most of the time. The good news is that you generally get a variety of levels which allows you to focus in on the ones that help you the most -- they also give you the chance to get pieces-parts from different levels and to revisit explanations that were too in-depth for you earlier but that the simpler explanations may have put you into a position to better understand later.

    Oh, I think you meant "completely independent upon the 10 A". In this case that is very true -- the short isolates the left hand of the circuit from the right because they have no means to communicate with each other. Consider the following tweak:

    Edit_2017-04-21_1.jpg


    The actual short circuit current is Isc1 + Isc2, but here clearly the left side of the circuit cannot influence the right side (and vice-versa).

    So now you can calculate Isc1 and Isc2 separately in which case Isc1 HAS to be 10 A and you quickly discover that Isc2 has to be 0 A.
     
  10. KevinEamon

    Thread Starter Member

    Apr 9, 2017
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    haha I'm sure that trick will be very handy when I learn, how and when to use it... but...what now...? completely independent upon the 10 A source.? I think you just caused some type of arc current to fluctuate in my brain. It seems to have automatically rebooted itself and now there's just a flashing error message, where my vision should be. I also have a dull pain behind my right eye. Lul... Ammm what the heck..?
     
  11. WBahn

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    Ah, I think I misinterpreted what you were trying to say. Now I see that you were probably saying that the behavior of the controlled source is completely dependent upon the presence of the independent source and ability of it to interact with the part of the circuit that has the dependent source in it. In this you are correct. I thought you were talking about the behavior of the dependent source in the circuit with the short. In this circuit, the independent source can have no influence on the dependent source (which means that the independent source is effectively turned off).
     
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  12. MrAl

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    Jun 17, 2014
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    Hi again,

    Thanks for clarifying that. I suspected that was what you meant but i hate to assume too much in problems where there comes a theoretical question as i dont want to presume to know everything.

    Also, i think you are right about my first post which may have been a little too short. I was trying to keep with the simplicity of this circuit. To the contrary though i think you are over complicating it and unfortunately that i believe is leading you to a faulty conclusion. I will explain of course.

    The simplest explanation to this i think is a simple example. Feel free to disagree and explain why you disagree, if you do disagree to this example. I trust your judgement so i will consider any argument carefully as usual.

    The example is a lone capacitor, with NO other connections whatsoever, and that means no resistor, no inductors, no voltage sources, no curernt sources, etc., because we have a single capaictor with two leads as typical, and absolutely nothing else, and it's floating in deep space in a region where it is far from any electrical activity of any kind including radiation of any kind.
    The key question is, "What is the voltage across the capacitor?".
    Now isnt it true that it could be any voltage? After all, we dont know what activity the capacitor has had BEFORE the time we looked at it (and assume we could measure it somehow without affecting the voltage).
    So is the answer:
    A. "It could have any voltage"

    or:
    B. "It has zero voltage".

    In deep space, the answer might actually be A because we dont know the activity it has seen before the measurement, but in circuit analysis, the answer is B. That's because in circuit analysis for the lack of any other information we assume zero initial conditions.

    So just because the cap can have any voltage across it and still satisfy some sub law of nature, we dont say it has any voltage across it because that's not how we do circuit analysis. Circuit analysis is based on things that we know. If we dont know something then the circuit is not sovlable unless we can prove that it never affects the final result.

    But that illustration brings up two more points about circuit analysis.
    1. Energy was a consideration and takes precedence.
    2. Time became a consideration in the deep space case.

    Now we get to the dependent source.

    Take a lone voltage controlled voltage source. What is the initial energy in a voltage controlled voltage source? It's zero. How about a current controlled current source? It's zero too.
    What this says then is that the initial condition of a dependent source is zero because if it was anything else it would be able to produce a voltage or current all on it's own, which it cant.

    Maybe more to the point is that energy considerations trump Kirchoff's laws. If we dont have any energy there is no sum of voltage drops, and there is no sum of currents.

    To prove this all we have to do is make sure that the dependent source does not get any energy from the 10 amp source no matter what happens, then see where the circuit equilibrium ends up. This is very hard to do in a circuit simulator because the circuit simulator could easily apply 1e-32 volts or amps to the dependent source during startup, and that throws the whole solution in the dumpster because then the circuit takes off in a direction that it would never have gone with zero energy. In order to at least try to stop the energy from reaching the dependent source during startup we could place a small capacitor across the 1 ohm resistor. During startup the initial voltage will be zero so that might prevent a false start. I wont guarantee this will work, but the algebra itself should work. If we start with zero energy, we must end up with zero energy in the dependent source because there is nothing that can power it. It does not matter if it obeys Kirchoff or not, it must first obey the supreme law of the universe: if it starts with zero energy and has no added energy later, the final energy must be zero, otherwise know as the conservation of energy (formally stated differently of course).

    Also, i think we can show that the value of R2 can be anything other than maybe a short circuit, but even that may work in this circuit. One way would be to show that R2 drops out of the solution.

    Did you actually do a nodal analysis or just tried to visualize what might happen?
    If you did not, try that next and see what you get for the final result.
     
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  13. KevinEamon

    Thread Starter Member

    Apr 9, 2017
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    Fascinating. Here comes me to drop down the level of the discussion again. See that diagram in post 9? Does that single wire make a complete circuit. Say we put some current into isc1 and isc2. Would any current mix between the 2 across that single branch. Or would it just drop to zero. I've started to realize that the ground is arbitrary concept but that whole bottom wire looks like the ground. Basically can the current pass along that single wire or does it need two wires
     
  14. WBahn

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    Your analysis based on energy considerations has one huge problem -- dependent sources do NOT get their energy from the independent sources in a circuit -- they have their OWN source of energy; that's why they are called "sources". The argument that they will output zero voltage/current in the absence of a SIGNAL that is due to an independent source has merit, particularly for linear circuits (that would not be the case for a nonlinear dependent source, in general, if for no other reason that a nonlinear dependent source could have a constant term). But the assumption that a signal can't exist in a circuit without an independent source is not, in general, valid.

    In some situations making the assumption that the initial conditions should be taken to be zero is reasonable, but it is NOT the proper way to perform circuit analysis in general. You need to justify that such an assumption is reasonable for the specific situation. To take your capacitor example, imagine walking up to a capacitor sitting on a counter. Is it reasonable to assume that it is uncharged just because we don't know the time history of it? Reminds me of an experiment my high school physics teacher did every semester -- he would charge up a fairly small capacitor to a few hundred volts and sit it out on a counter with a sign that said, "Do not touch." He had to charge it up several times a day because of all the people that would pick it up and get zapped, painfully but harmlessly, by it. They were operating under your approach which is that it is reasonable to assume that any capacitor that we don't know the history of must have an initial voltage of zero.

    Or consider another example -- you've got a superconducting magnet dewar sitting in the corner that hasn't been hooked up to a power source for months. Is it reasonable to assume that there is no current flowing in the magnet and that it is safe to open circuit it? That would be one quick way to meet your maker.

    This circuit is an example in which making such an assumption is not reasonable precisely because of the kind of problem that it is drawn from -- namely finding the transient response after a step change in the circuit. The switch grounding out the independent source is closed NOW, but it wasn't always closed -- in fact it was OPEN just a moment before this circuit arrangement came into existence. Therefore there could (and probably was) current flowing in the dependent source just before the switch was closed. Consider the case if, say, the 1 Ω resistor had been an inductor instead. Then we would have had to carry across the current in it from before the switch closing to after which would have resulted in an exponentially decaying current in the CCCS even after the independent current source was removed (shorted).

    Now, IF you wanted to make a claim that the current in the CCCS in THIS circuit is zero because (1) there is no mechanism for the independent source to provide a signal to the CCCS, (2) the circuit is linear, and (3) there are no reactive elements in the part of the circuit containing the CCCS, then I think you have a strong case for concluding that there is no current flowing in the CCCS. But just relying on (1) is NOT sufficient.
     
  15. WBahn

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    No. No current can flow through that wire in the bottom between the two circuit segments because there is no way for current to get back to where it started. All the wire does is force the voltage (whether it be zero or something else) to be the same in the bottom wire of the left half as it is in the bottom wire of the right half. But that's not actually important in this case. Here the purpose of splitting the wire that Isc flows in into two pieces is to simplify the analysis by showing that the two halves can't interact and, therefore, can be solved separately.
     
  16. KevinEamon

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    Apr 9, 2017
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    Oh Yeh I see that would be one big node. Hmmm... Now I'm following this discussion. The module I'm doing is linear circuit. So I don't need to worry about a dependant having potential. Interesting to see how it can though. Well I say that, though there's is some of these questions were both before an after the switch is changed. On both sides of zero the capacitor is connected to a source. I'm just making sure all the analysis is done for now. I'm back at uni on Monday so I get a quick refresher from the lecturer b4 class.
     
  17. MrAl

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    Jun 17, 2014
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    Hi again,

    I am going on the assumption that we can all SEE the circuit as drawn. That means i only want to take the time to explain the things that pertain to that circuit and any addition or removals will be explained, but i did not feel tht i had to explain everything because the circuit is visible to everyone. if you want to add to that it's fine though.

    The capacitor in deep space example was to show that in circuit analysis we assume zero initial conditions for circuit elements unless they are shown to be otherwise beforehand. If you still want to push the fact that we can find a capacitor on desk somewhere that has a charge like your instructor taught, then also show me one time when you saw a resistor and capacitor low pass filter with input source and assumed that the cap had some initial voltage. It's true that if it does have initial voltage the analysis is a little different, but we dont assume that unless it is drawn on the schematic or pointed out elsewhere.

    A dependent source can not produce any current or voltage unless it has a non zero input of some kind. If the input is zero, then the output must be zero. Of course we are talking about a linear dependent source because that is the kind that is in the original circuit.
    If you found a dependent source sitting on a desk somewhere with no connections to anything else, what would the output be? With zero input it must be zero output.
    The point of putting a capacitor across the input was to force the circuit to have to act in real time. If we had this very circuit and the 10 amp source was not turned on yet, the input to the dependent source would have to be zero because there was never any energy EVER introduced into the circuit yet. Thus the input must start at zero. Since energy can not be created, the dependent source can not produce any energy until it is given at least a small amount of energy.

    Also, the time domain solution with the capacitor is of the form (R1 is the 1 ohm resistor):
    vR1=E*e^(-t*K)

    and with R2=0.25 then K=0 so we have:
    vR1=E

    where E is the initial voltage across R1 at t=0.

    So if you want to say that at t=0 then E=1v for example, then you should have to show how that 1v got there.
    And as i said, if you see a regular low pass filter made from one R and one C, you dont say that the solution is that the capacitor can have any voltage across it because it is not shown on the schematic. So the solution is not:
    vC=can be any voltage

    Invariably, we will always see:
    vC=Vs*(1-e^(-a*t))

    It's only when we see some indication that there could be an initial voltage that we start to assume the other form. But i should point out that this is with a capacitor too, not with a dependent source.

    So if you want to say that the current in the circuit 'can be anything' then at least show how the dependent source input got some kind of stimulation. If you can do that, then you should try to show how all dependent sources in every circuit ever made can have an input that is non zero before the power is turned on :)
     
  18. MrAl

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    Jun 17, 2014
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    Hello again,

    Are you saying also that IS2 has current flowing in it?
    This makes things even simpler i think.

    A simpler explanation is that there are never any initial conditions for dependent sources, only for capacitors and inductors.
    For example for a capacitor we write: C=1uf, ic=2v.
    There's no such statement for a dependent source that i know of: I=5*x, ic=2v.
     
    Last edited: Apr 23, 2017 at 9:39 AM
  19. WBahn

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    Where did I say that Isc2 actually has current flowing in it? I showed a way to redraw the original circuit so that the total short circuit current is made up of two parts, one due to the left side and one due to the right side. The left side is trivial. The right side can easily be shown to be zero.
     
  20. MrAl

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    Jun 17, 2014
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    Hi again,

    I wasnt really saying you did, i was just asking. But i assumed you would say there was current flowing if the resistor was made equal to 0.25 ohms, or that you would suggest that it could be any current flowing again as in the other circuit.

    So what do you think? If the resistor was made 0.25 ohms, what do you think the solution would be for this circuit, which is almost the same as the other circuit? That was the only condition we did not agree on i think (when the 2 ohm resistor was made equal to 0.25 ohms).

    My suggestion is that the current IS2 is zero even when the 2 ohm resistor is made equal to 0.25 ohms.
     
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