KVL, KCL - mistake in problem?

Discussion in 'Homework Help' started by StasKO, Jul 29, 2012.

  1. StasKO

    Thread Starter Member

    Apr 28, 2012
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    hi guys,

    i have this problem which seems very easy as it only uses basic kvl and kcl.
    but the answer im getting to is different from the one in the book.
    plus both of the answers (the book's and mine) seems to not settle with kcl and kvl in other junctions and loop which makes me think that there is a mistake in the problem itself.

    the goals are to find the currents Ix and Ie2 (both of them are marked in the right side of the circuit).

    there is also another question that asks if the power output of the two current sources will change if the points a and b will be connected with a short circuit (just wire no resistance). the answer says that only Is2 (the source just under the points) will change its power because it will have a different voltage drop on it. that doesnt makes sense to me because of kvl in that loop - connecting a wire between a and b will not change the voltage source next to the points so kvl in that loop will remain the same.

    please help me understand this problem

    thank you!

    ps:
    if its not clear - the dependant current source (Is2) is αIx (and not α minus Ix).

    EDIT: on the resistor R2 the voltage drop is 17.5v (i just noticed that the point is not seen)
     
    Last edited: Jul 29, 2012
  2. WBahn

    Moderator

    Mar 31, 2012
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    Consider a two-terminal black box that has a current source and a voltage source connected in series within it.

    Q1) To the world outside the black box, what does the black box look like?

    a) A current source.
    b) A voltage source.
    c) Neither (i.e., something that is neither purely a current source or a voltage source).

    Answer that, and we will go from there.
     
  3. StasKO

    Thread Starter Member

    Apr 28, 2012
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    c. neither
     
  4. WBahn

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    Mar 31, 2012
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    Okay, let's test that answer.

    I have a two black boxes. One box, Box A, has a 1A current course in series with a 100V voltage source and the other, Box B, only has a 1A current source. The single port on each box has the current source pushing current out of the positive port and, for the one with the voltage source, the positive terminal of the voltage source is connected to the positive box port.

    Can you devise a test that would let me distinguish Box A from Box B?

    I give you one of the boxes and ask you to measure the open circuit voltage and the short circuit current. What do you get if I've given you Box A? What do you get if I've given you Box B?

    I now ask you to hook it up to a 10Ω resistor. What are the port parameters that you measure (by 'port parameters', I mean the voltage across the box ports and the current flowing into the positive box terminal)?
     
  5. StasKO

    Thread Starter Member

    Apr 28, 2012
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    if its box A:
    oc voltage is 100 volts. sc current is 1A. with 10 ohm resistor then current is still 1A and voltage drop on resistor is found by using ohm's law.

    if its box B:
    0 volts oc voltage (actually im not sure about this). with 10 ohm resistor connected then current is 1A and voltage drop on resistor is the same..

    i guess the conclusion is that the outside world doesnt care if in the box there's only a current source or a current source in series with voltage source.

    but how does it relate to my question?
    it is the voltage source who gets short circuited... not the the voltage source AND the dependant current souce...

    plus i also asked if you can check for a error in the problem itself because it seems that there is one..

    thnx
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    The question (at least the second part) asked how the power in the two supplies would change if the voltage supply were replaced with a short. So, once you know that the presence of the voltage supply can't be seen by the circuit outside of a black box containing the voltage supply and the current supply, what do you know about the effect that shorting it has the Is1 supply?

    For the the Is2 supply, things are unfortunately presented poorly, If you short 'a' to 'b', then if you walk up from the bottom of the battery to the top, you get an increase of 5V if you go through the battery and 0V if you go through the shorting wire. So which is it? The answer is that you have a mathematical contradiction that can't exist in the real world -- because you are assuming the battery is ideal, you assume it can deliver any amount of current needed to maintain the 5V. But the short is also assumed to be ideal, and you assume that it will have 0V across it no matter how much current is flowing. Essentially, you have a division-by-0 problem and there is no answer. Instead, either imagine turning the voltage supply 'off' by turning the voltage down to zero, remove the supply and replace it with a short, or assume a small internal resistance within the battery. All three will yield the same intended result, namely that the voltage from node 'a' to node 'b' is identically 0V.

    In light of that explanation, go back and revisit you answers above. In particular, what is the open-circuit voltage of an ideal current source?

    As for the error in the problem, you have given neither the book's answers or your own, so how am I supposed to determine if either is correct?
     
  7. StasKO

    Thread Starter Member

    Apr 28, 2012
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    so how i understand it is that because the voltage supply goes unnoticed by the outside world than all of the voltage drop occurs on the current supply. but i still having problem of understanding what really happens when measuring open circuit voltage when only an ideal current source is connected. on one hand i think the oc voltage is 0 volts because its an oc which means no current.. but on the other hand the current source does "want" to push current which means that there is a voltage across the oc. i think i just dont understand in a more of an intuitive way what is a current source..

    about the question itself - my answer is that Ix = 0.5A and the book's is 2A (this question is when the a and b terminals are NOT shorted)

    sorry if i make you a headache :) and thnx
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Just as the voltage/current for a ideal voltage source that has been shorted is a problem (because it involves and undefined, infinite current flow), the voltage/current for an open-circuited ideal current source is a problem because it involved an undefined, infinite voltage.

    In either case, the problem is usually not solved strictly as given. The best approach is to show that you recognize this, briefly explain the issue, and then make a reasonable modification to the problem that satisfies what you believe is the intent of the problem (which may not be easy to figure out).

    I'm on travel tomorrow, so I'll take the problem with me and work it on the airplane and hopefully post tomorrow evening from the hotel.
     
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  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As a check I have Ix=2A and IE2=1A [when a and b are open].
     
  10. StasKO

    Thread Starter Member

    Apr 28, 2012
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    my full solution is this:
    1- KVL to the loop containing R5, R6, E3, E2: result is 5 volts drop on R6.
    2- KCL to upper-left node results in 22.5v on R1
    3- KVL to loop containing R6, E3, R2, R1, E1, R4: results in 5v on R4
    4- ohm's law on R4 gives Ix=0.5A

    with the answer that Ix=2A, KVL is not satisfied in the outer loop containing R6, E3, R2, R1, E1, R4

    i might be wrong here somewhere.
     
  11. t_n_k

    AAC Fanatic!

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    I believe the circuit annotations are incorrect at some point - hence the different answers depending on how one derives the answer. Your approach is sound as is mine for the method I used.

    I used the relationships

    IR3+Ix=3.5
    10+10*Ix=20*IR3

    The first equation is based on the current in R2 being IR2=VR2/R2=17.5/5=3.5A which 'must' be the same as the current IE1 and hence the sum of the currents Ix & IR3. Using your value of Ix=0.5A would give IR3=15/20=0.75A and hence Ix+IR3=1.25A which is inconsistent with the annotation of VR2=17.5V ... and so forth.

    A useful check might be to add all the powers contributed / absorbed from the sources and compare the result with the total resistor power dissipation. That should indicate a discrepancy.

    In any case - dud question I think.
     
    Last edited: Jul 31, 2012
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  12. StasKO

    Thread Starter Member

    Apr 28, 2012
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    yeah your solution is exactly as the book's..

    well so i guess we can conclude that there is a mistake somewhere in the question itself..
    weird enough but this question actually appeared in a semester exam a few years back.. and is still given for today's students (like me) as a way to prepare for the exam.. it seems like no one has yet noticed the difficulty of this circuit..

    well thnx anyway
     
  13. t_n_k

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    Mar 6, 2009
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    Well that just shows you how it's possible even for the teacher [or teachers] to make an error and then perpetuate it.

    Interestingly, it's worth deciding what would change as far as annotation of circuit values if everything was correct.

    Suppose we still require that VR2=17.5V and therefore that Ix=2A. What would the outcome be for the various circuit parameters?

    It's possible to show that with

    VR2=17.5
    and
    α = 2.875

    then ...

    Ix=2A [as supposedly 'correct' in the original solution]
    IE2=4A [rather than the erroneous 1A]
    VR5=-5V [rather than +10V]
    and
    αIx=5.75A

    It's worth noting that source E4 has no effect on the outcome irrespective of its value.

    A good test for your skill would be to show how things would change if Ix was actually 0.5A as you found.:)
     
  14. hexram

    New Member

    Feb 29, 2012
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    You are right, StasKO; there IS a mistake somewhere in the question itself: Too much incompatible information is given. If you omit one of VR2 or VR5, then the problem is solvable BUT each solution is DIFFERENT...
    Anyway, a series connection of an ideal voltage source and an ideal current source acts just like an ideal current source alone; voltage across an ideal current source is unknown as is the current through an ideal voltage source because in both cases the parameter depends on whatever load is connected to the ideal sources. Problem is that the current through the voltage source is determined by the current source, but the voltage across the current source is an unknown and the result of the sum of a known voltage (the one in the voltage source) to an unknown voltage (the one in the current source) is unknown WHEN THEY ARE IN SERIES. So, it makes no difference to have E4 in the circuit or not in terms of what the rest of the circuit elements get as resulting current and/or voltage.
     
  15. WBahn

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    Mar 31, 2012
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    Sorry I haven't gotten back sooner -- they put me up in a nice hotel, which means you don't get anything for free, such as internet or use of a business center.

    As others have pointed out, there is a problem with the additional information given. You can't just solve the circuit without any additional information because you have an additional unknown, namely alpha. So you need one piece of additional information. They gave you two, and the two are not consistent. Hence you have three possible outcomes:

    Use the circuit plus the 17.5V across R2 and come up with one solution having a particular value for alpha. Use the circuit plus the 10V across R5 and come up with a difference solution having a different specific value for alpha. Use both of the annotated voltages and come up with a solution that is simply wrong and that is going to fail KVL/KCL at some point.
     
  16. StasKO

    Thread Starter Member

    Apr 28, 2012
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    thnx everybody!!

    the only way i found to correct the circuit is by changing R3 to 5Ω (i tried to not change the values of the components, but didnt manage to fix the circuit when "playing" only with the voltage drops) .
    IE2 = -0.5A (which means that in the short wire to the left of IE2 there's no current)
    αIx = 3A which gives α=6
    the same thing could have been achieved by changing R4 to 100Ω


    EDIT: i think i found a fix without changing component's values (please check):
    since im assuming that Ix = 0.5A, then current through E1 = 1.25A
    this makes the voltage drop on R2 = 6.25v (instead of 17.5v as shown) and on R1 = 56.25v (current direction is downwards)
    to satisfy KVL, voltage drop on R6 = 50v (current direction is downwards) and on R5 = 55v (instead 10v as shown)
    this results in IE2 = -9.5A and αIx = -11.5A which gives α = -23
     
    Last edited: Aug 2, 2012
  17. t_n_k

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    Nicely done.:)
     
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  18. WBahn

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    Yes, good job.

    We frequently learn more from a bad problem then from a good problem. If forces us to really consider what is going on and to defend our reasoning with solid analysis.
     
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  19. t_n_k

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    Here's a final little 'puzzle' [mainly for StasKO's interest] - suppose the controlled source 'α' value was set to zero. What would be the outcome and would the 5V source E4 now have an influence on the resulting value of Ix?
     
  20. StasKO

    Thread Starter Member

    Apr 28, 2012
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    umm good puzzle...

    if we set α to zero then Is2 will be set to zero which means that now there will be an open circuit instead of the dependent current source.

    this will change KCL since there wont be any current in the branch where Is2 was. this in turn will change the voltage drops on the resistors and eventually the currents in the circuit, including Ix.

    E4 will still have no effect on the circuit (as i understand it, the voltage drop that was on Is2, is now the voltage drop on the open circuit)

    im really not sure that this is correct..
    ive always had difficulties with the "if we change x, will y change" kind of questions..
     
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