Circuit Analysis Problems

Discussion in 'Homework Help' started by Chupa, Mar 27, 2008.

  1. Chupa

    Thread Starter New Member

    Mar 27, 2008
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    Hey all,

    I'm having some problems with these question sheets for my Analysis of Circuits course.

    3. Find the current i in the following circuit using superposition

    4. Find the voltage v in the following circuit using superposition

    5. Find the voc, isc and Req for the following subcircuit and hence determine the Thevenin and Norton equivalents

    7. Find v in the following circuit using source transportation

    (the numbers are irrelevant... either way, the attached images should be in order)

    The answers have been supplied but I have yet to to be able to achieve them.

    For #3, while I've tried the only two possible superpositions, I get a value of -8A (Ans: -6A)
    For #4, I get 45V (Ans: -45V)
    For #5, I'm really not sure how to get Voc and Isc. I understand how Req is attained but I get the wrong value.


    Thanks :)
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    For number 3 you only done half the superposition calculations.
     
  3. Chupa

    Thread Starter New Member

    Mar 27, 2008
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    Only half?
    Well for the first one I removed the current source and for the second one I removed the voltage source...
    ...is there anything else I'm forgetting?
     
  4. veritas

    Active Member

    Feb 7, 2008
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    For the voltage source calculation, I get i = 2A, and for the current source calculation, I get i = -8A. Which of your steps varies from that? (on the first problem)

    *edit*
    keep in mind that when your "remove" a voltage source, you replace it with a short, not an open.
     
  5. veritas

    Active Member

    Feb 7, 2008
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    for the second problem, the current values I calculate from +Vo to -Vo are:
    5A for the 15A source
    -4A for the 12A source
    -16A for the 24A source

    how do yours compare?
     
  6. Chupa

    Thread Starter New Member

    Mar 27, 2008
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    Sorry for the late replies

    3) For the Vs calc I get 2A as well, fair enough. But I get -10A for the Is calc.
    Is the short circuit loop (thanks to the missing Vsource) significant in terms of the current division calcs?

    4) Am I supposed to calculate from +Vo to -Vo? If the current from the 24A leads in to the -Vo... shouldn't that technically mean the same as leaving from +Vo? Or am I wrong there?
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    When you add the 2A to the -8A, the result is -6A ... or as you said, the "correct" answer.
     
  8. veritas

    Active Member

    Feb 7, 2008
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    He says he got -10A for the current source loop, and it sounds like he opened the voltage source instead of shorting it. Yes, the shorted loop is significant, because the circuit basically turns into a current divider through a 20 and 30 ohm resistor. Current through the 30 ohm resistor:
    -i = 24A * 20 / (30 +20) = 8 A

    Generally when you have a have a + and - terminal, you consider the current flowing from the (+) terminal to the (-). Therefore, current flowing into the (-) terminal is in the negative direction.
     
  9. veritas

    Active Member

    Feb 7, 2008
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    For your third problem, do you understand how to convert a Norton equivalent to a Thevenin, and vice versa? Once you find the Norton equivalent, Voc is the voltage across the resistor, and from the Thevenin equivalent, Isc is the current through the resistor when you short the output.
     
  10. veritas

    Active Member

    Feb 7, 2008
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    To solve your third problem, you have to convert portions of the circuit back and forth between their thevenin and norton equivalents, until you have it down to a single resistor and voltage or current source.

    The attachment shows the method of conversion.
     
  11. Chupa

    Thread Starter New Member

    Mar 27, 2008
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    (3rd) Hmm, I understand how I can convert the 9V Vsource to its Thevenin equivalent / source transformation. However how would I convert the Isource if I wanted to get a Vsource and in doing so, attaining Veffective (provided they actually end up in series)?

    The example in our notes uses superposition to attain the Thevenin Voc and Req. I tried that and ended up with something quite funny when you negate the Isource. Since it's an OC, the the two 4ohm resistors appear to be isolated particularly if you take into account the Req of them.

    If I perform the superposition by negating the Vsource, I get v1 as (2ohms * -8A) = -16V already. Does this mean the v2 (with the Isource negated) should be 0V in order to attain Voc = v1 + v2?

    Also, the answers as supplied are: Voc = -16V, Req = 2Ohm, Isc = -8A
     
  12. veritas

    Active Member

    Feb 7, 2008
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    Ah, I was looking at it all wrong before; that current source next to the output dominates the output.

    You're completely right about the 4 ohm resistors being isolated when you remove the I-source; this is what makes this circuit easier to solve. You can basically ignore the voltage source entirely.

    First, find Vthevenin by finding the voltage across the output.
    *hint* use superposition *\hint*

    Then, remove all sources (short Vsource and open Isource), and find the resistance between the output terminals.
     
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