From one circuit to another one

Discussion in 'Homework Help' started by juanjuan, Mar 23, 2015.

  1. juanjuan

    Thread Starter New Member

    Oct 24, 2013
    17
    0
    Using:

    [​IMG]
    where i3=Iab

    i1+i2=i3

    (Now, I'm using that Vb=0)
    1V-i1.1\ohm = 0 => i1=1A
    3V - i2.2\ohm = 0 => i2= 3/2A

    So,
    i3= 1A+3/2A = 5/2A

    Now, using Ohm Law: V=I.R
    Vab=Iab.Rab => 5/3V = 5/2A . Rab => Rab = 2/3 \ohm
     
  2. WBahn

    Moderator

    Mar 31, 2012
    24,454
    7,653
    Correct, but I think you got a bit lucky. Notice that you ignored the current in the upper branch even though you have stated that you believe it to be non-zero. If you are correct and it isn't zero, then that current changes i1 and i2, which would potentially change your results for i3. Don't ever ignore part of a circuit unless you can adequately articulate why it is justifiable to do so.

    So know have the short-circuit current, the open-circuit voltage, and the equivalent resistance of the entire circuit, except for the resistor we removed, as seen between the points where that that resistor was connected ('a' and 'b'). What does the Norton theorem say that this portion of the circuit is equivalent to? If you don't know Norton, then what does Thevenin say that it is equivalent to and, given that, what does source transformation allow you to do in order to convert the Thevenin equivalent into what you need for the answer?
     
  3. juanjuan

    Thread Starter New Member

    Oct 24, 2013
    17
    0
    Of course I didn't use the current in the upperside because it was 0.

    Norton says that I can replace some resistors and sources for just 1 current source and 1 resistor.

    OK, so now I apply Norton's theorem and the problem is solved.That's nice

    What do you think about i0=0. I think that cannot be true, because I have a source and a resistor, so there has to be a current. Why do you say that i0 can be 0?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    24,454
    7,653
    Think about what you have just said -- you don't believe that i0 can be zero, yet you have no problem proceeding with an approach that depends on it being zero. Does that sound like good, defensible engineering practice to you?

    Imagine that you go to a doctor and he has a blood test done and comes back and says that the test shows your iron levels to be way too low and he asserts that that simply cannot be true and the test results must be wrong and your actual iron levels must be considerably higher than the test shows for this reason and for that reason. None-the-less, he prescribes a treatment based on the test result even though it is potentially fatal if, in fact, your iron levels are not as low as the test showed. Would you be comfortable following that doctor's advice? Do you believe that doctor should be practicing medicine?

    If not, then ponder the fact that most incompetent doctors are limited to killing people one at a time, while incompetent engineers can and do kill people in job lots.

    As for how i0 and be zero even though you have a source and a resistor consider the following circuit in which none of the resistor values are known:

    thev5.png

    In general, if the resistance of the top resistor is R0, what is I0 (as defined in your earlier diagram) in terms of Vd and Ve? I'm looking for an expression that just has Io, Ro, Vd and Ve in it.

    Given the actual sources in the diagram, what are the actual voltages at nodes C, D, and E?

    What is I0 for this circuit with these voltage sources?




    , fr
     
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