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Homework Help: Superposition principle(circuits)

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    i found this circuit somewhere here in this forum


    it asks to find vo
    2. Relevant equations

    3. The attempt at a solution

    i found that vo is 6V using kirchhof's law, the current source i transformed it to voltage soure, and after some calculations i found that it's 6V

    am i right?

    but I cant find the 6 v using the superposition principle, can anyone help me out?

    when i transformed to voltage source i said that 10ohm and 8 ohm resistors are in series? so the resistor of both is 18 ohm

    then this 18 ohm resistor is in parallel with 6 ohm i found the new resistor and said that the current would be

    I = 15/3 = 5 A

    the voltage will be 5*(18||6+3)

    then having the first source and making the same step i find another voltage then i add them up and I don't get 6 V to be the Vo
  2. jcsd
  3. Apr 17, 2010 #2
    First find the voltage contributed by the current source by making the voltage source a short circuit.

    Then find the voltage contributed by the voltage source by making the current source open-circuit.

    Add the voltages contributed to the resistor from each.
  4. Apr 17, 2010 #3
    you mean something like this?

    [PLAIN]http://img405.imageshack.us/img405/4199/91234315.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  5. Apr 17, 2010 #4
    No. You draw two different schematics.

    In the first one, there is no current generator you just leave it off. That's what it means for it to be "open circuit".

    In the second one, you have the current generator but you replace the voltage generator with a wire. Just a plain wire, like it's not there. That's what it means to short it out.

    You solve the two problems and get two voltages. You add them up and that would be the voltage on the resistor in the original circuit. That's what superposition means, you add solutions.

    A current generator with zero current is an open circuit (it's not there.)
    A voltage generator with zero volts is just a wire. It's there but zero volts.
  6. Apr 17, 2010 #5
    i cant really understand 100% the whole procedure, so for the first source i should have this

    [PLAIN]http://img269.imageshack.us/img269/5896/ph1q.jpg [Broken]

    here Itot = E /Rtot

    but what's the Rtot? 24 Ohms or 14?

    if we have Itot

    V1 = Itot*Rtot

    then for the second one we have

    [PLAIN]http://img263.imageshack.us/img263/2521/ph2r.jpg [Broken]

    here Itot = 15/Rtot

    and Rtot is 9

    so V2 = Itot*9

    then Vtot = V1+V2


    Our teacher didnt explain this method at all, the book has no explanation and im having a bad time with it

    i cant understand why we should use this principle when there is kirchhoff rules

    using kirchhof rules i get 6 V is this result correct?
    Last edited by a moderator: May 4, 2017
  7. Apr 17, 2010 #6
    No, not quite. When you replace the voltage generator with a short, you still need the resistor connected to it. That part of the circuit is there, its only the voltage source that gets replaced by the wire.

    The second circuit you drew is ok because when the current generator goes open circuit, you no longer flow current through the resistor on the left.

    Your first post said you can't get the answer using the superposition principle. I am using the superposition principle for the sources, calculating the contribution of each source one at a time. You should use the method that makes the most sense to you.
  8. Apr 17, 2010 #7
    thanks for your help i understand it now

    one last question

    when we have a current source like this


    the one at the beginning and we want to make it voltage source
    V = 3*10 = 30 Volt right?

    but when we make it to voltage source we will have this

    [PLAIN]http://img710.imageshack.us/img710/2458/87121027.jpg [Broken]

    now about this E = 30

    is this the voltage that the circuit element gives to the circuit with the resistance of 10 Ohm

    or is it just the voltage that it gives to the circuit without the resistance of 10 ohm?

    if it is the second then the voltage that it gives will be 30 - Itot*10 right?

    but i dont think it's the second what do you think?
    Last edited by a moderator: May 4, 2017
  9. Apr 17, 2010 #8
    It's fully equivalent in all cases. All you have to do is check these two: do you get 30 volts from both? Yes.

    Do you get 3 amps from both if you short them? Yes you do because 30 volts onto 10 ohms gives three amps.

    You can still use superposition of sources in your new circuit. You'd replace one of the voltage sources with a wire and compute using the other source, then vice versa. Ad the contributions just like before. You should get the same answer.
  10. Apr 17, 2010 #9
    thanks a lot for your help :)
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