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Evaluating complex circuit

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the drop in potential across each resistor:

    8wjbrm.jpg

    2. Relevant equations

    V = IR
    Kirchhoff's circuit laws

    3. The attempt at a solution

    V1 + V2 has a current of I1 that goes through R1 and R3 which are in a parallel circuit. V3 has a current of I2. The two current combine so they go through R2 with I3 (I1 + I2)

    So I came up with:

    I3 = I1 + I2

    (V1 + V2) - I1*[(1 / R1) + (1 / R2)]^-1 - R2*I3 = 0

    I'm not sure if the above is right or how to find the third equation. Any help is appreciated.
     
  2. jcsd
  3. Mar 21, 2010 #2

    ehild

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    Gold Member

    A nice drawing always helps. Show the currents with arrows in the circuit diagram.


    ehild
     
  4. Mar 21, 2010 #3

    Filip Larsen

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    This does not look right, no.

    You should probably think about what the potential over your three top branches are (hint: you can directly read this potential on the diagram).

    After that you only have the R2 branch left, which then consist of a simple circuit with two voltage sources (remember to get the sign of the voltage right here) and one resistance.
     
  5. Mar 21, 2010 #4
    I've tried that on a sheet of paper (which is how I came up with those 2 equations) but I still don't know what to do. =\

    Sorry but I don't really understand this...the potential in the top three branches is V1 + V2? Not sure where to go from there.
     
  6. Mar 21, 2010 #5

    Filip Larsen

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    Look at the bottom branch with V3. What does this (ideal) voltage source do to the potential over each of the other three branches?

    If you still have trouble answering this it may be that you are confused over the V1+V2 branch. If so, try think about V3 again but now assume that that V1+V2 is zero (ie. that it for a short while has been replaced by a piece of wire). Now you have one voltage source V3 and three resistors in parallel and you should be able to see what the potential over each of the three resistors is. If this still seems puzzling try simplify further by removing two of the resistors so you end up with V3 and one resistor.
     
  7. Mar 21, 2010 #6
    Ah I think I get it now...so the V3 will counter-act the V1+V2 in the R1 and R3 branch. So the voltage drop in those two branches will be V1+V2-V3. And the voltage drop in the R2 branch would be V1+V2+V3? :smile:
     
  8. Mar 21, 2010 #7

    Filip Larsen

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    No, that is not correct. The voltage over R1 and R3 (and also V1+V2+R2) is determined solely by V3. Almost by definition (of an ideal voltage source) will V3 force the potential difference over those two branches to be V3. No matter what components each branch contains the voltage over that branch will be fixed by V3. And since two of the branches only contains a single resistance, the voltage over those resistances are given directly as V3.

    This, however, is correct.
     
  9. Mar 21, 2010 #8
    Thanks for the help!

    If it isn't too much trouble could you clarify why V3 will force the R1 and R3 branches to be V3?
    I'm still having trouble understand why V1+V2 won't effect R1 and R3 =\.
     
  10. Mar 21, 2010 #9

    Filip Larsen

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    Look at Kirchhoffs voltage law which says that the sum of the voltage difference (with sign) over each component in any loop in the circuit is zero. Now apply it to the loop that contains V3 and, say, R1. If we call the voltage from left to right over R1 for U1 (to avoid confusion with V1) and the voltage from left to right over V3 for V3, then Kirchhoffs voltage law for this loop, when we traverse it counter clock-wise on the diagram, says that V3 - U1 = 0 which is the same as saying U1 = V3.

    You should note, that this only works so simple as it does in this case because V3 is assumed to be an ideal voltage source which keep the given voltage difference V3 no matter how much current (and therefore power) it has to deliver. This is obviously a theoretical construction as no real voltage source can truly provide limitless power. It is possible to make more realistic models of a voltage source (simple models just include a resistance in series) that as a result will make the voltage drop the more current you draw from it. To employ such a source in your circuit you would replace the V3 source with this more complicated sub-circuit so that you then can "include" into the analysis of the voltage-current behaviour of the complete circuit.
     
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