A bridge circuit full of volts

Discussion in 'Homework Help' started by kiwi00, Dec 9, 2015.

  1. kiwi00

    Thread Starter New Member

    Jul 30, 2015
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    Hi! I need a bit of guidance in my assignment. So we've learned about doing some meshes, but the problem we've been left with has voltage sources instead of the common resistances.

    Do I solve it with meshes? The mesh goes like R(I1+I2) but since the given are voltages... can I still use that formula and interchange it to V/(I1+I2)? [I feel like it won't work like this though lol]

    Do I use series parallel? I'm not sure what to parallel though [I'm still a bit confused on this]

    [​IMG]
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Are the 8V, 9V, and 10V the voltages between the indicated nodes and you are being asked to find the voltages between the nodes indicated by V1, V2, and V3.

    If so, then just remember that voltage is conservative (in something like this) meaning that the voltage measured between any two points is independent of the path chosen to go from one to the other.
     
  3. RBR1317

    Active Member

    Nov 13, 2010
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    Kirchhoff's Voltage Law is applicable here. "For any lumped electric circuit, for any of its loops, the algebraic sum of the branch voltages around the loop is zero." A loop is a closed path whose starting node is the same as its ending node. So begin by identifying all loops in the circuit.
     
  4. kiwi00

    Thread Starter New Member

    Jul 30, 2015
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    Ah, I forgot to include the question above but, yes, we were asked to find V1, V2, and V3.

    [​IMG]
    So will it be...
    L1: V1 - 10V + V3 = 0
    L2: (10V - 8V) - V2 = 2V - V2 = 0
    L3: (8V - 9V) - V3 = -1V - V3 = 0

    V1 = 11V
    V2 = 2V
    V3 = -1V

    Ummm... Is this correct?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You don't indicate the polarity of V1. Assuming that the top is the positive, then you are correct.

    But you are making it far more difficult than it is.

    Label the nodes:

    BattBridge.png

    The voltage at a node (relative to some presently unspecified and arbitrary reference) is indicated by a single subscript. So we have Va, Vb, Vc, and Vg.

    The voltage difference between two nodes is indicated by a double subscript according to the convention Vxy = Vx - Vy. Make note that Vyx = -Vxy.

    So let's catalog what we want to know:

    V1 = Vc - Vg = Vcg
    V2 = Vc - Vb = Vcb
    V3 = Vg - Va = Vga

    Now let's catalog what we do know:

    9 V = Vb - Vg = Vbg
    8 V = Vb - Va = Vba
    10 V = Vc - Va = Vca

    Finally, let's write what we want to know in terms of what we do know:

    V1 = Vc - Vg
    V1 = Vc + (Va - Va) + (Vb - Vb) - Vg
    V1 = Vc - Va + Va - Vb + Vb - Vg
    V1 = (Vc - Va) + (Va - Vb) + (Vb - Vg)
    V1 = Vca + Vab + Vbg
    V1 = Vca - Vba + Vbg
    V1 = 10 V - 8 V + 9 V
    V1 = 11 V

    You can do the same for the other unknown voltages.

    Once you understand this, then you can do it much, much faster.

    You get to pick your reference voltage, so let's pick Node G and define it to be at 0V.

    Vg = 0 V

    By inspection, the voltage at Node B is 9V higher than Node G, so

    Vb = 9 V

    By inspection, the voltage at Node A is 8V lower than Node B, so

    Va = 1 V

    By inspection, the voltage at Node C is 10V higher than Node A, so

    Vc = 11 V

    Now we can write down the answers directly:

    V1 = Vc - Vg = 11 V - 0 V = 11 V
    V2 = Vc - Vb = 11 V - 9 V = 2 V
    V3 = Vg - Va = 0 V - 1 V = -1 V

    You should practice by picking another node and specifying it's voltage to be some arbitrary value (for instance, say that Va = 42 V). This write down the other node voltages by inspection and then calculate the desired voltage differences. You should get the same answers.
     
  6. kiwi00

    Thread Starter New Member

    Jul 30, 2015
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    0
    Ugh, I haven't also drawn the polarity on V1, sorry bout that. And, yeah, the top is positive lol. Thanks a lot for the detailed explanation! :)
     
  7. RBR1317

    Active Member

    Nov 13, 2010
    229
    48
    For maximum simplicity, I would choose different loops as follows:

    V1 = +10 -8 +9 = +11
    V2 = +11 -9 = +2
    V3 = -9 +8 = -1
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I wouldn't choose loops at all. I would simply follow a path between the two points of interest that goes through nothing but known voltages (which I think is what you are actually doing -- or, from a different perspective, you could say that I'm using loops implicitly).
     
  9. RBR1317

    Active Member

    Nov 13, 2010
    229
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    That is exactly what I was doing. And one must choose loops in order to apply Kirchhoff's voltage law. But don't confuse a loop with a mesh. All I did was start at the positive point of an unknown voltage and follow a loop of known voltages to the negative point of the unknown voltage, adding the voltage drops (+ sign encountered first) subtracting the voltage rises (- sign encountered first). Also, once an unknown voltage became known I then used it as a known voltage. Fortunately this problem simplified to loops of all known voltages, otherwise I'd have to write equations from the loops, then solve the equations.
     
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