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How do i solve the following circuit?

  1. Jul 25, 2011 #1
    I have the following circuit and i want to look for the current and voltage across R1:
    2zeytjq.png

    Let's ignore the numeric value for v1, v2 and R1 and generalize them.

    If i apply Mesh Analysis i get the following formulas:
    [tex]v_1 = (i_1 + i_2) R_1[/tex]
    [tex]v_2 = (i_1 + i_2) R_1[/tex]

    And if i apply Crammer's rule i get:
    [tex]\Delta = R_1^2 - R_1^2 = 0[/tex]
    [tex]\Delta_1 = v_1 R_1 - v_2 R_1[/tex]
    [tex]\Delta_2 = v_1 R_1 - v_2 R_1[/tex]

    So, if i want to get i1, i2:
    [tex]i_1 = \frac{\Delta_1}{\Delta} = \frac{v_1 R_1 - v_2 R}{0}[/tex]
    [tex]i_2 = \frac{\Delta_2}{\Delta} = \frac{v_2 R_1 - v_1 R_1}{0}[/tex]

    And if v1, v2 are equal it comes even worse!:
    [tex]i_1 = i_2 = \frac{(0)(0) - (0)(0)}{0}[/tex]

    As for the voltage across R1:
    [tex]v_R1 = (i_1 + i_2) R_1 = (\frac{v_1 R_1 - v_2 R}{0} + \frac{v_2 R_1 - v_1 R_1}{0}) R_1 = \frac{0}{0} R_1[/tex]
     
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  3. Jul 26, 2011 #2
    I am not sure what you are trying to do here. V1 and V2 are in parallel - they will always be the same. Putting voltage sources in parallel is a bad idea! The current through the resistor is simply V/R.

    Plugging in zeros for everything just makes the current undefined.
     
  4. Jul 26, 2011 #3

    EWH

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    If V1 does not equal V2, then the non-grounded node of the circuit will have two different voltages at the same time, which seems impossible for ideal voltage sources and current analysis. In real life there would be a big current through the wire, limited only by the internal battery source resistances and the resistance of the wire. If V1=V2, then just remove one or the other. (Perhaps finding the equivalent parallel source resistance if the problem calls for it.)
     
  5. Jul 26, 2011 #4
    I have actually done this experiment myself with two or more batteries. I know that the answer happens to be the average of all the input voltages. However, i can't get the formulas to come out as the averages of the sources (regardless of the methods i use for solving the equations). This is the reason why i created this topic.

    I believe that result happens to be the average of the inputs because each source is trying to make the voltage difference to be the amount each battery of them supply. In doing so, they play a tug of war until they settle onto a value that represents all of them which is also the definition of average.

    It is generalized that components parallel to a voltage source will have the same voltage. However, it doesn't mean that two or more voltage sources can't be in parallel. Although, it is true that it doesn't make sense to have multiple voltage sources in parallel; i want to be able to calculate and analyze the effects of it.

    If you actually look at it closely, i didn't use zero or any specific numeric value for the components. Ignore the drawing saying 0V because i could not get rid of those labels on the schematics. I used general variables and the equations without plugging any values gave me 0/0.
     
    Last edited: Jul 26, 2011
  6. Jul 26, 2011 #5

    berkeman

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    Staff: Mentor

    You cannot put two ideal voltage sources in parallel, because they have zero internal output resistance, so the "tug of war" current would be infinite.

    You can put two real voltage sources in parallel, but you need to be careful, because the lower the output resistance of the voltage sources, the higher the fighting current will be. This can easily lead to a fire or the bursting of batteries. Please do not do this kind of experiment without understanding the basics of circuit operation.
     
  7. Jul 26, 2011 #6
    Thanks for the explanations and allow me to be a little sarcastic. It is not that we "cannot" put two ideal voltage sources in parallel. But that the result in putting two ideal voltage sources in parallel could be hazardous (like having infinte output current which you pointed out). On the other hand, because real voltage souces have internal resistance and they want the output voltage to be the same as their own; they end up messing around with the internal resistance of the parallel voltage source(s).

    And since I wanted to see how it worked out mathematically I created a modified circuit. It includes R1, R2 as the internal resistance of the voltage sources and R3 as the reistance of the load:
    2r2ay44.png
    [tex]v_1 = i_1 R_1 + (i_1 + i_2) R3 = i_1 (R_1 + R_3) + i_2 R[/tex]
    [tex]v_2 = i_2 R_2 + (i_1 + i_2) R3 = i_1 R3 + i_2 (R_2 + R_3)[/tex]
    [tex]\Delta = (R_1 + R_3)(R_2 + R_3) - R_3^2 = R_1 R_2 + R_2 R_3 + R_1 R_3[/tex]
    [tex]\Delta_1 = v_1 R_2 + v_1 R_3 - v_2 R_3[/tex]
    [tex]\Delta_2 = v_2 R_1 + v_2 R_3 - v_1 R_3[/tex]
    [tex]i_1 = \frac{v_1 R_2 + v_1 R_3 - v_2 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3}[/tex]
    [tex]i_2 = \frac{v_2 R_1 + v_2 R_3 - v_1 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3}[/tex]

    From here we can see:
    [tex]i_3 = i_1 + i_2 = \frac{v_1 R_2 + v_2 R_1}{R_1 R_2 + R_2 R_3 + R_1 R_3}[/tex]
    On the top of the fraction we can see that v1 messing with v2's resistance and vice-versa.
    While on the bottom of the fraction we can see that it is R1||R2||R3, which indicates a simple voltage to current source transformation:
    2yxhztl.jpg

    So assuming that both voltage sources have the same internal resistance R1 = R2 = R and let the limit go to zero:
    [tex]\lim {R \to 0} = i_3 = \frac{R (v_1 + v_2)}{R^2 + 2(R)R_3}[/tex]
    Without simplifying any further, we can see that the bottom part dominates which leads to infinite current.

    However, if we do a proper simplification:
    [tex]\lim {R \to 0} = i_3 = \frac{v_1 + v_2}{R + 2R_3} = \frac{v_1 + v_2}{2R_3}[/tex]
    Voala! The average of both voltage sources and the voltage across R3 is simply:
    [tex]v_3 = \frac{v_1 + v_2}{2R_3}R_3 = \frac{v_1 + v_2}{2}[/tex]
     
    Last edited: Jul 26, 2011
  8. Jul 26, 2011 #7

    berkeman

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    Staff: Mentor

    And to add one more thing to your understanding of this...

    Real power supplies typically can only pull *up* on their voltage outputs. If you look at the equivalent circuit of a LM7805 linear voltage regulator, for example, you will see that the pass transistor can source current into the load, but there is no mechanism to sink current from the load if another power supply pulls the output up higher than 5V.

    So in that situation, the highest power supply would determine the combined output voltage.
     
  9. Jul 26, 2011 #8
    I am not quite sure how transistors work. But do the voltage supplies on the transistors end up parallel and not in series for the output voltage to be a "pull up"?
     
  10. Jul 26, 2011 #9

    EWH

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    With LTSpice, it throws an error of "overdetermined circuit" if there are no source resistances. With e.g. V1=10V V2=5V, R=1K, and V_s for both V1 and V2 = 0.1 Ohm, then:
    V(n001): 7.49963 voltage
    I(R1): 0.00749962 device_current
    I(V2): 24.9963 device_current
    I(V1): -25.0037 device_current

    With real batteries, one will be trying to charge the other and hydrogen could be produced, perhaps bursting the battery and even causing a fire or explosion. Also 25 A could easily melt most wires, again potentially leading to a fire.

    With a more reasonable situation V1=9.2V V2=9.1V, R=1K, and V_s for both V1 and V2 = 0.25 Ohm:

    V(n001): 9.14886 voltage
    I(R1): 0.00914886 device_current
    I(V2): 0.195426 device_current
    I(V1): -0.204574 device_current

    Which still means that ~95% of the current is not going through the resistor.
     
  11. Jul 26, 2011 #10

    berkeman

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    Staff: Mentor

    I don't quite understand your question. A transistor like the pass elelment NPN transistor in a LM7805 linear regulator is a "pullup" element controlled by a feedback circuit. If the output voltage is below the 5V that is desired, the pullup transistor is turned on harder to try to pull up the output voltage.

    If the output voltage is above the desired 5V, all that the pullup transistor can do is shut off, and hope that the load pulls down the 5V until it is back in regulation.

    BTW, never anthropomorphise transistors. They hate it when you do that. :tongue2:
     
  12. Jul 26, 2011 #11
    Well you see, we are talking about the effects of voltage regulation when two voltage sources are in parallel. Then you come up with the analogy of transistors as an explanation on how the behavior on the parallel voltage sources work. What i understood from your analogy is that there is no such thing as a "pull downs" but only "pull ups". In other words, what i understood from you is that in transistors a lower voltage can be amplified and not de-amplified.

    But that is when i say, is the configuration on the transistor exactly like having two power sources in parallel? Like i mentioned, i don't know much about transistors. All i know is that there are three terminals. One terminal has a fixed voltage value the collector. The other terminal is the base which only if the current flowing in there is high enough then current flows on the emitter terminal.

    But then i don't know exactly the internals of a transistor is configured. But in my opinion, only if some part of the transistor can be substituted by two parallel voltage sources then we can use the transistor as a explanation on how parallel voltage sources work.

    The reason i say that is because it is hard to say whether there really is just a "pull up" or "pull down" dominated by only one of the parallel voltage sources. Let's take for example V1>V2 and V1||V2. In V1's point of view, there is a pull down but in V2's point of view there is a pull up. However, because both V1 and V2 have the exact same terminals, when we try to measure one source's pull we end up measuring the other one too.
     
  13. Jul 27, 2011 #12

    berkeman

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    Staff: Mentor

    All I was pointing out is that there are different kinds of "voltage sources" in real life. Batteries (depending on what type) will be able to both source and sink current. Power supplies will generally only be able to source current (think of a battery with a diode in series with the output). Power amplifiers are able to both source and sink current, because their outputs have both pull-up and pull-down transistors.
     
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