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Circuitry Question

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data


    http://i967.photobucket.com/albums/ae152/timmy_064/circuitryprob.jpg


    2. Instructions: Find the unknowns : V1, V2, V3, i1, i2, i3, i4



    i wasn't able to solve this problem due to the circuit containing 2 batteries. its confusing. can someone help me. plz. thanks.

    EDIT: Since there are two batteries with different voltages I was wondering what the actual circuit voltage is? thts all i need to know and ill figure the rest out
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 8, 2010 #2
    I1 and I4 are the same.

    There are two loops to use Kirchhoff's loop laws. You'll also need Kirchhoff's current law in some point.
     
  4. Apr 8, 2010 #3
    do u know what the actual circuit voltage is for this circuit since there are 2 batteries with different voltages? thanks for your reply btw.
     
  5. Apr 8, 2010 #4

    collinsmark

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    Hello VJ_1991

    Welcome to Physics Forums,

    I don't think there is such a thing as some sort of overall "actual circuit voltage", particularly in this case where there are multiple voltage sources. The voltage between any two given nodes depends upon which nodes you are "probing" (i.e. which nodes you are taking into consideration; i.e. between which particular nodes you are finding the potential difference).

    As willem2 suggested, the way to solve this problem is to use Kirchhoff's laws.
     
  6. Apr 8, 2010 #5
    What do you mean with "The actual circuit voltage". Kirchhofs loop rule is only about the voltage differences across the various batteries and resistances.

    Another way of doing this kind of problem:

    1 replace any series and parallel resistors (not necessary here)
    2 assign a potential of 0 to some point of the circuit. (for example the wire at the bottom)
    3 assign an unknown potential V_1, V_2, ... to all other points where 3 or more wires come together (only one point here where the 3 resistors connect)
    4 Work out the currents in all the wires between all those points as a function of the unknown voltages
    5 use Kirchhofs current law in all the points of the circuit from step 3 to get one equation for each unknown voltage (only one equation here)
    6 solve the equation(s)
     
  7. Apr 8, 2010 #6
    is it possible to use voltage division in this circuit? if so, which battery will i use( 20 V or 32V)? or will i use the difference between the two batteries when calculating for voltage division?

    PS: Thanks for your help collinsmark and willem2
     
  8. Apr 8, 2010 #7

    collinsmark

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    The way I would approach this problem is to use Kirchhof's current law.

    Define two different current loops in the circuit (say, i1 and i2 -- you can even use them the way the are already defined on your circuit diagram if you want to [or you an define your own if you prefer]). For each current loop, go through and add up the respective voltage drops, setting everything to 0.

    You'll end up with 2 (simultaneous) equations and 2 unknowns. Solve for the unknowns using substitution, linear algebra, or whichever method you prefer to use for solving simultaneous equations.

    There is another way to solve this problem using Thevenin equivalent circuits. After performing some Thevenin equivalent transformations, you can end up with some voltage dividers, so to speak (you'll have to do some work to get there though). Whatever the case, you can end up solving the problem that way. I've always preferred using Kirchhof's current law, but Thevienin equivalent approach works too.

    [Edit: Technically, there are also Norton equivalent circuits/transformations that are sort of the same thing as Thevenin equivalent circuits/transformations, but in the other direction (above, I just grouped everything under the name Thevenin -- but technically there are both Thevenin and Norton equivalents). If you use these transformations to solve this problem, you can avoid solving simultaneous equations. But it might take more notebook paper, because you'll be redrawing the circuit a lot, and there is still a lot of math involved.]
     
    Last edited: Apr 8, 2010
  9. Apr 8, 2010 #8
    i agree, this is the method i've learned in school. btw im a gr.12 high school student.

    so the two equations i came up with are:

    Eqn 1: -V4 + V1 + V3 = 0

    Eqn 2: -V3 + V2 + V5 = 0

    Note: V4 = 32V and V5 = 20V( i assigned the two batteries V4 and V5)

    are these equations correct?
     
  10. Apr 8, 2010 #9

    collinsmark

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    Yes, those equations will work and are valid. The next thing you need to do solve for the currents which define each loop. You can do this by bringing in the resistor values, and by noting that V = iR.

    A couple of things to be careful of:
    (1) Be careful of how you handle the voltage drop across R3 (i.e. V3). It involves the currents in both of your loops.
    (2) The direction in which you have defined your second current loop is opposite that of i2, as labeled on the diagram. There's no problem with that. Just don't forget the negative sign when you give your answer for i2.
     
  11. Apr 8, 2010 #10
    thanks, it helped a lot. just another quick question, does i1 = i2 ?
     
  12. Apr 8, 2010 #11

    collinsmark

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    No. However, because the sum of all the currents going into a particular node must add up to zero, it is correct to say that i3 = i1 + i2. (where i1, i2, and i3 are defined according to your attached circuit diagram).
     
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