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Batteries in Parallel

  1. Mar 5, 2007 #1
    anyone help me out with this probelem?
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2007 #2
    What you want to do is to attack this problem by using nodal equations, as you can see from the drawing this circuit has only one node placed in the junction between the three resistors, you can start by assuming the following:

    1. By KCL the current at the node is equal to 0

    Since we don't know anything about the direction of the currents, we can safely let the node voltage determine the direction.

    2. Let the [tex]V_{x}[/tex] denote the node voltage
    Let [tex]I_{1}[/tex] denote the current from 28.3 V source
    Let [tex]I_{2}[/tex] denote the current from 14.15 V source
    Let [tex]I_{3}[/tex] denote the current in the 16.5Ohm resistance

    3. Applying the nodal equations we get the following:

    [tex]I_{1}+I_{2}+I_{3}=0[/tex]

    Note that the current can take any direction. Now we replace currents with voltage drops divided by resistor values:

    [tex]\frac{V_{x}-28.3}{30.6}+\frac{V_{x}-14.15}{4.32}+\frac{V_{x}}{16.5}=0[/tex]

    Solving for [tex]V_{x}[/tex] we get 12.9

    Now we can solve for the current in the leftmost resistor:

    [tex]I_{3}=\frac{V_{x}}{16.5}=0.784A[/tex]

    DONE!
     
  4. Mar 5, 2007 #3
    thanks a lot for the help...
    much obliged!
     
  5. Mar 6, 2007 #4
    No problem, I hope you understood the steps towards the solution.
     
  6. Mar 6, 2007 #5
    Yeah, you use KCL bc all the current is conserved, and you find the actual current that enters the bottom resistor, after that you can find the current across any of the resistors, using the way you set it up
     
  7. Mar 6, 2007 #6

    russ_watters

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    Staff: Mentor

    In the future, it is better to not just do the problem for the OP but to give little hints to push him/her in the right direction...
     
  8. Mar 6, 2007 #7
    Correct and remember if the batteries were switched the other way around, then you would use [tex]V_{x}+V_{battery}[/tex] instead

    I won't argue about that, but i thought that making the complete description of the steps including the description of what rules i applied would help him better, so he/she will know what to do next time he/she is faced with the problem.
     
  9. Mar 7, 2007 #8
    do you think you can "push me in the right direction" on this one, I know they're looking for the difference between A and B, but i forget how to do the voltage drop and i'm clueless when there's more than one battery
     

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  10. Mar 8, 2007 #9

    Hehe, shure. First you have to determine the current flowing through the loop, note that the current will be the same at all places. When You've determined the current in the loop, start by writing the equation expressed as voltage drops , you should start at point a and move towards point b, note that the first voltage drop across the battery will be negative, since you're moving from -to +, and it will be positive across the resistors and positive across the battery that is near the b point, since the battery is going from + to -.
     
  11. Mar 8, 2007 #10
    do you mean a positive drop...I know the current is 3.614mA, and that it drops almost 16V in between the two batteries, but what do I do about the negative voltage on the other side of the bottom right battery, add it to get a negative number?
     
  12. Mar 8, 2007 #11
    I tried 40.86V and that was wrong, is it 25V - 15.86V = 9.14V as the difference
     
    Last edited: Mar 8, 2007
  13. Mar 8, 2007 #12
    As all the components are in series you can rearrange the circuit to group all the cells and all the resistors.
    This should make it easier to work out the circuit voltage and hence the current.
     
  14. Mar 8, 2007 #13
    got it....take the current of -3.614mA and multiply it by the 2 resistors added together on the bottom to get -15.86V

    Thanks,

    You'll be hearing from me again this week about RC Circuits.....totally new to me
     
  15. Mar 8, 2007 #14
    LOL, ok ;)
     
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