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Find the current in a circuit - Do I use Kirchhoff's?

  1. Jan 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Using the following values for the circuit in the figure, compute the current through R3.

    |-----------R1--------|-------R3---------|
    |..............................|.........................|
    +..............................|.........................+
    V1............................R2.......................V2
    -..............................|..........................-
    |..............................|..........................|
    |-----------------------------------------|

    V1 = 1.500V V2 = 1.082
    R1 = 2787 ohm R2 = 7219 ohm R3 = 10 ohm

    (I hope my ghetto diagram makes sense. Ignore the ......... I had to use them for spaces. The dashes represent the wires.)


    2. Relevant equations
    Kirchhoff's 2 equations I assume


    3. The attempt at a solution
    I have tried to use kirchhoff's equations to make a system of equations.

    I tried using the "node" equation at the top in between all the resistors, with two "loop" equations. I also tried a system with three "loop" equations. I keep getting bad answers like 2(I3)(R3)=0

    Any help/clues are greatly appreciated.

    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 17, 2007 #2

    berkeman

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    Staff: Mentor

    Show us the equation that you write for the KCL at the middle node, at the top between R1 and R3. Assume that the voltage of the bottom node at the bottom of R2 is 0V.
     
  4. Jan 18, 2007 #3
    Can't you just use ohm's law? V/I*R
     
  5. Jan 18, 2007 #4

    mjsd

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    Homework Helper

    |-----------R1-------[V]-------R3---------|
    |..............................|................... .......|
    +..............................|.................. .......+
    V1............................R2.................. .....V2
    -..............................|................... .......-
    |..............................|.................. ........|
    |------------------[GND]------------------|

    I have labelled your diagram with V and GND at the relevant node. V is an unknown at this stage which denotes the potential at that top node with respect to the Gound (GND, this is your reference node) below.

    Now try using KCL at [V] and ohms law (to express current in terms of Voltages and Resistances that are given) to write down an equation of V. Solve V and then you can go back and work out I3 using ohms law again.
     
  6. Jan 18, 2007 #5
    Im sorry I still dont understand how to do it....

    KCL at V is I3+I1=I2 right? Using Ohm's law to express this in terms of Voltages and Resistance gives me (V3/R3)+(V1/R1)=(V2/R2). Solving for V2, or [V] gives me V2 = (R2V3/R3)+(R2V1/R1). Now I work out I3 using Ohms Law, I3 = V3/R3.

    Do I solve V2 = (R2V3/R3)+(R2V1/R1) for V3/R3 to find I3? If I do that I need to know term V2....

    I have spent so much time on this question over the past days, I can barley think about it any more. Unfortunalty threre are no tutors available to me until next week after it is due.
     
  7. Jan 19, 2007 #6

    berkeman

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    Staff: Mentor

    I don't think your last equation is correct. It should be more in the form of:

    (V-V1)/R1 + (V- ...)/R.... + ... = 0

    And then solve for V.
     
  8. Jan 19, 2007 #7
    Thanks for sticking with me.

    What I did was try to get I3 + I1 = I2 in terms of voltages and resistance. I3 = V3/R3, I1 = V1/R1, I2 = V2/R2.

    (V-V1)/R1 + (V- ...)/R.... + ... = 0 Is this a loop equation around the outer loop? (V-V1)/R1 + (V-V2)/R3 - V2 + V1 = 0?
     
  9. Jan 19, 2007 #8

    berkeman

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    No, I3 is *not* V3/R3. It is (V-V2)/R3. Do you see the difference?

    The best way, IMO, to do the KCL equations is at each node, set the sum of all currents OUT of the node to zero.
     
  10. Jan 19, 2007 #9
    No Im sorry I dont see the difference... V-V2 is the voltage at teh top node minus the voltage from V2, basically the voltage drop across R3. Thats what I meant by V3... maybe my notations are inconsistent...

    The best way, IMO, to do the KCL equations is at each node, set the sum of all currents OUT of the node to zero. In my book they have an example where they say the direction of current is arbitrary. I dont know the direction of the current at the top node. You say sum of all currents out of the node set to zero, but that is not what is really happening is it?
     
  11. Jan 19, 2007 #10

    berkeman

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    Staff: Mentor

    Yes, your notation will not help you solve the problem. Stick with the listed voltages for the solution. Making up a new voltage difference will not simplify the equation any.

    Were you able to solve for V?
     
  12. Jan 19, 2007 #11

    mjsd

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    Homework Helper

    you must be consistent everywhere, otherwise it defeats the purpose of setting V and GND (the reference, which is set a zero volts). All you need to do is write everything in terms of V. For example, if you want to find out I3 (which is say defined to be flowing OUT of node V, OUT/IN is really your choice) going through R3, you go and look at the circuit... ohms law tells you that voltage across R3 divided by R3 is the current through it. At this point you also have to be careful about sign, because you have defined it flowing out of node: that branch of diagram looks like
    .........I3 ===>
    [V]------ R3 ------( + V2 -)-------[GND =0 volts]
    ........+ V3 -

    ==> indicates reference direction of I3.

    So to find I3 (as in the direction defined), you first say to yourself, since conventional current flows from Higher to lower potential... to get I3 (in that direction) must imply V3 (as defined in diagram) equals to V-V2 (that's your potential difference across R3!) and hence using Ohms law, I3 = V3/R3 = (V-V2)/R3.

    you do the same for all branches and use KCL to write one big equation in variable "V".. and then solve for it... (NB everything has been done with respect to the GND chosen)

    your equation would look something like what berkeman said.

    by the way this is only one way to do it, there are other ways. In general however, node voltage analysis always works.
     
  13. Jan 19, 2007 #12
    Ok, I think I have it now. Thank you all very much.

    I solved it the way suggested here, but in the book they dont use a [V] for similar examples. But after solving it your guys way with [V] I was able to do it the books way without introducing a [V]. Very similar methods.
     
  14. Jan 20, 2007 #13

    mjsd

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    If it is a physics book, I am not surprised. Only an Electrical Engineering book tends to introduce node-voltage analysis and mesh current analysis formally.... anyway, the only fundamental thing here is KCL and KVL.
     
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