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CIRCUIT ANALYSIS: 7 resistors, 2 Indep. Volt Source, V.C.C.S, V.C.V.S. - find I

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data

    For the circuit below find [itex]I_1[/itex] and [itex]I_2[/itex]:

    [​IMG]


    2. Relevant equations

    KVL
    KCL
    Ohm's Law


    3. The attempt at a solution

    I tried the problem many times, but I always get crazy answers. It seems that every time I need one new equation to have a system of solveable equations, I have to add a new variable and hence I need another equation. It's a vicious cycle that when I get up to 13 variables for all of the V's at the resistors and different I's at nodes 1-4, I get a crazy answer like -1.35 mA for [itex]I_1[/itex]. Does that seem right?

    Any suggestion on what to do about the one-more variable, one-more equation problem? I tried a super-node between nodes 2 and 3. Didn't help though.
     
  2. jcsd
  3. Jan 16, 2007 #2

    AlephZero

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    Start by finding the voltages. You can write down the voltage at nodes 4, 3, 2 (two different ways) and 1 (in that order) without knowing any currents.

    The two ways of getting the voltage at node 2 gives you a relation between Va and Vb, so eliminate one of them.

    Then start finding the currents in terms of the voltages.

    You don't need to add any more nodes.
     
  4. Jan 17, 2007 #3
    Ok, so I get these for the node voltages(4,3,2,2,1):

    [tex]V_4\,-\,0[/tex]

    [tex]V_3\,+\,5V[/tex]

    [tex]V_2\,-\,0[/tex]

    [tex]V_2\,-\,2\,-\,V_4\,=\,0[/tex]
    [tex]V_2\,=\,V_4\,+\,2[/tex]

    [tex]V_1\,-\,0[/tex]

    Are these right? If not, how am I supposed to make these voltage equations?

    Also, I am stuck again, I don't know where to go from here (even if the voltage EQs are correct)!

    Should I use KVL or KCL? I tried to add 2 current variables at Node 2. I used KCL there and I made some EQs for the currents there (that I had an R for) using [itex]i\,=\,\frac{V}{R}[/itex]. I am seriously stuck now though!!!

    Can someone walk me through the most logical way to proceed from here, I am really confused. Thanks
     
  5. Jan 17, 2007 #4

    AlephZero

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    I don't understand exactly what those expresssions are.

    The way I would do this is by looking at the circuit and thinking about what you know, not trying to apply the K. laws in a mechanical way.

    It's "obvious" from the circuit diagram that

    V4 = VA

    V3 = V4 = VA (the current source has no internal resistance so no voltage across it)

    V2 = V4 - 2 = VA - 2 (from the 2V voltage source)
    and also V2 = VB so VB = VA - 2

    V1 = V2 - 2VA = -2 - VA

    Now try and find a node which doesn't have the unknown currents I1 and I2 flowing into it: there is one, node 3. Find all the currents flowing into node 3 by Ohms law. By KCL they add up to zero. That will give you an equation for VA.

    Now you know all the voltages, you can use Ohms law and KCL to find the other currents.
     
    Last edited: Jan 17, 2007
  6. Jan 18, 2007 #5
    Okay Vinny, we'll use nodal analysis or KCL for the problem if that is fine with you. And for the moment, let's avoid supernodes but stick with the conventional nodes.

    To get you started, do this for me: Note down the 4 equations corresponding to the 4 nodes using KCL, i.e., the sum of currents entering/leaving the node equals zero. And do that with only the variables V1, V2, V3, V4, I1, I2 and no other variables.

    Not true. The current source could present a voltage drop/gain without any internal resistance.
     
  7. Jan 18, 2007 #6
    OK, I have added 3 currents to the diagram (in green).

    [​IMG]

    [tex]V_4\,=\,V_a\,=\,V_3[/tex]

    [tex]V_2\,=\,V_4\,-\,2\,=\,V_b[/tex]

    [tex]V_b\,=\,V_a\,-\,2[/tex]

    [tex]V_1\,=\,V_\,2\,-\,2\,V_a[/tex]

    Just what you said above. Now, I use [itex]i\,=\,\frac{V}{R}[/itex] to get the new currents in green.

    KCL: [tex]I_3\,+\,I_4\,+\,I_5\,+\,\frac{V_b}{4K\Omega}\,=\,0[/tex]

    [tex]I_3\,=\,\frac{(5\,V)}{4000\Omega}\,=\,0.00125\,A\,=\,1.25\,mA[/tex]

    [tex]I_4\,=\,\frac{V_2\,-\,V_3}{2000\Omega}[/tex]

    [tex]I_5\,=\,\frac{V_1\,-\,V_3}{4000\Omega}[/tex]

    Now if I combine those four equations above:

    [tex]\frac{V_2\,-\,V_3}{2000\Omega}\,+\,\frac{V_1\,-\,V_3}{4000\Omega}\,+\,\frac{V_4\,-\,2}{4000\Omega}\,=\,-1.25\,A[/tex]

    How do you proceed?
     
  8. Jan 18, 2007 #7
    Ok, lets try NODE 1 first:

    [tex]\left(\frac{-V_1}{1000\Omega}\right)\,+\,\left[\left(\frac{V_3\,-\,V_1}{4000\Omega}\right)\,+\,\left(\frac{V_4\,-\,V_1}{3000\Omega}\right)\right]\,=\,I_1[/tex]

    Is that correct?
     
  9. Jan 18, 2007 #8
    As I have said earlier, to claim that V3 = V4 is incorrect. Read my earlier post.

    Well yes, that is right. But you do know that the above expression gives only 1 nodal equation, that is, the nodal equation for node 3. You would have to do the same thing for the other 3 nodes.

    This is not right- you're forgetting V3.

    These are correct.

    And yes, combine these terms into an equation. Note that Vb = V2. Also, at the moment, let's forget about V2 = V4-2. Now write down again the nodal equation for node 3.
     
  10. Jan 18, 2007 #9
    That's right. :)
     
  11. Jan 18, 2007 #10
    Cool! NODE 2 now:

    [tex]I_1\,+\,\left(\frac{-V_2}{2000\Omega}\right)\,+\,\left(\frac{V_2\,-\,V_3}{2000\Omega}\right)\,=\,I_2[/tex]

    Is that right?

    Is [itex]V_b[/itex] still equal to [itex]V_2[/itex]?
     
  12. Jan 18, 2007 #11
    A small mistake here, check the equation again. Yes, Vb = V2. Two more nodal equations to go (nodes 3 and 4). Keep it up!
     
  13. Jan 18, 2007 #12
    The mistake fixed?

    [tex]I_1\,+\,\left(\frac{-V_2}{2000\Omega}\right)\,+\,\left(\frac{V_3\,-\,V_2}{2000\Omega}\right)\,=\,I_2[/tex]
     
  14. Jan 18, 2007 #13
    That's right, now move on to the other 2 equations.
     
  15. Jan 18, 2007 #14
    OK, for NODE 3:

    [tex]\left(\frac{V_1\,-\,V_3}{4000\Omega}\right)\,+\,\left(\frac{V_2\,-\,V_3}{2000\Omega}\right)\,+\,\left(\frac{5\,-\,V_3}{4000\Omega}\right)\,=\,\left(\frac{-V_2}{4000\Omega}\right)[/tex]

    And for NODE 4:

    [tex]\left(\frac{-V_4}{4000\Omega}\right)\,+\,\left(\frac{V_1\,-\,V_4}{3000\Omega}\,+\,I_2\right)\,=\,\left(\frac{V_2}{4000\Omega}\right)[/tex]

    Are those right? Or did I mess up the path with the 5V independent voltage source and 4Kohm resistor?
     
  16. Jan 18, 2007 #15
    Very good. Now, I would like to have these equations simplified a bit. As an example, for node 4, you wrote:

    [tex]\left(\frac{-V_4}{4000\Omega}\right)\,+\,\left(\frac{V_1\,-\,V_4}{3000\Omega}\right)\,+\,I_2\,=\,\left(\frac{V_2}{4000\Omega}\right)[/tex]

    I want it simplified to become:

    [tex]\frac{1}{3}V_1 - \frac{1}{4}V_2 - \frac{7}{12}V_4 + I_2 = 0[/tex]

    Specifically, I have ignored the '000 in the R's (the V's are still as before but the I's are now in milliamperes) and arranged the equations such that on the left side are the unknowns, ordered V1, V2, V3, V4, I1, I2 and on the right side, the constants. There's a reason for doing all these of course. :)

    Do the same for the other 3 equations and we will proceed from there.
     
  17. Jan 18, 2007 #16
    NODE3: [tex]V_1\,+\,3\,V_2\,-\,4\,V_3\,=\,-5[/tex]

    NODE2: [tex]2000\,I_1\,-\,2000\,I_2\,-\,2\,V_2\,+\,V_3\,=\,0[/tex]

    NODE1: [tex]-4000\,I_1\,-\,\frac{19}{3}\,V_1\,+\,V_3\,+\,\frac{4}{3}\,V_4\,=\,0[/tex]

    Right?
     
    Last edited: Jan 18, 2007
  18. Jan 18, 2007 #17
    Okay, that's close enough. Let me edit a bit...

    Node 1: [tex]-\frac{19}{12}V_1 + \frac{1}{4}V_3 + \frac{1}{3}V_4 - I_1 = 0[/tex]

    Node 2: [tex]-V_2 + \frac{1}{2}V_3 + I_1 - I_2 = 0[/tex]

    Node 3: [tex]V_1 + 3V_2 - 4V_3 = -5[/tex]

    Node 4: [tex]\frac{1}{3}V_1 - \frac{1}{4}V_2 - \frac{7}{12}V_4 + I_2 = 0[/tex]

    I prefer to have the equations in the manner above, with the I's in milliamperes. Now, note that there's a voltage source across nodes 1 and 2, similarly a voltage source between nodes 2 and 4. Due to that, we ought to form a supernode, that is, to combine nodes 1, 2 and 4 into a single supernode. The result is the elimination of the unknown variables I1 and I2. To do that, try combining equations 1, 2 and 4 above into a single equation such that the unknowns I1 and I2 disappear.
     
  19. Jan 18, 2007 #18
    OK, I did N1 + N2 + N4.

    [tex]-\frac{5}{4}\,V_1\,-\,\frac{5}{4}\,V_2\,+\,\frac{3}{4}\,V_3\,-\,\frac{1}{4}\,V_4\,=\,0[/tex]

    Is this what the supernode would look like?

    [​IMG]
     
    Last edited: Jan 18, 2007
  20. Jan 18, 2007 #19
    Good. You now have the following equations:

    Node 3: [tex]V_1 + 3V_2 - 4V_3 = -5[/tex]

    Supernode: [tex]-\frac{5}{4}V_1 - \frac{5}{4}V_2 + \frac{3}{4}V_3 - \frac{1}{4}V_4 = 0[/tex]

    Since there are 4 unknowns but you have only 2 independent equations, you will need 2 other equations. These equations will come from within the supernode. Observe the two voltage sources within the supernode (nodes 1, 2, 4). The voltage sources will define/constrain the nodal voltages. To illustrate, the 2V source between nodes 2 and 4 gives us V4 - V2 = 2. Try getting the other equation for the 2Va voltage source. And let's not forget to write Va in terms of the unknowns V1, V2, V3, V4.
     
  21. Jan 18, 2007 #20
    [tex]V_a\,=\,V_4[/tex]

    [tex]V_2\,-\,2\,V_1\,=\,2\,V_a[/tex]

    [tex]V_2\,-\,2\,V_1\,=\,2\,V_4[/tex]

    So, now it's four EQs and four variables?
     
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