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Where can we apply mesh and nodal analyses?

  1. Jun 30, 2011 #1
    Hello,:smile:

    Actually I'm just curious to know when and where we can apply mesh and nodal analyses. Perhaps, you can help me to understand this with some simple circuit diagrams. Thanks

    1: To apply nodal analysis we must have at least two nodes. Am I correct? Moreover, we also need one extra node which would be used as a reference node - so in total we need three nodes.

    2: To apply mesh analysis we must have at least two meshes. In mesh analysis we need to subtract the current of loop from the current of the other. So, if we have a single loop, then we have only one current, therefore having at least two meshes is a must.

    Cheers
     
  2. jcsd
  3. Jun 30, 2011 #2

    vela

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    In that these methods are simply the application of Ohm's law and KVL or KCL, you can use them with any circuit, no matter how simple. With only two nodes or one loop, you'll end up with one equation and one unknown, so circuits that simple are, in a sense, trivial. You'd need more nodes or loops to see the benefit of each method.
     
  4. Jun 30, 2011 #3
    Thank you, vela.

    I don't think we can apply each analysis everywhere. What am I missing here?

    For example, I don't think we can successfully apply mesh analysis on this circuit (at least this is what I have been told): http://img810.imageshack.us/img810/1671/imgef.jpg

    Help me please.

    Cheers
     
  5. Jun 30, 2011 #4

    vela

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    With that circuit, you can't blindly follow a recipe to get a system of equations using mesh analysis because there's no relationship between the current and voltage for the current sources. On the other hand, you don't need the mesh equations for the left and right loops because you already know the current in those loops are 2Ix and I0. The mesh equation for the middle loop and an equation for expressing Ix in terms of the loop currents are enough for analysis of the circuit.
     
  6. Jun 30, 2011 #5
    Many thanks, vela.

    But I don't see how I can apply mesh analysis to center loop (the one having I2 current). There is no voltage source. Could you help me please with it?

    Cheers
     
  7. Jun 30, 2011 #6

    vela

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    Every loop doesn't have to have a voltage source. Why do you think it does? KVL applies to any loop around a circuit.
     
  8. Jul 1, 2011 #7
    Yes, you are correct. The loop doesn't have to have a voltage source to apply KVL. The voltage source would equate to 0V.

    Q1:
    Do you find the following statements correct?
    1: To apply nodal analysis we must have at least two nodes.
    2: To apply mesh analysis we must have at least two meshes.


    Q2:
    I have tried to apply mesh analysis to the circuit I mentioned above. I wasn't successful.

    Please have a see here: http://img853.imageshack.us/img853/5296/imgil.jpg

    By the way, I had the wrong equations for loop 2 and loop 3. For loop 2 it should be:
    4(I2 - (-2Ix)) + 2(I2 - I3) = 0

    For loop 3 it should be:
    2(I3 - I2) = 0
    I3 - I2 = 0 [We know the value of I3 because it's equal to Io which is 1A in this case]
    Using I3 = Io = 1A
    1 - I2 = 0
    ERROR...

    We have these equations now:
    For loop 1: 4(-2Ix - I2) = 0

    For loop 2: 4(I2 - (-2Ix)) + 2(I2 - I3) = 0 [But we have been given the value of current flowing through 2 ohm resistor which is Ix which means that (I2 - I3) = Ix ERROR, it would mean that (I2 - 1 = Ix) ]

    Please help me out. I have a test tomorrow.

    Cheers
     
  9. Jul 1, 2011 #8

    vela

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    Your equations for loop 1 and loop 3 are wrong because you're not accounting for the voltage across the current sources. Reread what I said in post #4.
     
  10. Jul 2, 2011 #9
    Hi vela,

    This is what you said in the post #4:
    For loop 1 the current is: 2Ix
    For loop 3 the current is: Io

    Ix = I2 - I3, and I3 = Io

    vela, I'm confused now. Please tell me how to proceed. Thanks

    Cheers
     
  11. Jul 2, 2011 #10

    vela

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    For loop 2, you found

    4(I2 - (-2Ix)) + 2(I2 - I3) = 0

    So you have 3 equations and 3 unknowns, I2, I3, and Ix. Just solve.
     
  12. Jul 2, 2011 #11
    Thank you, vela.

    I have tried to solve it. I'm sorry if I'm grossly wrong. Please help me. I think the value I have got for I2 is incorrect.

    4 ( I2 - (-2Ix)) + 2(I2 - I3 ) = 0, Ix = I2 - I3, and I3 = Io

    4 ( I2 - (-2 (I2 - I3))) + 2( I2 - I3 ) = 0

    4 ( I2 - (-2I2 + 2I3)) + 2I2 - 2I3 = 0

    4 ( I2 + 2I2 - 2I3) + 2I2 - 2I3 = 0

    4I2 + 8I2 - 8I3 + 2I2 - 2I3 = 0

    Let Io = 1A

    4I2 + 8I2 - 8 + 2I2 - 2 = 0

    14I2 - 10 = 0

    14I2 = 10

    I2 = 10/14 = 5/7 A
     
  13. Jul 2, 2011 #12

    vela

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    You made a sign mistake when you said Ix=I2-I3. It should be Ix=I3-I2.
     
  14. Jul 2, 2011 #13
    Thanks a lot, vela. I'm extremely sorry if this is frustrating you. Please excuse me for this. I think I'm still going wrong.

    4 ( I2 - (-2Ix)) + 2(I2 - I3 ) = 0, Ix = I3 - I2, and I3 = Io

    4 ( I2 - (-2 (I3 - I2))) + 2( I2 - I3 ) = 0

    4 ( I2 - (-2I3 + 2I2)) + 2I2 - 2I3 = 0

    4 ( I2 + 2I3 - 2I2) + 2I2 - 2I3 = 0

    4I2 + 8I3 - 8I2 + 2I2 - 2I3 = 0

    Let Io = 1A = I3

    4I2 + 8 - 8I2 + 2I2 - 2 = 0

    14I2 + 6 = 0

    14I2 = -6

    I2 = -6/14 = -3/7 A
     
  15. Jul 2, 2011 #14

    vela

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    Check your algebra. :smile:

    I know I could tell you exactly where the mistake lies, but I feel it's important for students to develop their error-spotting skills.
     
  16. Jul 2, 2011 #15
    vela, thanks a lot. I have been learning a lot from you and some other members. I'm very much tired now. If you feel like then please tell me where the mistake lies. I don't want to frustrate you any more by creating some new errors in the process of removing one!:wink: Will come back to the forums, I think, day after tomorrow now. Bye

    Cheers
     
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