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RL circuit analysis

  1. Jan 14, 2016 #1
    Hello! It's a pretty long time time since I did my last circuit analisys, so I feel a bit rusty.

    I have to calculate the evolution of current in the inductor:

    image.jpg
    L is generic and u(t) is the Heaviside step function. Do note that at t = 0 the inductor behaves like a short circuit (I don't know the exact english term for a "full" capacitor/inductor).

    Firstly I calculate the current in the inductor at iL(0-).
    Hence the circuit becomes:

    image.jpg

    I found out that iL(0-) = iL(0+) = 6,6667A

    At t = 0+, the second source starts to supply voltage. So I have to analise the circuit to find iL(+∞)

    image.jpg

    Now I am stuck because I got to some results which are different from the ones LTSpice returns.

    Using the KVL clockwise I get the system:

    -10 + i1 - (i1 - i2) = 0 (above mesh)
    -i2 + (i1 - i2) + 1 = 0 (left below mesh)
    -1 + 2i3 = 0 (right below mesh)

    Which solved, gives:

    i3 = 0.5 A
    i2 = 10 A
    i1 = 19 A


    But according to LT Spice:

    i3 = 0.5 A
    i2 ≈ -2.66 A
    i1 ≈ -6.66 A


    So I don't get where I mistakened :/


    Second thing. After getting the right iL(+∞) I still have to get the equivalent resistence to calculate τ = L/R

    Would it be a good idea to find the thevenin equivalent circuit between a and b?

    image.jpg

    Thank you very much :)
     
  2. jcsd
  3. Jan 14, 2016 #2

    gneill

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    Staff: Mentor

    Hi Frank,

    Please remember to use the formatting template provided in the edit window when you start a thread in the homework sections. This is a forum rule.
    I think you mean that prior to t = 0 the circuit has reached a steady state with u(t) = 0 and the inductor is behaving as a short circuit. I'm not aware of any particular english term for an inductor or capacitor that operating in steady state conditions.
    I think you'll want to check your mesh equations. I can see some sign issues.
    This is your setup for solving, right?
    Fig2.PNG
    I think that ##i_1## should be a bit smaller in magnitude. Check your LTSpice again.
    Remove the inductor and find the Thevenin equivalent circuit at the terminals where it was connected. That's A and B on the circuit diagram that I provided above.

    Edit: I corrected an labeling error in my figure and added indications of the potential drops encountered by each current.
     
    Last edited: Jan 14, 2016
  4. Jan 14, 2016 #3
    Sorry I had thought that I had to follow those guidelines but not to keep the statements themselves. My fault.

    Yes I meant steady state, thanks :)

    Yes, you're image is right.
    However, I got the right numbers but I still have to get used again to sign conventions (it's passed quite a long time), so I'm not completely sure about the reason. In particular I am using the passive sign convention because this is what I was used to, but I don't know whether the active sign convention would be more practical. So the current is positive if it enters from the positive terminal in passive components and viceversa in active ones. Practically, as far as resistors are concerned, I use + from left to right, and from top to bottom, and - from right to left and bottom to top.

    Thus for i1:
    - 10 + i1 - (i2 - i1) = 0 (I got the sign error here)

    While for i2:
    -i2 + (i2 - i1) + 1 = 0 (here i2 should be positive but I don't get why)

    Thanks for the tip :)
     
  5. Jan 14, 2016 #4

    gneill

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    Staff: Mentor

    You're going clockwise around the circuit and summing potential drops. When ##i_2## passes through the resistor in the bottom left corner (R1 in your original diagram) it creates a potential drop of ##i_2~1Ω##. So the term in your equation is ##+i_2##.
     
  6. Jan 16, 2016 #5
    Thank you I solve the problem :)
    One more thing if you can.

    I did the problem again using the nodal analysis in this way:

    image.jpg

    With this equation:

    1 - Vx = Vx + 10 - 1 + Vx

    Which gives the correct result. As I was writing it I didn't know the direction of the current, so I could set the direction of the right branch rightwards as well as leftwards right? It is just a matter of sign? In this other case, the equation should have become

    Vx - 1 = Vx + 10 - 1 + Vx ???

    This would have given a wrong result so I think I am missed something in the middle :frown:
     
  7. Jan 16, 2016 #6

    gneill

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    Staff: Mentor

    When doing nodal analysis it is best to assume that all currents are flowing into the node or all currents are flowing out of the node. Either direction will do, whatever is your preference. Then write the node equation so that all the branch currents sum to zero. So you should have terms on the left hand side and a zero on the right hand side of your equation.

    Most of the errors that occur doing nodal analysis are due to people trying to puzzle out current directions and writing the wrong expressions to take into account the current direction. If you always assume that the directions are either out or into a node, the process of writing the term for each branch is always identical, no thought involved. Always doing it the same way will reduce your error rate to practically nil.

    Suppose you are writing a node equation for some node A and you want the term for current flowing out of A and to node B. That is, you want the term for the branch AB with resistance RAB. Then the term is (VA - VB)/RAB. The voltage for the node you're at always comes first in every term, every time.

    For current sources connected to nodes you are forced to honor their given direction of current flow. So choose its term sign accordingly to suit your in/out choice.
     
  8. Jan 17, 2016 #7
    Thank you very much indeed. You have been very very helpful :D
     
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