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Inductors and branch currents

  1. Jan 13, 2016 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    image.jpeg Find the current passing through the 5mH inductor in steady state.

    2. Relevant equations
    I=V/R

    3. The attempt at a solution
    Since it says steady current, so no change in current which implies that the inductors act as conductors so the current should just split equally
    So I thought it should be 2A
    But that's not the answer.
    Any help? Please?
    Thank you
     
  2. jcsd
  3. Jan 13, 2016 #2

    gneill

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    When the power is first applied the current through each inductor will grow at different rates, heading towards steady state. The division of the total current between them will remain the same though.

    I think I'll suggest using mesh analysis, with two loops. Solve for the mesh currents and the individual inductor currents. Laplace transformations will make this pretty easy. Then let time → ∞ in the current expressions.
     
  4. Jan 13, 2016 #3

    Suraj M

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    Thank you for replying gneill but I haven't learnt Mesh transformations or Laplace transformation
    what are they? I'd like to learn though.
    Is it not just 2A?
     
  5. Jan 13, 2016 #4

    gneill

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    You'll cover them in one or more of your courses eventually. Mesh analysis is a particular method of applying KVL to the loops of a circuit that tends to be simpler for writing out the equations. You can use KVL and KCL as usual instead. Laplace transforms are a way to "transform" equations into what is called the "Laplace domain" (as opposed to the time domain), and makes solving differential equations much easier. Without them you can write out the loop equations as differential equations and solve them in the usual ways.
    No. While the total current will be 4 A as you found, it will will not be evenly split between the two inductors.
     
  6. Jan 13, 2016 #5

    Suraj M

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    Do you mean during the growth of the current? Oh
    Yes you'll get 8/3 that is the answer given
    So is that right?
     
  7. Jan 13, 2016 #6

    Suraj M

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    And I don't think I'll be learning mesh transformations in the future also :-(
     
  8. Jan 13, 2016 #7

    gneill

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    Hmm, let's see.... yes, that looks like it should be correct for the amperage through the 5 mH coil.
     
  9. Jan 13, 2016 #8

    Suraj M

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    Our teacher directly applied I inversely proportional to inductance, that's what confused me.
    Thank you gneill :-)
     
  10. Jan 13, 2016 #9

    gneill

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    Did your professor supply a justification for that operation?
    You're welcome.
     
  11. Jan 14, 2016 #10

    Suraj M

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    Not exactly
    I'll ask him once again if we can do what he did and for a reason:-)
     
  12. Jan 14, 2016 #11

    cnh1995

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    I think you are right here. The drop across inductor coils could be calculated if the resistances of the coils were given, but you are assuming ideal inductors. If the source were AC, your teacher's method would work fine but in case of dc, as di/dt is 0 in steady state, I believe the current should be same through both the coils.
     
  13. Jan 14, 2016 #12

    gneill

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    In the end, there is no reason why ideal conductors in parallel need to share current evenly; the current could be split any way at all without altering the steady state situation of the rest of the circuit. It's only our intuition to opt for a symmetrical solution that urges us to make the even split assumption.

    It turns out that the currents will not be equal. Even though we make the approximation that transients die out after about five time constants and we then imagine the inductors to be simple ideal conductors, all through the transient the currents will be different and there is nothing to disturb that division over time.

    I've just thought of a way to show this with just a bit of calculus and not too much circuit analysis.

    First write the expression for the potential across the parallel inductors with respect to time. This can be done easily because we can find the time constant for the circuit using the net inductance and circuit resistance, and we know the mathematical form of the voltage: ##v(t) = 20 e^{-\frac{t}{\tau}}## where v(t) is in volts.

    Now this voltage must appear across both inductors since they are in parallel. Write the equations ##v = L \frac{dI}{dt}## for each and solve for ##I## for each. Let t → ∞ to find the final currents in each.
     
  14. Jan 14, 2016 #13

    Suraj M

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    Is that growth equation for v correct? I thought it should be
    Code (Text):
    ## v(t) = 20(1-e^{-\frac{t}{\tau}}##
    and even then that would be the voltage across the cell and not the inductors right? Or am I getting it wrong?
     
  15. Jan 14, 2016 #14

    gneill

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    The voltage across the inductors will start at 20 V and decay exponentially to zero. That is, when the cell is first connected the inductors behave as open circuits (current cannot change instantaneously) so the full 20 V of the cell will appear across the inductors to prevent current flow. As the current begins to flow the potential across the coils falls. Eventually it reaches zero and the coils behave like short circuits.
     
  16. Jan 14, 2016 #15

    cnh1995

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    Brilliant! :smile: Solving the integral, it turns out that the current division is not equal.But how can this be proved intuitively? I mean, current division in parallel resistors is easy to understand intuitively but what is the physics behind this current division in inductors?
     
    Last edited: Jan 14, 2016
  17. Jan 14, 2016 #16

    cnh1995

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    Voltage across both the inductors is same. So, 5di1/dt=10di2/dt.
    Hence, current in 5mH rises faster than the current in 10mH inductor(as if there's a race of currents!) in the same time. Hence, since the transient dies out in the same time for both, final current in 5mH is greater than that in the 10mH inductor.
    Also, the above equation becomes 5i1=10i2 after removing 'dt' from both the sides and then integrating both sides. So, current in the inductor is indeed inversely proportional to the inductance( for parallel combination). Am I right?
     
    Last edited: Jan 15, 2016
  18. Jan 15, 2016 #17

    Suraj M

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    I guess that's what Gneill said at the beginning of this conversation So it must be right.
     
  19. Jan 15, 2016 #18

    gneill

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    Yup. :smile:
     
  20. Jan 15, 2016 #19

    Suraj M

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    Thanks Gneill :-)
     
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