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Calculating current through an IR circuit

  1. Oct 1, 2013 #1
    Hi, if I have a DC indoctor-resistor circuit with and inductance (L) of 700μH and a resistance (R) of 0.8Ω, I realise that I can use:

    IL = (Vsource/R) * (1-e-t*(R/L))

    to calculate the current through the inductor if I have a constant voltage source, but would it be possible to calculate how the current build (and also decay) might look in this or a similar method if my source voltage looked more like:

    http://www.dransfield.talktalk.net/VoltageTrace.jpg [Broken]

    So far I can't work out how to make this work with a changing voltage, as the equation appears to assume that you always start at 0

    Also, on the decay side, the equation:

    IL = (Vsource/R) * (e-t*(R/L))

    Seems to assume that you always start at current I=V/R, which is not always necessarily true

    I hope I've explained myself well enough, and thanks for any help in advance!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 1, 2013 #2

    tiny-tim

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    Hi LADransfield! Welcome to PF! :smile:

    LdI/dt + RI = V(t)

    so use an integrating factor to simplify the LHS into a single integral, and then use area-under-a-graph :wink:

    (btw, it's not called an IR circuit, it's an RL circuit)
     
  4. Oct 1, 2013 #3
    Ah, thanks very much, I know it was something similar, but got the wrong letter!

    In this case, which "I" would I use in RI? Would this also be the induced current, or just the dI/dt?
     
  5. Oct 1, 2013 #4

    tiny-tim

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    uhh? :confused:

    I is I

    dI/dt is d/dt of I

    (and the current through R is the same as the current through L … where else would it go?)
     
  6. Oct 1, 2013 #5
    I was under the impression (from here and a few other places) that the total current was I = V/R , but there were different amounts in the inductor and resistor as the current has to build over time in the inductor (iL), but the current in the resistor (iR) was I-iL

    I am expecting the current to rise sharply due to the high voltage, then remain a little more constant through the lower (hold) voltage

    Simulation and testing shows that this happens, though creating a more one-dimensional approach is proving a little more tricky
     
  7. Oct 1, 2013 #6

    tiny-tim

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    Hi LADransfield! :smile:
    From your link …
    "once the Storage Phase has finished, the Current, iL, that flows through it is stable, iL = V / R, no Self Induced e.m.f. is produced and vL is zero. The Inductor acts like an ordinary connecting wire, its Resistance is zero."​
    … iL only = V/R when steady-state has been reached

    until then, the total voltage = the voltage drop across the resistor (IR) plus the "induced" LdI/dt across the inductor.

    It's the same I across both, for the same reason you have the same current through a water pipe that goes through various attachments …

    water in = water out, and electric current in = electric current out. :wink:
     
  8. Oct 1, 2013 #7
    In that case, how do I calculate the induced current at any time-step using a changing voltage input?

    The equation you specified doesn't appear to calculate the induced current, which is what I'm after?
     
  9. Oct 1, 2013 #8

    tiny-tim

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    the current I is the current through the inductor

    it's "induced voltage" (LdI/dt) across an inductor, not "induced current"

    (eg if you look up "induced current" in wikipedia, you'll be redirected to a page which doesn't mention "induced current")

    an inductor is a voltage source (emf source), not a current source :wink:
     
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