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LR Circuit with resistors in parallel

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data


    RL.png


    2. Relevant equations


    V = IR

    I = V/R (1 - e^(-t / τ))

    τ = L / R


    3. The attempt at a solution

    I'm so thrown off by these two problems. Every example I have done has nothing in parallel and it is simple, but my homework question involves having to do a much more difficult problem.


    I've tried just plugging things in, but I cannot figure out which values of R go in.

    If I(t) is the current in the inductor branch, then I think I(t) = I_max * (1 = e^(-t/τ))
    From looking at t -> ∞, I calculated the total equivalent resistance as 25/3 ohms and got that the total current would be 4.2 Amperes, then the current going through the inductor branch is 1.2 Amperes (at infinity).

    I tried 1.2 (1 - e^(-t / (τ) ))
    and using τ = 4.2 / (25/3)
    and I've tried using τ = 4.2 / 10


    I tried setting up loop-rule equations for the outer loop and the lower half + a sum of currents equation, but I am fairly confident that doing this is not the best way to go about the problem. (and solving that is far beyond ap physics)

    I understand equivalent resistance + equations for inductance + ohm's law, so no need to have to reexplain those things. (unless I did something wrong with them)
     
  2. jcsd
  3. Apr 3, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Your current Imax doesn't look right. Can you show your calculation?

    Are you familiar with Thevenin equivalents? This would be an excellent tool here.
     
  4. Apr 3, 2012 #3
    I'm not familiar with those, but I know they don't appear on any of the notes I have, nor are they in the syllabus.

    Anywho, equivalent resistance is 25/3. (1 / (1/10 + 1/5) ) + 5 = 8.333...

    V = IR
    I = V/R

    I get that total current (at the battery) = 35 / 8.3333...
    Itotal = 4.2 Amperes

    At t-> infinity, the inductor acts like a wire, so I can ignore it for calculating I through the branches.

    the 4.2 amperes splits off in two branches, 2/3 of it going to the to and 1/3 going to the bottom, which should give the two branches the same potential difference.

    So 4.2 * 1/3 = 1.4 Amperes

    I typed 1.2 above, but my work on paper says 1.4, so I think that was just a typo. I am still not sure where to go with plugging in the R values.

    My guess is that the time constant needs to include all of the R-values, because I would think that the other resistances play a factor in how quickly the current is going to change.
     
  5. Apr 3, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Okay, total resistance as "seen" by the battery. Good.
    Yup. That's good.
    Yes, good again.
    Okay, putting aside the Thevenin equivalent thing but borrowing a bit of its power, suppose you were to pull the inductor out of the circuit and look into the network at its connection points. What equivalent resistance would you see? NOTE: For this exercise you want to "suppress" the sources. That means you can replace the voltage source with a short circuit (a piece of wire). if there had been a current source you could have replaced it with an open circuit -- that is, simply removed it.

    This equivalent resistance will be this resistance that the inductor sees from its source of voltage or current. It's the resistance that goes into the time constant calculation.
     
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