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Time constant in RL circuit

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data
    9nvRTj1.png

    2. Relevant equations
    τ = L/R

    3. The attempt at a solution
    So I already know the initial current and final current (-4mA and -1mA, respectively), my question is how I would find the resistance value to use to solve for the time constant.

    After the switch has been closed for a very long time, the 200mH inductor will become a short circuit, and the current from the 75 V source will bypass the 50kΩ resistor (because it will look for the path with least resistance) after travelling through the 75kΩ resistor and go through the short circuit from the inductor, and then the short circuit from the switch being closed. What would then be the resistance for the time constant equation?
     
  2. jcsd
  3. Nov 7, 2014 #2

    NascentOxygen

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    Staff: Mentor

    After the switch closes, the inductor current time constant involves the Thévenin equivalent resistance of the circuit to the right of the inductor.
     
  4. Nov 7, 2014 #3
    So even though the current bypasses the 50k Ω resistor, the R I would use would still be 75k // 50k?

    Or, would it be better to say that after doing the source transformation, the current bypasses both the 50k AND 75k resistor, and still have 50k // 75k as the R value?
     
  5. Nov 8, 2014 #4

    NascentOxygen

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    Staff: Mentor

    The inductor bypasses all of the 50k resistor's current only in the steady state, not while current is varying. So, no, the 50k does not have all its current taken by the inductor during the time it is of interest to us in this exercise.
     
  6. Nov 8, 2014 #5
    Okay I see, thank you.
     
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