Inductor current calculation in time.

Discussion in 'Homework Help' started by Winten, Nov 12, 2010.

  1. Winten

    Thread Starter New Member

    Nov 12, 2010
    2
    0
    I am having problems calculating currents in time. Please help me to understand it, for example we have this circuit:

    [​IMG]
    And during time in seconds.
    0-1 (switch up)
    1-5 (switch down)
    5-6 (switch up)
    6-6,5 ( switch down)

    So in current circuit phase, then switch is up: I=5A, and at end time switch is down I=7,5A.

    Time 0->1
    I(change) = 2.5(1-1/(2.71^(-1)) = 1,58A

    At zero time I = 2,5A, so I(1sec) = 2,5 + 1.58 = 4,08A (please tell if this is OK)

    Now at 1-5 switch is down. At 5 sec its getting up again, so I(end) = 5A, and current I = 4,08A.

    Ant tau = 1/3 (please tell if this is ok, either)

    I(change) = (5 - 4,08)(1- 1/2.71^(5/ (1/3)) = ~ 0,92

    So I at 5sec is I(5sec) = 4,08 + 0,92 = 5

    (it gives me the same 5) something must be wrong, please help me out.
    Also current should have decreased in this time.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Rather than use "switch up" and switch down", try "switch open" and "switch closed". That is correct terminology, and not dependent on the orientation of the drawing.

    For your figuring, Rtotal with the switch open is 3 ohms. It becomes 2 ohms with the switch closed.

    Your currents are incorrect - it's 5 amps open and 7.5 closed.
     
  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Why at zero time I=2.5A?
    Is it an initial condition?

    If it is an initial condition it will change with time and thus the current at 1 sec is not 2.5+1.58.
     
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