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Finding the current for each time interval (Electric circuits)

  1. Sep 22, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and its solution are attached as TheProblemAndSolution.jpg.

    2. Relevant equation
    ν = L di/dt

    3. The attempt at a solution
    I don't understand how the solution gets i = 10.0 A for t = 2 ms as well as to how it gets i = 30 – (10 * 10^3) t) (A) for 2 ms < t < 4 ms. More specifically, I see that the equation when 0 < t < 2 ms has i = 10.0 A if one plugs in 2 ms in the t variable like so, 5 * 10^3 (2 * 10^(-3) s) = 10.0 A. However, what I do not get is, why does the equation for the interval (0, 2) work for the point t = 2 ms (does the reasoning relate to the fact that 1.9999 (periodic 9) can be proved to equal 2? As to when 2 < t < 4, something seems off with the first equality but I cannot figure out exactly what. Just to note, I don't see anything wrong with Fig. 2-12 and the way it relates to the equations in the work so I don't think anything is wrong but, to reiterate, I don't get why i = 10.0 A at t = 2 ms and I'm confused as to how to get the equation for when 2 ms < t < 4 ms because the first equality is messed up which throws me off for the rest of the work.

    Any help in understanding these things would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Sep 22, 2012 #2

    TSny

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    You have the right idea about the current at t = 2 ms. The current will be a continuous function of time, so the value at t = 2 ms is the limit of the expression for the first time interval as t approaches 2 ms.

    There is a typo in the first line of the expression for i in the second time interval. It should read something like i = 10 + ∫vdt = 10 + (1/.003) ∫-30dt . The second and third lines look ok.
     
  4. Sep 24, 2012 #3

    s3a

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    I'm still having doubts as to whether (t = 2 ms, i = 10.0 A) should be an included point or not.

    Is it correct to say that it is BECAUSE equating the 0 < t < 2 ms and 2 ms < t < 4 ms equations as follows will yield an input t such that each equation's output is i = 10.0 A?:

    5 * 10^3 t = 30 - 10 * 10^3 t
    15 * 10^3 t = 30
    t = 30/(15*10^3) [Plugging this value in either equation yields i = 10.0 A.]
     
  5. Sep 24, 2012 #4

    TSny

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    I think of it this way. For a circuit that has inductance, the current cannot change discontinuously even when the voltage changes discontinuously. [To change the current discontinuously would require a "dirac-delta" type of impulsive voltage which is rather fictitious and is not what you are dealing with here.] So, the current will be a continuous function of time. So, the current at t = 2 ms must be the limit as t approaches 2 ms in the interval 0 < t < 2 ms.

    From calculus, we know that if you take the integral from 0 to t of a function with a finite discontinuity, then the integral will be a continuous function of t. Here, the current at any time t is the integral of V from 0 to t. Even though V is discontinuous at t = 2 ms, the current will be continuous at t = 2 ms.
     
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