Help with an Inductor question with voltage vs time graph

Discussion in 'Homework Help' started by gengm, Apr 7, 2010.

  1. gengm

    Thread Starter New Member

    Apr 7, 2010
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    Hello everyone, new to the boards here and I'm looking for some help with an Inductor question.
    Here is the question.
    [​IMG]

    I'm given the equation for power stored, W = (1/2)Li^2
    In which i is a function of the voltage in which i = 1/L * integral of V*dt
    1st off I switched the L = .5H, then at given time 100ms, the Voltage is 30V

    So I did i = 30V/.5H = 60 for the current
    then inputed it into the W equation and got .5*.5*(60^2) = 900J
    but 900 isn't the correct answer.
    Can someone point out where I went wrong?
     
  2. Ghar

    Active Member

    Mar 8, 2010
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    72
    Where did you get i = 30V/0.5H ?
    What happened to the integral?
     
  3. gengm

    Thread Starter New Member

    Apr 7, 2010
    23
    0
    Considering they want the energy stored at a certain time, wouldn't the voltage at that time be the only thing relevant?
    then from that equation i = (1/L)*V
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    That equation refers to current and you don't know the current at any time until you calculate it.
    Current is the integral of voltage, meaning voltage for all time up to the time of interest matters.

    Specifically, you get:

    <br />
i = \frac{1}{L} \int_0^{40ms} 80 dt  \quad+\quad  \frac{1}{L}\int_{40ms}^{100ms}30 dt<br />
     
    Last edited: Apr 7, 2010
    gengm likes this.
  5. gengm

    Thread Starter New Member

    Apr 7, 2010
    23
    0
    Alright, that's where I got mixed up there. Assumed that at that time 100ms, thought that only mattered.

    Much thanks Ghar.
     
  6. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    No problem
     
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