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Current from square wave into inductor in series with cap and resistor in parallel

  1. Apr 10, 2011 #1
    1. The problem statement, all variables and given/known data
    In the circuit of Fig. 16(a), the voltage v has the periodic waveform shown in Fig. 16(b) with a period of 4 us and an amplitude of 20 V.

    [​IMG]

    2. Relevant equations

    i = Cdv/dt

    v = Ldi/dt

    3. The attempt at a solution

    Assuming that x is constant (at its average value), draw a dimensioned sketch of the waveform
    of iL(t) and determine its maximum and minimum values.

    x = 5 (as duty cycle is 1/4 => 20/4)
    Therefore at begining of cycle v = 20 => inductor has drop of 15v across it

    using v = Ldi/dt
    di/dt = 15/2e-3
    = 7500 Amp per sec
    i = 7.5mA

    So in one microsecond the current in the inductor goes from 0 to 7.5mA agreed? I've presumed the charge is linear, is this correct? Why?

    Now after 1us the v is 0 volts for 3us. This means the inductor's magnetic field will collapse into the capacitor or resistor or both?

    Thanks
    Thomas
     
  2. jcsd
  3. Apr 10, 2011 #2
    Re: Current from square wave into inductor in series with cap and resistor in paralle

    Nope. The current isn't starting up at zero.

    You need to find the average current too (you have a resistor with an average voltage already so this is easy.)

    The inductor current won't collapse. It will be a triangle wave up and down with the applied voltage riding on top of a steady DC current.
     
  4. Apr 10, 2011 #3
    Re: Current from square wave into inductor in series with cap and resistor in paralle

    Okay the current to begin with is 20/R = 5/1000 = 5mA. Average current at iR = 5/1000 = 5mA through the resistor. Correct? Why is the shape straight lines (not exponentially stuff)
     
  5. Apr 15, 2011 #4
    Re: Current from square wave into inductor in series with cap and resistor in paralle

    ?????????????
     
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