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

  • Thread starter thomas49th
  • Start date
  • #1
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Homework Statement


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.

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Homework Equations



i = Cdv/dt

v = Ldi/dt

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
 

Answers and Replies

  • #2
1,679
3


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.
 
  • #3
655
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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)
 
  • #4
655
0


?????????????
 

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