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

by thomas49th
Tags: current, inductor, parallel, resistor, series, square, wave
 P: 648 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. 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
 P: 1,781 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.
 P: 648 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)
P: 648

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

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