(adsbygoogle = window.adsbygoogle || []).push({}); 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

**Physics Forums - The Fusion of Science and Community**

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

Have something to add?

- Similar discussions for: Current from square wave into inductor in series with cap and resistor in parallel

Loading...

**Physics Forums - The Fusion of Science and Community**