Re: inductor starting
Hello All.
If you remember your basic lecons in the University you will probably understand.
1-The relation between current and voltage in an inductor is:
Eq.1
V = L * dI/dt
or
Eq.2
I = (1/L) * Integral ( V dt ) + I0
I= current crossing the inductor
V= Voltage across the inductor
I0= initial condition, continuous current crossing the inductor at t<0, t=0 and t>0.
2-Looking at the Eq.2 you can see that at the time t=0, the moment when the voltage V was applied to the inductor the current crossing the inductor will be the integral of the applied voltage plus a continuos current (stead state).
3-Assuming for simplicity that second term I0 is equal to zero (there was no continuous current crossing the inductor wen the voltage was applied).
4-Now, looking at the first term you can see that for any value of the applied voltage V the integral will be zero at t=0 (when the voltage was applied).
5-The current will be the summ of the 2 terms (first and second of Eq.2), that means, the current is zero at t=0.
6-The impedance offered by the inductor is:
Z=dV/dI
Z= impedance of the inductor in function of the time.
V= applied voltage, V=0 for t<0, V=V for t>=0.
I= current crossing the inductor given by Eq.2
7-According to item 5, the current is equal to zero at t<0 and at t=0. So dI, at t=0, i.e., the difference of the current just before t=0 and at t=0 is zero. And will grow linearly with the time. That means that the impedance will be infinity at t=0 (since dI=0 at t=0), and will decrease following 1/x relation with time.
Hope it helps.
S.