As shown in this image, how does a negative current appear at the beginning, when the voltage starts to increase ?
I understand the current lead lag mathematics behind it. I just can't figure out, what is going on with the electronics inside the wire. How can an increasing voltage, cause a reverse current ?
Thanks to Andy aka for pointing out the right solution with correct references. I realised that the answer to the anomaly lies in the equation of the inductor.
Since for an inductor v = L * di/dt
Therefore current = Integral (v/L * dt)
Integral of Vsin(wt)/L = -Vcos(wt)/L
And the TRICKY part is to take integrals between 2 points A and B
So the current at a time t = -Vcos(wt)/L – (-Vcos(0))/L = -Vcos(wt) + V/L
The cos(0) is the DC component that "lifts" the entire current wave above the 0-line so the current keeps flowing only in positive direction (assuming a 0 resistance path). This is mentioned here –
This is probably why the SMPS of my CPU cabinet makes a loud noise right when I turn it on. The sound persists for about a minute and then fades away. Because the voltage supply starts with a 0V at t=0 and the ac current flows only on the positive side of the graph above the 0 level, giving a higher dc value.
I tried reading a few more online tutorials on circuits to understand the very basics of RL and RC AC circuits.
Every RL and RC ac circuit has 2 currents. One is the steady-state current I(ss) and another is the transient current I(tr).
The phasor method calculates only the steady-state current of it. Which many believe to be the "real" behavior of the circuit, while it is not.
This link shows the calculation of the transient current in a RL ac circuit –
I have created a partsim simulation of the RL ac circuit, which shows the whole thing nicely –