Question

I have a basic question regarding inductor ac behavior. This is regarding the very first quarter/half cycle when an ac sine wave voltage is applied across an inductor (which is not yet ‘charged’). As per normal graphs shown, as the voltage starts increasing from 0, the current is shown starting at negative maximum and decreasing from there. However, it is not clear how the current can be at a negative maximum when the inductor is not yet ‘charged’. What exactly is the behavior of current during the first cycle of the applied ac voltage?

Answer 1

You are right is noticing that if there is a voltage applied initially there is no current in the inductor at that instant. The current will begin to increase in response to the voltage exponentially with a time constant L/R where L is the inductance and R the resistance of the inductor. Thus the current build up over about 3 time constants to its max value and during this time the current and voltage are establishing the phase relationship that you expect for an inductor which will continue until the power is removed.

The usual plot of voltage and current for an AC circuit is usually done to represent the situation after the circuit has been on for a long time and is referred to as the steady state response.

Answer 2

In a theoretical inductive circuit (no resistance - which is an impossibility), the AC voltage will lead the Alternating Current by 90 degrees. So for that 1st qtr cycle the current hasn't caught up with the voltage,and never will in such a circuit.

Answer 3

Thanks for the answers! Yes, my question was with respect to an ideal inductor with no resistance. The conclusion (as indicated by ‘radar’) that the current will not catch up with the voltage came as a surprise as well as relief! I was trying to find out at what point the current waveform would become the same as normally shown in technical material.

Analyzing qualitatively, it appears that if an ac sine wave voltage is applied across an inductor, the current waveform gets shifted up such that the ‘negative’ peak aligns with the X-axis. In other words, the current stays mainly positive. Does this sound correct?

If, instead, we consider the applied voltage to start from its positive or negative peak (that is a +ve or –ve cosine waveform), then the current waveform is as normally shown.

Answer 4

You are right in that the current waveform shifts with respect to the voltage waveform. My drawing is not the best, but I will try and show the "steady state" wave forms. They will both still be sine waves and their frequency is the same, only the "phase" or timing has changed.