Understanding Inductors in Initial Phase: Voltage and Current Behavior Explained

[Bassalisk]

Everybody on the internet talks about inductors when they are in steady state. But nobody talks about inductor, when its in initial phase.

Here is what I am talking about:

We made a clear statement why voltage leads the current in inductor. And I understood that part quite well. But consider this:

You have a solo inductor(ideal) connected to a sinusoidal wave voltage generator. What happens with the voltage and current in that important first half of the sine wave [0,pi].

How does then current behave? After what time will it go into steady state, and have a pi/2 voltage over current lead?


This question mainly popped up because, at capacitor, current leads the voltage by pi/2. A simple minded one would thought that at 0 time, there is already current when voltage is 0.

 

[berkeman]

Everybody on the internet talks about inductors when they are in steady state. But nobody talks about inductor, when its in initial phase.

Here is what I am talking about:

We made a clear statement why voltage leads the current in inductor. And I understood that part quite well. But consider this:

You have a solo inductor(ideal) connected to a sinusoidal wave voltage generator. What happens with the voltage and current in that important first half of the sine wave [0,pi].

How does then current behave? After what time will it go into steady state, and have a pi/2 voltage over current lead?


This question mainly popped up because, at capacitor, current leads the voltage by pi/2. A simple minded one would thought that at 0 time, there is already current when voltage is 0.

The following differential equation is always true for the inductor:

v(t) = L \frac{di(t)}{dt}

The transient and steady-state solutions of this equation are what gives you the current lagging the voltage.

So for example, if you put a step voltage across an inductor, the current in the inductor slowly ramps up (because the derivitave of a ramp is a constant...).

 

[Bassalisk]

The following differential equation is always true for the inductor:

v(t) = L \frac{di(t)}{dt}

The transient and steady-state solutions of this equation are what gives you the current lagging the voltage.

So for example, if you put a step voltage across an inductor, the current in the inductor slowly ramps up (because the derivitave of a ramp is a constant...).

Hmm... So what you are saying here is, that pi/2 lag/lead(depends what you are looking) is slowly building up, over let's say 1 period? Are there any equations which determine the point in time where you can consider a circuit to be in a steady state?

 

[berkeman]

See the simulation by vk6kro at the bottom of the first page in this long thread:

https://www.physicsforums.com/showthread.php?t=15489

You can see how the current waveform is offset in the direction of the first voltage sinusoid half-cycle at first, and is starting to settle out some number of cycles later. You should be able to just solve the differential equation above for i(t) given the initial conditions of i(0)=0 and v(0)=0, and apply v=sin(wt) at t=0. That's what that simulation plot is displaying.

 

[Carl Pugh]

Another complicating factor is where in the cycle you apply voltage to the inductor.
If a sine wave voltage is applied to an inductor when the voltage is at its maximum or minimum value there is no transient.
The maximum transient occurs when a sine wave voltage is applied when the voltage is 0 volts.

 

[jim hardy]

The math will show that an ideal inductor (with a perfect iron core and perfect copper windings) energized at the positive going zero crossing would have a sinewave current with DC offset such that the negative peaks will just graze zero.
In other words current would swing not from normal negative peak to positive peak but from zero to twice normal positive peak.
Flux would do the same.

Any real iron core inductor will of course settle out to its normal peak to peak sinewave current after a small number of cycles. But it will have a surprise in that first positive half cycle, and maybe the second and third one too. Here's why...
... Flux is integral of voltage.
Normally in half cycle of voltage interval, from a positive zero crossing to the following negative zero crossing, flux integrates from the nominal negative peak value to the positive, let us say from -Phi to + Phi.
Draw your sine and cosine waves - one is at peak when other at zero. If you close in at a voltage peak flux is already where it belongs, whch is at zero.

BUt at the condition you inquired, close at the voltage zero crossing , flux starts from not where it belongs, -Phi, but from zero . So the integration will attempt to take flux from zero to +2Phi.

Well, almost all cores are sized for their nominal Phi not twice that.
So shortly past ninety degrees the core will approach saturation, flux virtually stops increasing, and the inductor can no longer maintain the dPhi/dt necessary to support counter-emf . So counter-emf falls off and current goes way high in fact being limited mostly by resistance of the winding. This might happen again on a few subsequent half cycles as the core rids itself of the DC offset by walking down the B-H curve.

Here's a writeup on the effect in transformers.
www.ece.mtu.edu/faculty/bamork/ee5220/Inrush.pdf

This is a real phenomenon. It can cause nuisance tripping on fast circuit breakers. For that reason manufacturers of solid state relays make ones that close at 90 degrees instead of zero crossing and sell them for use on inductive loads.

From Wikipedia: " When a transformer is first energized a transient current up to 10 to 50 times larger than the rated transformer current can flow for several cycles. This happens when the primary winding is connected around the zero-crossing of the primary voltage."
http://en.wikipedia.org/wiki/Inrush_current

Whoever made that spice simulation - it sure shows the offset!
if you omit the resistor does current offset remain constant? Does Spice include hysteresis ad nonlinearity of iron?

old jim