# Do I understand impedance correctly? Can anyone fact check my post?

1. Apr 24, 2014

### Vishera

$$Let\quad \vec { Z } =R+Xj$$
If X>0,then the impedance is lagging (current lags behind voltage).If X<0,then the impedance is leading (current leads voltage or more accurately, voltage lags behind current). If X=0, the current and voltage are in phase and the load is called purely real. If x≠0, the current and voltage are not in phase and the load is called complex.

Some load is purely real if and only if the load is purely resistive.
Some load is purely imaginary if and only if the load is purely reactive.

Questions:
1. Does it make physical sense for a load to be purely imaginary or purely reactive?
2. How do we know that when X>0, current lags behind voltage and not the other way around?

2. Apr 24, 2014

### Staff: Mentor

I hadn't heard the terms leading/lagging impedance before, but perhaps they are standard terms somewhere. Your statements are basically correct.

The easiest way to see "why" is to look at the differential equations that relate voltage and current for inductors and capacitors:

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

$$i(t) = C \frac{dv(t)}{dt}$$

3. Apr 24, 2014

### vk6kro

It is quite possible for a load to be almost entirely reactive, although all loads have at least some resistive components.

You could have an inductor made with very thick wire, or a capacitor with thick metal plates, for example.

You can observe phase differences between voltage and current with an oscilloscope.
Unfortunately, this involves placing a resistor or a current probe in series with the reactive component to measure the current and this introduces a resistive component.
Keeping this small still gives an acceptable result.

You still don't know if the voltage leads the current or the current lags the voltage, but it probably doesn't matter.

An amazing free program is available if you have a PC. It is called LTSpice and you can find it via Google.
It takes a little getting used to, but it can teach you all you want to know about impedance and circuits.

4. Apr 24, 2014

### Vishera

Actually, it's funny that you mention that. We had a lab where we had to do exactly this. I'm not sure if our oscilloscope was special, but the leading waveform was the waveform that was on the left. So for example:

5. Apr 24, 2014

### vk6kro

Yes, that is a typical result.

You need to understand that this is what you get after a few cycles of input voltage.

When you first apply power to a capacitor, for example, there can't already be a current flowing in the capacitor, although the leading current graph seems to show that.
It takes a few cycles to settle down to a steady pattern.

6. Apr 24, 2014

### jim hardy

In my day it was the duty of the textbook author to take you through the assumptions that led to that result.

Firstly, these equations apply only to sine waves which, although most common, are a mathematical special case.

I was taught to represent the sine wave voltage as a rotating vector called a "phasor", which rotates counterclockwise.
Phasor's tail is at center of a circle, head lies on its diameter.
Zero degrees was defined as horizontal with head pointing right.
Next teacher explained how sine at any angle of rotation is equal to the vertical distance of phasor's head from horizontal diameter of the circle. That's just Pythagoras and 8th grade geometry....
So voltage at any instant = Asin(ωt) where A is amplitude of your voltage
and a positive angle is in the counterclockwise direction

Now - once you have defined your starting point, ie where is zero and what direction does your phasor rotate , all else falls out.

But it is SOOO easy to dismiss those essential first steps as just "textbook boilerplate" and brush over them,
that astute students like yourself ask "How'd we get here? Which way is up? "
Valid questions, both.

http://resonanceswavesandfields.blogspot.com/2007/08/phasors.html

hope I addressed the right question.

old jim