# Question about reactances and phase shift

Discussion in 'Electronic Basics' started by Patrick Leonard, Apr 26, 2004.

1. ### Patrick LeonardGuest

I'm studying electrical engineering and I've learned that inductors and
capacitors introduce a -/+ 90 degree phase shift between the voltage and
current. But lets look at an inductor with just a single loop - no tricks
here, just a wire that loosely forms a single turn (pick up your mouse cord
and try it). Such an arrangement most certainly does not produce a 90deg.
phase shift. I'm sure you can see where I'm going with this - at what point
does an inductance cause a 90-deg. phase shift in a circuit? Presumably,
for any given frequency, w, there would be a particular inductance
(ultimately, a particular geometry) that would be required for a full
90-deg. phase shift. Is what I've said so for correct? Does the magnitude
of the inductance have some correlation to the magnitude of the phase shift?
(It must...)

In that case, how does one calculate the 90-deg phase shift inductance?

If this is true, why is that inductors are modelled as |x|<90 in phasor
notation (ie, magnitude of reactance at an angle of 90) -- as this will
ALWAYS yield a 90-deg. phase shift in circuit analysis...despite the fact
that the inductor, may not in fact produce such a phase shift!

As noted, this is something I'm studying but don't really understand.
Please feel free to use math to explain.

PS. This is leading up to another question 2. ### John PopelishGuest

A perfect inductance of zero size (zero length of conductor) has a 90
degree phase shift between the applied voltage and the current through
it. REal inductors have series resistance as well parallel
capacitance and wire length that brings in the speed of light as waves
travel through the conductor. All these imperfections alter the
magnitude and phase of the current with respect to the applied
voltage.
To the extent that the inductance dominates the total impedance of the
path between the ends of the winding, the phase approximates the that
if an ideal inductor. At some low frequency, the resistance of the
wire has a higher impedance than the inductance does, and for all
frequencies lower than that, the inductor is better modeled as a
resistor than as an inductor. At some high frequency, the parallel
capacitance of the inductor has a lower impedance than the inductance,
and for all frequencies much higher than that the inductor is better
modeled as a capacitor than as an inductor. These sort of
imperfections ar part of the reality of all components, including
capacitors and resistors. Using these components effectively involves
learning their limitations.
If you don't understand these things in a conversational way, you
probably don't yet have use for the math.
Good.

3. ### BobGuest

Patrick,

Very good questions. It shows that you're thinking.

The phase relationship of the current and voltage in inductors and
capacitors is ALWAYS 90 degrees. They (*perfect* L's and C's) just don't
ever act any differently -- regardless of what type of signal is being
applied.

However, in the real world -- either by circuit design or due to intrinsic
flaws in the parts -- the net phase of the voltage and current may not be 90
degrees. For example:

Let's say you take a 3" piece of wire and loop it into a 1" (roughly)
diameter loop, and put a 1K ohm resistor in series with it. Now, connect a
sine wave (constant voltage) generator across the series combo.

Apply various frequencies -- let's say between 1Hz and 1MHz.

If you monitor the phase difference of voltage and current at the inductor
it would be 90 degrees. However, you would notice that the magnitude of the
voltage, across the inductor, would be VERY small with respect to the
applied voltage at the generator. This is the key idea, here.

The phase relationship of the voltage/current (V-I) at the generator will be
(almost) 0 degrees. This is because the majority of the voltage drop is
across the resistor, and the V-I phase of a resistor is 0 degrees. So, the
resulting (vector sum) of the I*R drop of the resistor, and the I*2*pi*f*L
drop of the inductor will be a voltage/current phase relationship of
(nearly) zero.

If you were to increase the generator's frequency to a larger frequency,
let's say where X(subL)=2*pi*f*L = 1K ohms, then the phase of the voltage
and current, at the generator, will be 45 degrees. Increase the frequency
even more and the phase will keep increasing, and will eventually approach
90 degrees.

Remember, while all these different frequencies are being applied, the phase
of the voltage/current at the inductor has ALWAYS been 90 degrees.

Keep thinking. It WILL pay off.

Bob

4. ### John JardineGuest

The concept and the way it's taught, must have caused whole legions of
potential EE's to choose other career paths. Esp the profligate use of
Yes; under perfect conditions and as single components, the inductors (and
capacitors) are capable of causing a time delay which equates to a quarter
cycle of a sinewave driving waveform.
Basically, there's *got* to be some kind of time delay (phase shift) as any
inductor must take time to charge or discharge it's magnetic field (amps per
second, slew or charge rate = Voltage/Inductance)
The potential quarter cycle delay never happens though, as there is *always*
some resistance kicking about which means that the delay will never reach
the magical quarter cycle point (nor will the voltage/current components
ever be completely in phase).
Surprisingly, if that single turn mouse wire is connected across the
power-station-fed mains voltage, it would have a current sent through it
that is -nearly- a quarter cycle, 90 degrees late), out of phase etc, with
the supply voltage (plus much smoke). Stick some serious resistance in
series and the voltage and current quickly move into step.
The true 90 deg nature of a cap' and inductor can be easily seen when their
phase shifts are measured -relative- to each other. The classic example
being a resonant ind'/cap'tuned circuit. The phase-to-phase relative
measurement nature can ignore circuit resistance .
Assuming the ultimate 90 deg aspect is true, then the phasor diagrams are a
neat way of telling the truth of a real world situation. You've just got to
make sure the numbers are correct.
regards
john

5. ### PatrickGuest

Heh, no kidding. If only my profs knew that too...

Okay, well, I sort of understand. So inductance really has no impact on
phase then - inductance only affects impedance?

You're going to hate this part. Lets say I have my coil of wire and I
slowly begin pulling it apart - the inductance drops, and the impedance
drops. IF the phase is ALWAYS 90 degrees in an inductor, at what point does
my inductor no longer behave as an inductor? Obviously the phase shift
doesn't disappear instantly...it'll slowly reduce from 90 as I pull the coil
apart. So once again I must conclude that phase shift IS a property of
inductance, and ultimately, coil geometry.

6. ### RatchGuest

Phase is ALWAYS present in ALL circuits where moving charges are
present. In a circuit where ONLY resistance is the passive component, the
phase is zero. In a series circuit driven by an AC current, and only a
"pure" inductor is present, the phase will be 90°. In a series circuit
driven by a AC current, and composed of a pure inductor and a resistor, and
whose frequency of the current and inductance of the coil make its inductive
reactance the same value as the resistor, the phase difference between the
voltage and the inductor will be 45° WHEN the voltage is measured across
BOTH the inductor and resistor. In the same circuit, if the voltage is
measured ONLY across the inductor, it will be 90° with respect to the
current. In the same circuit, if the voltage is measured ONLY across the
resistor, it will be 0° with respect to current. Since we are talking about
a series circuit, the current is the same in both the inductor and resistor.
Therefore current is taken as the reference. Although voltage is not a
vector quantity, its phase and magnitude is added vectorially and the result
is called a PHASOR.
You are talking about a variable inductor. As you decrease the
inductance and keep the frequency constant, the inductive reactance will
become less. As you lower the inductance, the reactance will become
insignificant compared with the other resistance of the circuit. The
inductor no longer behaves like an inductor when you pull the wire straight.
No matter how small you make the inductor with repect to the other
resistance, if you measure the voltage and current across ONLY the inductor
, it will show a 90° phase difference. Ratch

7. ### RatchGuest

In my last message, the ASCII degree symbol seems to have been translated to
0. Therefore 00, 450, and 900 should be read as 0 degrees, 45 degrees, and
90 degrees respectively. Sorry for the mixup. Ratch

8. ### John PopelishGuest

Why should I hate it if you have a coil of wire?
Everyone should have several. The net impedance is the vector sum of the resistance of the wire and
the inductance. So the inductive component of the coil drops, causing
the vector sum (the hypotenuse of a right triangle, where each leg
represents either the resistive or inductive component) to drop.
If you model that coil as an ideal inductor in series with an ideal
resistor, the inductor part maintains the 90 degree relationship. But
at some point, the voltage drop across that inductive part is so much
less than the voltage drop across the resistive part, that from the
outside, you only see the resistive drop. If that loop had been super
conductive, the 90 degree relationship would remain.
But you aren't talking about an inductance, but about a series LR
circuit. What you are describing is the normal properties of a series
resistor inductor, not the inherent property of inductance, itself.
It is a description of the limitations of what we can make with
reasonably good conductors like copper or aluminum. The guys that
actually make inductors out of superconductor don't have this flaw to
deal with (though they have others).  