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### Network # Ive got this problem with AC through capacitors (phase lag)

S

#### Steve Evans

Jan 1, 1970
0
Helo,

If Ive got a resistor, a battery and an ammeter, and I stick the
battery across the resitor, I get a certain number of milliamps
current going through the resistor.

If I increase the voltage thruogh the resitor, I get more current.

If I decreas the voltage, the resistor passes less current.

So far, so good.

Now, if I try to do the same thing with a capactor instead of a
resistor, I get an initial curent flow that tapers off over time as
the cap charges up and winds up with the same voltage acsross it as
the battery.

Again, so far so good.

Now, if I try to stick a AC voltage accross the capacitor; everything
goes tits-up. For part of the cycle, yeah, the current increases with
the applied voltage and for other parts of the cycle, it decreases
along with deminishing voltage.

Fine again. (well not really -see below)

The problem is, at *other* parts of the cycle, the current has started
to reverse direction whilst the voltage is still increasing! And the
bit further on again, the same thing happens vice-versa. Whats going
on here?? I thoght current was proportional to applied volts? Thiese
parts of the cylce are defying Ohms law! Can anyone explain?

And how can current *lead* voltage? YOu have to have a voltage to get
a current to flow in the first place!!

tnx,

steve

J

#### John Popelish

Jan 1, 1970
0
Steve said:
Helo,

If Ive got a resistor, a battery and an ammeter, and I stick the
battery across the resitor, I get a certain number of milliamps
current going through the resistor.

If I increase the voltage thruogh the resitor, I get more current.

If I decreas the voltage, the resistor passes less current.

So far, so good.

Now, if I try to do the same thing with a capactor instead of a
resistor, I get an initial curent flow that tapers off over time as
the cap charges up and winds up with the same voltage acsross it as
the battery.

Again, so far so good.

Now, if I try to stick a AC voltage accross the capacitor; everything
goes tits-up. For part of the cycle, yeah, the current increases with
the applied voltage and for other parts of the cycle, it decreases
along with deminishing voltage.

Fine again. (well not really -see below)

The problem is, at *other* parts of the cycle, the current has started
to reverse direction whilst the voltage is still increasing! And the
bit further on again, the same thing happens vice-versa. Whats going
on here?? I thoght current was proportional to applied volts? Thiese
parts of the cylce are defying Ohms law! Can anyone explain?

And how can current *lead* voltage? YOu have to have a voltage to get
a current to flow in the first place!!

The relationship between voltage and current for capacitors is:
The capacitor current (in amperes) is proportional to the time rate of
change of the voltage across it (in volts per second), and that
proportionality constant is the capacitance (in farads).

So the peaks of current when a sine wave of voltage is applied across
a capacitor occurs at zero volts, because that is when the voltage is
changing the most rapidly.

The current leads (by 90 degrees) the voltage only after many cycles
have passed, so that only sine wave rules apply. The current is then
driven by the energy stored during the previous peak. This is a sine
wave only rule and is a special case of the above formula.

But the more general formula applies all the time for any situation.

J

#### john jardine

Jan 1, 1970
0
Steve Evans said:
Helo,

If Ive got a resistor, a battery and an ammeter, and I stick the
battery across the resitor, I get a certain number of milliamps
current going through the resistor.

If I increase the voltage thruogh the resitor, I get more current.

If I decreas the voltage, the resistor passes less current.

So far, so good.

Now, if I try to do the same thing with a capactor instead of a
resistor, I get an initial curent flow that tapers off over time as
the cap charges up and winds up with the same voltage acsross it as
the battery.

Again, so far so good.

Now, if I try to stick a AC voltage accross the capacitor; everything
goes tits-up. For part of the cycle, yeah, the current increases with
the applied voltage and for other parts of the cycle, it decreases
along with deminishing voltage.

Fine again. (well not really -see below)

The problem is, at *other* parts of the cycle, the current has started
to reverse direction whilst the voltage is still increasing! And the
bit further on again, the same thing happens vice-versa. Whats going
on here?? I thoght current was proportional to applied volts? Thiese
parts of the cylce are defying Ohms law! Can anyone explain?

And how can current *lead* voltage? YOu have to have a voltage to get
a current to flow in the first place!!

tnx,

steve

Yes. The current is due the voltage. It's traditionally called 'leading' cos
it just looks that way on an oscilloscope screen. It's actually 'lagging'
but anyone saying this risks being taken out and shot.
Ohms law doesn't apply to caps and inductors. It's best to just look at 'em
as energy storage devices. They'll take current from the supply and charge
up during one part of a cycle and then stuff the current *back into* the
supply the next part of the cycle. Net energy consumption is therefore
zilch. Whereas a resistor constantly turns current and pro-rata volts drop,
into a heat loss.
Sinewaves are *extremely* non linear waveforms and for visualisation it's
very, very, difficult to usefully sketch, draw, or graph out what's
happening to voltages and current over 2 or 3 cycles. Many people give up
at this point and just take the maths at face value.
A very, very, useful learning aid though, is a Spice prog'. The R or C or L
(or mixtures) currents and voltages can be watched exactly, point by point,
from time=0. Also, perhaps more valuable from a learning POV is the use of a
square wave voltage source and the knowledge that a fourier transform can
neatly tie the markedly different sine/square output waveshapes together.
regrds
john

B

#### Bob Myers

Jan 1, 1970
0
Ohms law doesn't apply to caps and inductors.

Minor nit here: Ohm's Law most certainly DOES apply
to caps and inductors, and in fact to AC circuits in general;
the only difference is that it has to be used in the proper form,
which in the case of AC means that the voltage, current, and
impedance (which takes the place of pure resistance when dealing
with AC) have to be given as vectors, or complex quantities.
Do that, and E=IR becomes E=IZ, and everything works
just as it should.

In the simplest terms I can think of, to answer the original
poster's question: what you have to realize is that reactances
(capacitors and inductors, in AC circuits; reactance is the
"imaginary" portion of impedance, while resistance is the
"real" portion) never dissipate any power (which is all
that resistances CAN do). Instead, they sometimes are
storing energy, and sometimes releasing it back into the
rest of the circuit. The "current leads voltage by 90 degrees"
(or vice-versa, for an inductor) happens basically because
PART of the time (part of every cycle of AC), you've got
an element of the circuit which is now acting as a source
rather than a sink.

Bob M.

J

#### john jardine

Jan 1, 1970
0
Bob Myers said:
Minor nit here: Ohm's Law most certainly DOES apply
to caps and inductors, and in fact to AC circuits in general;
the only difference is that it has to be used in the proper form,
which in the case of AC means that the voltage, current, and
impedance (which takes the place of pure resistance when dealing
with AC) have to be given as vectors, or complex quantities.
Do that, and E=IR becomes E=IZ, and everything works
just as it should.

In the simplest terms I can think of, to answer the original
poster's question: what you have to realize is that reactances
(capacitors and inductors, in AC circuits; reactance is the
"imaginary" portion of impedance, while resistance is the
"real" portion) never dissipate any power (which is all
that resistances CAN do). Instead, they sometimes are
storing energy, and sometimes releasing it back into the
rest of the circuit. The "current leads voltage by 90 degrees"
(or vice-versa, for an inductor) happens basically because
PART of the time (part of every cycle of AC), you've got
an element of the circuit which is now acting as a source
rather than a sink.

Bob M.
Indeed yes. Ohm can be pressed into service nicely, -if- we pretend the
capacitor is a resistor having a fixed ohms value.
This though applies only when dealing with particular AC circuitry and using
en-masse pure sinewaves that have had no awkward beginnings and will have no
awkward endings.
I.e. steady state answers that result from .AC type analysis used to
characterise linear amps, filters, phasing networks etc. This is the case
that always applies in textbooks and classrooms.
But ... the capacitor as a component has ceased to exist, simplified to a
linear impedance vector that can never accumulate/lose charge, or generate a
current transient sufficient to cause fuses to blow. For calculating
convenience, ohms law is being applied to a subset of special cases from the
real world.

In Steve's post I read he was coming in from problems figuring through the
initial startup transient. (ie ".TRAN").
This messy aspect is usually glossed over but needed to account for the
effects of reactive components in real circuitry.
I'd think it would be a perverse person (masochist!) who would wish to
characterise or transform the non linear curve of the initial capacitor
charging current over maybe the first 1.5 cycles, into its near infinite
number of component sinewaves and associated phase angles and then
manipulate and sum each of these vectors to allow forcable time varying
application of ohms law to all the reactive impedances.
Basically I'm saying that Ohm shouldn't even be considered ( my "ohms law
doesn't apply"). Maybe considered only if some kind of steady (sinewave)
state can exist. In many cases this will not occur or even be possible.
I'd guess the most useful number till then is the slew=i/c item.
regards
john

B

#### Bill Bowden

Jan 1, 1970
0
Steve Evans said:
Helo,

If Ive got a resistor, a battery and an ammeter, and I stick the
battery across the resitor, I get a certain number of milliamps
current going through the resistor.

If I increase the voltage thruogh the resitor, I get more current.

If I decreas the voltage, the resistor passes less current.

So far, so good.

Now, if I try to do the same thing with a capactor instead of a
resistor, I get an initial curent flow that tapers off over time as
the cap charges up and winds up with the same voltage acsross it as
the battery.

Again, so far so good.

Now, if I try to stick a AC voltage accross the capacitor; everything
goes tits-up. For part of the cycle, yeah, the current increases with
the applied voltage and for other parts of the cycle, it decreases
along with deminishing voltage.

Fine again. (well not really -see below)

The problem is, at *other* parts of the cycle, the current has started
to reverse direction whilst the voltage is still increasing! And the
bit further on again, the same thing happens vice-versa. Whats going
on here??

Yes, but at other parts of the cycle, the source voltage
is still higher than the capacitor voltage, so the cap will
continue to charge even though the source is falling.

The capacitor will stop charging and begin discharging
*exactly* when the source voltage falls to the same
potential as the capacitor.

All you have to do is think about the difference in voltage
at various points along the waveform to determine the direction
of current.
And how can current *lead* voltage? YOu have to have a voltage to get
a current to flow in the first place!!

You don't really need a voltage to have a current flowing.
Think about a tank circuit with a capacitor and inductor
exchanging energy at some frequency. The current in the
tank circuit will be maximum when the voltage across the circuit
is zero, and visa versa.

-Bill

S

#### Steve Evans

Jan 1, 1970
0
Yes, but at other parts of the cycle, the source voltage
is still higher than the capacitor voltage, so the cap will
continue to charge even though the source is falling.

The capacitor will stop charging and begin discharging
*exactly* when the source voltage falls to the same
potential as the capacitor.

All you have to do is think about the difference in voltage
at various points along the waveform to determine the direction
of current.

tnx, bill, but that can't be quite right (with al due respect).
The source voltage and the cap voltage are tied together with jumper
wire with approx. 0 ohms resistance! Therefore, whatever the source
votage is, the capactor voltage *must* be the same!

Apart from that; *is* there *always* 90 degrees of phase lag/lead with
reactive components, *or* can it vary. (i;ve simulated it in spice and
have found varying amounts of phase diffrence dpending on hte values
of hte cap and coil concerned. ) I guess the amount of reactance
present determines jthe amount of phase shift (more reactance= more
phase shift?)

tnx,

steve

S

#### Steve Evans

Jan 1, 1970
0
So the peaks of current when a sine wave of voltage is applied across
a capacitor occurs at zero volts, because that is when the voltage is
changing the most rapidly.

Thnx, John.
that's not the whole problem though!
Just after that voltage zero-crossing point; the voltage is still
increasing at *nearly* maximum rate (on the pos. half-cycle), but the
current has already started to fall back! The real problem is at
*this* point when the voltage is still rising steeply, and yet the
current is now *falling* steeply! It doesn't make sense! Look at the
point on the graph where current and voltage overlap and youll see
what i mean.

tnx,

stve

J

Jan 1, 1970
0
john jardine said:
Indeed yes. Ohm can be pressed into service nicely, -if- we pretend the
capacitor is a resistor having a fixed ohms value.
This though applies only when dealing with particular AC circuitry and
using
en-masse pure sinewaves that have had no awkward beginnings and will have
no
awkward endings.

Not true. But straightforward extension, you obtain the Laplace transform
of your circuit (just replace all those 'j-omegas' with 's') and via
convolution (multiplication in the s domain) you can find the output for
relatively arbitrary input signals. The Laplace transforms for, e.g., step
functions are quite 'reasonable' (not at all 'awkard').

This is done quite commonly.
But ... the capacitor as a component has ceased to exist, simplified to a
linear impedance vector that can never accumulate/lose charge, or generate
a
current transient sufficient to cause fuses to blow. For calculating
convenience, ohms law is being applied to a subset of special cases from
the
real world.

Well, there's no such thing as an ideal capacitor anyway (and this fact
become very significant at high frequencies), Ohm's law assume a 'lumped'
network (no distributed effects -- this also becomes significant for many
designers), Ohm's law comes from Maxwell's equations that are only an
approximation of quantum electrodynamics (very significant to semiconductor
guys), blah, blah, blah --> very quickly this all becomes philosophical.
Circuit simulators use components that are meant to _model_ 'reality,' but
obviosuly the results are no good if your model isn't any good.

When doing a pure AC simulation, the net charge on the capacitor remains
unchanged so it'd be silly for an AC simulator to consider this. If you
want second by second charge, you'd use a transient analysis.
In Steve's post I read he was coming in from problems figuring through
the
initial startup transient. (ie ".TRAN").
This messy aspect is usually glossed over but needed to account for the
effects of reactive components in real circuitry.

It's not that messy. A sine wave that starts at time 0 when applied to a
capacitor gets you a voltage that has an exponential term and the (same)
sine wave term; this comes directly from the inverse Laplace transform.
With a small source impedance, however, the exponential term decays very
quickly and can be neglected.
I'd think it would be a perverse person (masochist!) who would wish to
characterise or transform the non linear curve of the initial capacitor
charging current over maybe the first 1.5 cycles, into its near infinite
number of component sinewaves and associated phase angles and then
manipulate and sum each of these vectors to allow forcable time varying
application of ohms law to all the reactive impedances.

Again, this is undergraduate electrical circuit analysis. It isn't
difficult.
Basically I'm saying that Ohm shouldn't even be considered ( my "ohms law
doesn't apply"). Maybe considered only if some kind of steady (sinewave)
state can exist. In many cases this will not occur or even be possible.

The question is, "What do you want to know?" AC analysis is quite useful
for many circuit problems. So is transient analysis. Both can be performed
easily by hand so long as you stick with resistors, capacitors, and
inductors in your circuit. On the other hand, once you start sticking
non-linear active devices into the mix, you either decide you're operating
in the small signal domain and everything is still straightforward, or else
(for networks consisting or more than a few components) you end up trying to
solve differential equations which rapidly becomes intractable except via
numerical methods.

J

Jan 1, 1970
0
Steve Evans said:
Apart from that; *is* there *always* 90 degrees of phase lag/lead with
reactive components

Assuming an _ideal_ capacitor or inductor, _in the steady state_ there's
always 90 degrees of phase lead or lag between the voltage across _that_
device and the current through _that_ device. Now, if you go and put an L
and C in series, across the entire thing there's certainly not (generally) a
90 degree phase diference, but across each component there is.
*or* can it vary. (i;ve simulated it in spice and
have found varying amounts of phase diffrence dpending on hte values
of hte cap and coil concerned. )

It could be:

-- Not small enough time steps. Your output should look like a nice
'smooth' waveform. During transient analysis, simulators choose time steps
such that certain errors are minimized. While the voltage and each time
_step_ will be correct (within that error bound), if your graphing a
straight line between voltages at those time steps, it'll often not be at
all 'nice' looking (sinusoidal). All simulators have a way of forcing them
to limit the maximum time step they take so that you can get a 'nice
looking' plot. (And unfortunately most simulators implement this feature in
a rather inefficient way, but that's a topic for another day.)
-- You're not looking at the steady state solution (you're looking too close
to time=0 -- run the simulation for a few hundred cycles and then look)
-- Numerical rounding in the simulator. But it should still be, e.g.,
89.xxx degrees.
I guess the amount of reactance
present determines jthe amount of phase shift (more reactance= more
phase shift?)

With a single element, you have no control over the phase shift. Start
stringing more elements together, and the reactances combine to determine
the amount of phase shift (but again, the phase shift of _each component_ is
still +/- 90 degrees).

Have you taken any circuits classes? Do you know what a phasor is (no, not
what they used in Star Trek)? A Laplace (or at least Fourier) transform?
It's difficult to go from the qualitative to the quantitatitve without their
use. (Not that it can't be done -- in many community college classes it
is! -- but it tends to force you to accept a lot more on faith rather than
being able to dervice the results yourself.)

J

#### John Popelish

Jan 1, 1970
0
Steve said:
Thnx, John.
that's not the whole problem though!
Just after that voltage zero-crossing point; the voltage is still
increasing at *nearly* maximum rate (on the pos. half-cycle), but the
current has already started to fall back! The real problem is at
*this* point when the voltage is still rising steeply, and yet the
current is now *falling* steeply! It doesn't make sense! Look at the
point on the graph where current and voltage overlap and youll see
what i mean.

You are not as good at perceiving rate of change as a capacitor is.
The slope of a sine wave is a cosine wave, and that cosine is exactly
what the current through a capacitor connected across an AC voltage
follows.

S

#### Steve Evans

Jan 1, 1970
0
Assuming an _ideal_ capacitor or inductor, _in the steady state_ there's
always 90 degrees of phase lead or lag between the voltage across _that_
device and the current through _that_ device. Now, if you go and put an L
and C in series, across the entire thing there's certainly not (generally) a
90 degree phase diference, but across each component there is.

thnx, Joel.
There\s obviously something still amiss with my understanding, then. I
have a friend whos a radio ham and he says that to eliminate
capacitive reactance, you "add a bit of inductance until it's
cancelled out." Is this the same as cancelling out the phase-shift? If
so, it suggests it is possible to have varying degrees of shift due to
varying degrees of inductance. If that weren't the case, you could
simply use *any* coil; its value wouldn't matter because whatever it
was, it would only change the phase by 90'.
It could be:

-- Not small enough time steps.

Yes, i know what you mean, but the timestep is fine, thnks.
With a single element, you have no control over the phase shift. Start
stringing more elements together, and the reactances combine to determine
the amount of phase shift (but again, the phase shift of _each component_ is
still +/- 90 degrees).

So are you saying if I have 3 caps and 3 coils in series the overall
phase shift will be zero because they all cancel eachother out? What
if they're all different values? From what you say if you get 90
degrees of shift (fixed) per element, that must be true, regardless of
the different values of those elements:

ac
source------0.nF------1uH------22nF------4uH-------33pF-------180nH---->

Regardless of the frequency of the source and the values of these
components, the overall phase shift is zero. Is that what you're
saying?

J

#### John Popelish

Jan 1, 1970
0
Steve said:
There\s obviously something still amiss with my understanding, then. I
have a friend whos a radio ham and he says that to eliminate
capacitive reactance, you "add a bit of inductance until it's
cancelled out." Is this the same as cancelling out the phase-shift? If
so, it suggests it is possible to have varying degrees of shift due to
varying degrees of inductance. If that weren't the case, you could
simply use *any* coil; its value wouldn't matter because whatever it
was, it would only change the phase by 90'.

The current through the capacitor is 90 degrees shifted with respect
to the voltage across it. The current through the inductor is 90
degrees shifted the other way with respect ot the voltage across it.
So the two currents are 180 degrees shifted with respect to each
other. This means the smaller current cancels some of the larger
current. When you select the right sized components those two
currents are also the same magnitude, and cancel each other,
completely. With real components (that have loss) The currents are
shifted slightly less than 90 degrees, so the cancellation is not
perfect.
(snip)
So are you saying if I have 3 caps and 3 coils in series the overall
phase shift will be zero because they all cancel eachother out? What
if they're all different values? From what you say if you get 90
degrees of shift (fixed) per element, that must be true, regardless of
the different values of those elements:

The phase shift between voltage across each reactive component will be
90 degrees with respect ot the voltage across that component. If a
component has internal series resistance, there is no way to measure
the voltage across only the reactive part of its impedance, so any
measurement includes the voltage drop across the resistive component,
also. But the current measurement is that of each of the series
parts. Likewise, if the component contains internal parallel
resistance, then the voltage measurement accurately represents the
voltage across the reactive part, but the current measurement includes
the resistive parallel current, so you will not see a perfect 90
degree phase shift. Most real components exhibit both series and
parallel resistive effects, so neither the voltage or current
measurements are looking at pure reactance.

B

#### Bill Bowden

Jan 1, 1970
0
Steve Evans said:
tnx, bill, but that can't be quite right (with al due respect).
The source voltage and the cap voltage are tied together with jumper
wire with approx. 0 ohms resistance! Therefore, whatever the source
votage is, the capactor voltage *must* be the same!

Maybe this will make more sense;

You can't change the voltage on the capacitor, only the current
can be changed. So, if the current is rising toward a peak,
the voltage is also rising according to CE=IT. Now, as
the current reaches a peak and the starts decreasing,
the capacitor voltage is still going up because the current
is still flowing the same way, even though it reached a
peak and is decreasing. So, maybe you can see the phase
shift is 90 degrees because the capacitor voltage will
reach a peak exactly when the current stops at zero and
starts going the other way. Therefore V is max when
I is minimum, and visa versa.

Does that make any sense?

-Bill

S

#### Steve Evans

Jan 1, 1970
0
The current through the capacitor is 90 degrees shifted with respect
to the voltage across it. The current through the inductor is 90
degrees shifted the other way with respect ot the voltage across it.
So the two currents are 180 degrees shifted with respect to each
other. This means the smaller current cancels some of the larger
current. When you select the right sized components those two
currents are also the same magnitude, and cancel each other,
completely. With real components (that have loss) The currents are
shifted slightly less than 90 degrees, so the cancellation is not
perfect.

thanks! That made some sense at least.

[snip real-world stuff]

tnx for that, but I'm only concerned here about perfect lumped
elements. Lemme get a proper handle on that before worrying about the
paracitics!

S

#### Steve Evans

Jan 1, 1970
0
You can't change the voltage on the capacitor, only the current
can be changed. So, if the current is rising toward a peak,
the voltage is also rising according to CE=IT. Now, as
the current reaches a peak and the starts decreasing,
the capacitor voltage is still going up because the current
is still flowing the same way, even though it reached a
peak and is decreasing. So, maybe you can see the phase
shift is 90 degrees because the capacitor voltage will
reach a peak exactly when the current stops at zero and
starts going the other way. Therefore V is max when
I is minimum, and visa versa.

Does that make any sense?

Yes, it does, thnx, Bill. I guess the source resistance has some
bearing on it too.
However, - and its a *big* however - I'm still finding I'm getting
varying degrees of phase shift. I've posted a simple spice schematic
to alt.binaries.schematics.electronic htat shows a 180' phase shift
between voltage and current in one capacitor! I want to see hwere I'm
going wrong so itll be interesting to read what everyonte has to say
THere are two pictures: one for the circuit and one for the waveforms
I get showing the current and voltage in antiphase.

tnx,
steve

J

#### John Fields

Jan 1, 1970
0
I've posted a simple spice schematic
to alt.binaries.schematics.electronic htat shows a 180' phase shift
between voltage and current in one capacitor! I want to see hwere I'm
going wrong so itll be interesting to read what everyonte has to say
THere are two pictures: one for the circuit and one for the waveforms
I get showing the current and voltage in antiphase.

S

#### Steve Evans

Jan 1, 1970
0
I'll play, but on abse; it's too much of a pita to be bouncing back
and forth.

OK. Thats cool. I'm subscribed now.

tnx,

steve

J

Jan 1, 1970
0
Hi Steve,

Steve Evans said:
There\s obviously something still amiss with my understanding, then. I
have a friend whos a radio ham and he says that to eliminate
capacitive reactance, you "add a bit of inductance until it's
cancelled out." Is this the same as cancelling out the phase-shift?

Yes, that's what he's saying.

Here's a little bit of 'Phasors 101' without any proof whatsoever (you'll
find it in any circuits text):

-- The impedance (Z) of a resistor is R
-- The impedance (Z) of an inductor is j*2*pi*F*L, where j is sqrt(-1), F is
the frequency you're worrying about, and L is the inductance.
-- The impedance of a capacitor (Z) is -j/(2*pi*F*C), same symbols as
before.
-- Ohms law is now V=I*Z, and all of V, I, and Z are typically complex.
-- Complex numbers are often expressed as a magnitude and phase. E.g.,
1.414+j1.414 is the same as '2 angle 45 degrees'

Now, notice that for an inductor or capacitor, V/I is some purely imaginary
number, that is, something with an angle of +/-90 degrees. This confirms
what you know that the voltage and current in a capacitor or inductor leads
or lags either other by 90 degrees.

If I do something like put a resistor and capacitor in series and power it
up from a 1V source, the current through _both_ will be
I=V/Z=1/(R-j/(2*pi*F*C)) -- notice that it'll now be some arbitrary complex
value. But the voltage across the resistor is just V=I*Z=I*R, so it'll just
be that same complex value multiplied by a scalar, implying that voltage and
current are still in phase. For the capacitor, V=I*Z=I*-j/(2*pi*F*C). Now,
whenever you multiple a complex number by +/-j[something], the result is a
complex number with a magnitude [something]*[magnitude of what your started
with] and angle [+/-90 degrees]+[angle of what you started with]. I.e.,
current and voltage across the inductor are still 90 degrees out of phase.
If
so, it suggests it is possible to have varying degrees of shift due to
varying degrees of inductance. If that weren't the case, you could
simply use *any* coil; its value wouldn't matter because whatever it
was, it would only change the phase by 90'.

What you're missing here is that while the phase across the an inductor and
capacitor is always +/- 90 degrees, you can only 'null out' the two when the
MAGNITUDE of the signal across them is equal as well. That is, if we're
dealing with voltage, (5 angle +90) + (5 angle -90) is, indeed, zero, but (3
angle +90) + (5 angle -90) is (2 angle -90), i.e., you still have something
that looks like a capacitor or inductor. (This means that if you start
with, say, a capacitor -- Z=-j*something Ohms -- and add an inductor --
Z=j*something_else Ohms -- as the inductor gets bigger and bigger, you'll
reach a point where the two are a short circuit. This is resonance. If you
keep adding inductance, the network then starts looking inductive.)

You can _not_ just 'add phases' across a bunch of components in series and
get the overall phase change through the network. You can (and _must_) add
_impedances_ .
So are you saying if I have 3 caps and 3 coils in series the overall
phase shift will be zero because they all cancel eachother out?

If the _impedance_ of all the caps and coils in series adds to zero, yes,
they will all cancel each other out. This will only happen when the sum of
the voltage magnitudes across the capacitors is equal to the sum of the
voltage magnitudes across the inductors.
What
if they're all different values?

Then the voltage magnitudes across the inductors won't sum to the same thing
as that across the capacitors, and you'll end up with something like that
likes a single inductor or capacitor (at one frequency).
From what you say if you get 90
degrees of shift (fixed) per element, that must be true, regardless of
the different values of those elements:

Again, only if you arrange things such that the magnitudes of the voltage
across each one is the same as well.
ac
source------0.1nF------1uH------22nF------4uH-------33pF-------180nH---->

In your network here, let's assume the frequency is 1/6.28 to keep things
simple. Hence, Z(res)=R, Z(ind)=j*L, and Z(cap)=-j/C. So, in your network,
the series impedance is -j/0.1e-9 + j*1e-6 -j/22e-9 + j*4e-6 - j/33e-12 +
j*180e-9 -- Obviously this won't sum to zero! If you pick the right
frequency, though, it will. (Bonus question: What is this frequency?)
Regardless of the frequency of the source and the values of these
components, the overall phase shift is zero. Is that what you're
saying?

At a certain frequency the phase shift is zero. Otherwise, no.

You know, if you really want to get into this, just sign up for your local
college's first engineering first 'circuits' (or 'networks') class. I took
mine the summer of 1995, and it was really a blast. ---Joel

S

#### Steve Evans

Jan 1, 1970
0
On Mon, 15 Nov 2004 17:41:16 -0800, "Joel Kolstad"

[snip]
What you're missing here is that while the phase across the an inductor and
capacitor is always +/- 90 degrees, you can only 'null out' the two when the
MAGNITUDE of the signal across them is equal as well. That is, if we're
dealing with voltage, (5 angle +90) + (5 angle -90) is, indeed, zero, but (3
angle +90) + (5 angle -90) is (2 angle -90), i.e., you still have something
that looks like a capacitor or inductor. (This means that if you start
with, say, a capacitor -- Z=-j*something Ohms -- and add an inductor --
Z=j*something_else Ohms -- as the inductor gets bigger and bigger, you'll
reach a point where the two are a short circuit. This is resonance. If you
keep adding inductance, the network then starts looking inductive.)

[rest snipped]

Okay Joel, I think I've finally gottit now. Some people seem to
explain things more clearly than others and I guess your one of those
good communicators! There is still a minor point or two I gotta work
out, though, but the worst is over now, I reckon. Many thnx for your time.

steve

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