Am I allowed to use KLV and KCL with a capacitor in the circuit?

AI Thread Summary
The discussion centers on the use of Kirchhoff's Laws (KVL and KCL) in circuits involving capacitors, particularly in the context of a circuit that appears to have a short circuit across a capacitor. Participants debate the applicability of these laws during transient states, such as when a capacitor is charging. While some argue that KVL and KCL are universally applicable, others suggest that they may not hold in non-conservative fields or during time-varying conditions. The importance of using Thevenin equivalents for analyzing circuits with capacitors is also highlighted, emphasizing the need for proper analysis techniques. Ultimately, the consensus is that Kirchhoff's Laws remain valid in lumped circuit theory, despite the complexities introduced by capacitors and other components.
Femme_physics
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Femme_physics said:
I.e.

http://img337.imageshack.us/img337/2701/15irk.jpg

Is this alright? Is this "legal"?

You seem to have a short circuit across the capacitor and so across the rest of your circuit apart from the top resistor ( presumably R1)

Kirchoff's Laws really only apply to a stable situation, so a charging capacitor would be analysed by taking a "snapshot" of the circuit operation.

The two parallel resistor-diode circuits can be analysed by getting the parallel resistor combination and putting it in series with one diode.
You can do this because both diodes will drop 0.7 volts, so you can join their anodes together and no current will flow between points of equal voltage.

For example, let R1 = 1000 ohms, R2 = 2000 ohms, R3 = 3000 ohms Diode voltage = 0.7 volts.
Parallel R2 and R3 = 1200 ohms call this R4

So voltage across R1 and R4 = 15 - diode voltage = 14.3 volts.
Current in R1 and R4 = 14.3 Volts / (1000 ohms + 1200 ohms) = 0.0065 amps

Voltage across R4 if C wasn't there = IR = 0.0065 * 1200 = 7.8 volts
R4 plus diode voltage = 8.5 volts

So this is the maximum voltage that C could charge to, if it didn't have a short circuit on it. :)
 
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You seem to have a short circuit across the capacitor and so across the rest of your circuit apart from the top resistor ( presumably R1)

Oh yes, there supposed to be a button there to cause it. You're right, the button does cause a short circuit.

The two parallel resistor-diode circuits can be analysed by getting the parallel resistor combination and putting it in series with one diode.
You can do this because both diodes will drop 0.7 volts, so you can join their anodes together and no current will flow between points of equal voltage.

For example, let R1 = 1000 ohms, R2 = 2000 ohms, R3 = 3000 ohms Diode voltage = 0.7 volts.
Parallel R2 and R3 = 1200 ohms call this R4

So voltage across R1 and R4 = 15 - diode voltage = 14.3 volts.
Current in R1 and R4 = 14.3 Volts / (1000 ohms + 1200 ohms) = 0.0065 amps

Voltage across R4 if C wasn't there = IR = 0.0065 * 1200 = 7.8 volts
R4 plus diode voltage = 8.5 volts

So this is the maximum voltage that C could charge to, if it didn't have a short circuit on it. :)

So suppose instead of a shortcut there is a switch there or whatever, so there is no short-circuit in fact. Is my method valid in this case?

I appreciate you writing up a way to the solution, but I want to understand whether I have a glimpse of validity in my method before examining yours. :)
 
You have 3 unknowns but two equations, so you need another equation.

But the equations you have written look OK.

However as I tried to show, there is no need to use equations on a trivial problem.
 
You can use KCL and KVL on a capacitor. You represent it'd impedance with a phasor or in the laplace domain.
 
vk6kro said:
Kirchoff's Laws really only apply to a stable situation, so a charging capacitor would be analysed by taking a "snapshot" of the circuit operation.

Huh? :confused:
I though Kirchhoff's laws *always* apply.
Femme_physics said:
I.e.

http://img337.imageshack.us/img337/2701/15irk.jpg

Is this alright? Is this "legal"?

Is it legal and valid?
Yes.
 
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I agree. Kirchoff's laws and Ohm's law apply to 100% of the cases.

Adding a capacitor, inductor, diode...etc does not change that.

Accepting these laws seems to take time...but it shouldn't.

KVL, KCL and V=IR

Just accept it now.
 
psparky said:
I agree. Kirchoff's laws and Ohm's law apply to 100% of the cases.

Adding a capacitor, inductor, diode...etc does not change that.

Accepting these laws seems to take time...but it shouldn't.

KVL, KCL and V=IR

Just accept it now.

Not really.

If you have a capacitor that is still charging, you might apply some equations to it for one instant in time, but then it changes as the capacitor charges up some more.

So, it is better to use logic and derive a Thevenin equivalent circuit which can then charge up the capacitor.

Incidentally, phasors are only relevant for AC circuits. This is DC.
 
vk6kro said:
Not really.

Yes. Really.


If you have a capacitor that is still charging, you might apply some equations to it for one instant in time, but then it changes as the capacitor charges up some more.

At any point in time, you have voltages and currents to which KVL and KCL apply.


Incidentally, phasors are only relevant for AC circuits. This is DC.

The theory of phasors holds true in DC as well.
With DC, you would need to calculate the inverse Laplace (or Fourier) transform to get the proper result.
 
  • #10
I like Serena said:
Yes. Really.




At any point in time, you have voltages and currents to which KVL and KCL apply.




The theory of phasors holds true in DC as well.
With DC, you would need to calculate the inverse Laplace (or Fourier) transform to get the proper result.

You know...I have to say.

"I like Serena"
 
  • #11
psparky said:
You know...I have to say.

"I like Serena"

That ought to be: I like I like Serena. :wink:

--I like ILSe
 
  • #12
Basically, while KLV and KCL always stand true, they won't have me to find a solution with respect to time... which is what I have in my case. But I will post it in the homework section later.. I have more relevant concerns when it comes to electronics, so I'll post this exercise later. Thanks a lot, everyone! :)
 
  • #13
Femme_physics said:
Basically, while KLV and KCL always stand true, they won't have me to find a solution with respect to time... which is what I have in my case. But I will post it in the homework section later.. I have more relevant concerns when it comes to electronics, so I'll post this exercise later. Thanks a lot, everyone! :)

That's right. It is a question of using the right tool for the job.
In this case, you work out a Thevenin equivalent of the circuit apart from the capacitor and then use this to charge the capacitor.

In the example above, the Thevenin voltage is 8.5 volts and the Thevenin resistance is about 545 ohms.
So the time constant with a 500 μF capacitor is (545 ohms * 0.0005 Farads) or 0.272 seconds.

So, the capacitor voltage will rise to 0.636 times 8.5 volts (5.4 volts) in 0.272 seconds and it will eventually reach 8.5 volts in about 3 seconds.

I wonder if many of the queries you make could be solved if you became familiar with LTSpice. This is free and very easy to use. Have you tried it?
 
  • #14
Femme_physics said:
Basically, while KVL and KCL always stand true

KVL represents conservation of energy in what we call a "conservative field" and KCL represents conservation of charge in a system that doesn't allow too much charge to build up at any particular node.
 
  • #15
rbj said:
KVL represents conservation of energy in what we call a "conservative field" and KCL represents conservation of charge in a system that doesn't allow too much charge to build up at any particular node.
Which is why I Like Serena and Psparky are wrong when the say that Kirchoff's Laws are universal. In non-conservative fields such as time-varying electric/magnetic fields, the voltage between points A and B is not uniquely defined--it depends on the path followed between the points.

The sum of the voltages around the loop driven by a time-varying magnetic field is non-zero, as shown by Prof. Lewin here:

OP should ignore this post.
 
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  • #16
gnurf said:
Which is why I Like Serena and Psparky are wrong when the say that Kirchoff's Laws are universal. In non-conservative fields such as time-varying electric/magnetic fields, the voltage between points A and B is not uniquely defined--it depends on the path followed between the points.

The sum of the voltages around the loop driven by a time-varying magnetic field is non-zero, as shown by Prof. Lewin here:

OP should ignore this post.


you mean your post? for circuits, KVL and KCL are within \epsilon of being precisely true.
 
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  • #17
rbj said:
you mean your post? for circuits, KVL and KCL are within \epsilon of being precisely true.
Lumped circuit-theory is already based on approximations of Maxwell's equations (e.g. capacitor current I = C*dV/dt is derived by ignoring the effect of the time-varying magnetic field in Faraday's Law). So within lumped circuit theory, I'd say KVL and KCL are always true. That's why I said OP could ignore my post. What is this epsilon you mention?
 
  • #18
gnurf said:
Which is why I Like Serena and Psparky are wrong when the say that Kirchoff's Laws are universal. In non-conservative fields such as time-varying electric/magnetic fields, the voltage between points A and B is not uniquely defined--it depends on the path followed between the points.

The sum of the voltages around the loop driven by a time-varying magnetic field is non-zero, as shown by Prof. Lewin here:

OP should ignore this post.


Here we go again. KVL and KCL are universal and always work. Period.

There are no EM fields in circuit theory and therefore no non-conservative fields either.

There are no magnetic fields in the inductors of circuit theory. There are no electric fields on a schematic. That's physics, not circuit theory.
 
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  • #19
Antiphon said:
Here we go again. KVL and KCL are universal and always work. Period.

It's truly unbelievable that people try to argue this.

Like I said...for some reason it takes a while to sink in.
 
  • #20
Antiphon: Here you go again--I must have missed something! Did you read my follow-up comment, though? I tried to communicate that Kirchoff's Laws apply in lumped circuit theory precisely for the reasons you mentioned.

If I understand you correctly, you're saying that Kirchoff's laws are "universal" because they should only be applied to lumped circuits. So when Prof. Lewin replaces the DC source with a time varying magnetic field to induce a voltage in the loop, he re-applies KVL to a problem that should no longer be solved with KVL? Is this what you're saying?

OP, sorry for messing up your thread.
 
  • #21
psparky said:
It's truly unbelievable that people try to argue this.

Like I said...for some reason it takes a while to sink in.
If EM seems that simple to you, then you either know a lot or more likely nothing.
 
  • #22
vk6kro said:
For example, let R1 = 1000 ohms, R2 = 2000 ohms, R3 = 3000 ohms Diode voltage = 0.7 volts.
:)
For the voltage drop across a standard LED, I have seen specs ranging from 1.5 to 5V. Because this depends on the LED used, this parameter should be supplied in this question. I suspect you were thinking of a silicon diode having a typical drop of 0.7 V.
 
  • #23
gnurf said:
Antiphon: Here you go again--I must have missed something! Did you read my follow-up comment, though? I tried to communicate that Kirchoff's Laws apply in lumped circuit theory precisely for the reasons you mentioned.

If I understand you correctly, you're saying that Kirchoff's laws are "universal" because they should only be applied to lumped circuits. So when Prof. Lewin replaces the DC source with a time varying magnetic field to induce a voltage in the loop, he re-applies KVL to a problem that should no longer be solved with KVL? Is this what you're saying?

OP, sorry for messing up your thread.

Yes I saw it but it was at odds with the post above it and I mistook you for two authors.

I humbly recommend editing the post in a case like this but I missed it, sorry.

Edit: professor Lewin violates the rules of the lumped circuit abstraction deliberately to provoke thought among his students.

In short: any schematic that has any fields on it has left the realm of the lumped circuit theory and stepped back into physics.

In circuit theory there is voltage and current. There isn't an electric or magnetic vector (E,H) there isn't an electric or magnetic flux (D,B), there are no constitutive relations (permeabilities, permittivities).
 
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  • #24
Ouabache said:
For the voltage drop across a standard LED, I have seen specs ranging from 1.5 to 5V. Because this depends on the LED used, this parameter should be supplied in this question. I suspect you were thinking of a silicon diode having a typical drop of 0.7 V.

Ah yes, thanks. I didn't notice the LED symbols.

It doesn't make any difference to the calculation method, though, unless the LEDs were different colors, in which case they wouldn't have the same voltage.

If they were different voltages, we wouldn't be able to do this:

http://dl.dropbox.com/u/4222062/diodes.PNG

which makes the calculations simpler.
 
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  • #25
RE Lewin's video:

i don't buy it.

The professor did not represent with his drawing the circuit he evaluated.

Had he drawn voltage sources in the wires representing ∫e(dot)dl where flux couples them his drawing would be accurate . And Kirchoff would prevail.

Anybody who's worked around magnetics knows your voltmeter leads are one turn. Short them together around an energized transformer core and observe meter.
He measured the voltage arriving at his meter not the voltage between A and D.

But he's an entertaining lecturer.

old jim
 
  • #26
psparky said:
It's truly unbelievable that people try to argue this.

Like I said...for some reason it takes a while to sink in.

KVL and KCL are not universal, they result from solutions of Maxwell's equations with specific conditions, and solutions to his equations are not generic/universal. Even my crappy EM professor taught us how KVL comes from solutions to maxwell's equations; I don't think this should be a controversial subject.

Also, ohm's law is not universal. Ohm's law is a linear relationship, and it is useful in many applications, but it takes one counter example to show that i-v relationship is not always linear and not described by Ohm's law.

For example, shockley's equation:
I = I_{s}e^{\frac{Vd}{nV_{T}}}

You can't really manipulate that equation to resemble Ohm's law. The universe is actually nonlinear in many ways, and so its not far fetched to accept that ohm's law is not universal.

Here are some cool lecture notes I found for showing the current flow in a diode, and application of maxwell's equations to do it.

http://ocw.mit.edu/courses/electric...evices-spring-2007/lecture-notes/lecture8.pdf

Replace the lecture8.pdf up to 12 to get them all.

I feel bad for the OP tho, your thread is kind of off track, but I think you got your answer that KVL and KCL can be used with a capacitor.
 
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  • #27
I feel bad for the OP tho, your thread is kind of off track, but I think you got your answer that KVL and KCL can be used with a capacitor.

The thread is right on track.

She originally asked "Am I allowed to use KLV and KCL with a capacitor in the circuit?"

Some of you guys are more or less saying we can't used KVL when calculating drag on an airplane at 600 mph. Ya, we get that.

The KVL, KCL, and V=IR refer to circuits. There is never a case in a CIRCUIT where these are not true.

Show me a circuit where the sum of the voltage drops in a loop don't add up to be the voltage sorce.

Show me a circuit where current in doesn't equal current out of a node.

Show me a circuit where V does not equal IR. Not sure I get the point of your "shockley example". Resistance is not in the equation unless I am missing something.
 
  • #28
psparky said:
The thread is right on track.

She originally asked "Am I allowed to use KLV and KCL with a capacitor in the circuit?"

Some of you guys are more or less saying we can't used KVL when calculating drag on an airplane at 600 mph. Ya, we get that.

The KVL, KCL, and V=IR refer to circuits. There is never a case in a CIRCUIT where these are not true.

Show me a circuit where the sum of the voltage drops in a loop don't add up to be the voltage sorce.

Show me a circuit where current in doesn't equal current out of a node.

Show me a circuit where V does not equal IR.

Thank you in advance.

https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/10RootFinding/Newton/diode.png

Solve for the voltage and current in the diode using only KVL, KCL, and ohm's law :D
 
  • #29
DragonPetter said:
https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/10RootFinding/Newton/diode.png

Solve for the voltage and current in the diode using only KVL, KCL, and ohm's law :D

Comon!

KCL, KVL and V=IR are all true! You just proved my point.

Just because the voltage source is not strong enough to turn on the circuit...that does not mean the laws are cancelled!

It is a FACT you may not be able to solve all circuits with those three tools...but those three tools are at your disposal at any times...and there laws always hold true!

AHHHHHHHHHHHHHHHHHHHHH!
 
  • #30
psparky said:
Comon!

KCL, KVL and V=IR are all true! You just proved my point.

Just because the voltage source is not strong enough to turn on the circuit...that does not mean the laws are cancelled!

It is a FACT you may not be able to solve all circuits with those three tools...but those three tools are at your disposal at any times...and there laws always hold true!

AHHHHHHHHHHHHHHHHHHHHH!

The diode is not "on" but it is conducting a very small current, just look at the i-v curve of a diode. Ohm's law does not apply to the i-v relationship of all circuit elements and so therefore its not universal.

Edit: I know wikipedia is not supposed to be an authority to reference but maybe this will get the point across:

http://en.wikipedia.org/wiki/Electrical_resistance

Read here "For a wide variety of materials and conditions, V and I are directly proportional to each other, and therefore R and G are constant (although they can depend on other factors like temperature or strain). This proportionality is called Ohm's law, and materials that satisfy it are called "Ohmic" materials.

In other cases, such as a diode or battery, V and I are not directly proportional, or in other words the I–V curve is not a straight line through the origin, and Ohm's law does not hold. In this case, resistance and conductance are less useful concepts, and more difficult to define. The ratio V/I is sometimes still useful, and is referred to as a "chordal resistance" or "static resistance",[1][2] as it corresponds to the inverse slope of a chord between the origin and an I–V curve. In other situations, the derivative dV/dI may be most useful; this is called the "differential resistance"."
 
  • #31
Gotta run for an hour...I'll be back...not done with this.

If the diode is on at .5 volts...and has a trickle current...it also has a resistance.

V=IR~!
 
  • #32
psparky said:
If the diode is on at .5 volts...and has a trickle current...it also has a resistance.

V=IR~!

That's where you're mistaken. Ohm's law is linear, the I-V relationship of the diode is not. It only has an incremental, small signal resistance and that is not defined by Ohm's law. In a diode, V != IR, but rather V = nV_{T}ln\left(\frac{I}{I_{s}}\right)and so ohm's law is not valid.

I'm not trying to argue that ohm's law is not useful or not essential to learn circuits, but its misleading to say it always holds true and is universal.

Another example: Replace the diode and resistor in the above circuit with an open. What is the resistance of an open circuit? Even if you let the resistance be infinite, ohm's law will not necessarily give you the correct voltage across the open.

Any time in a circuit when there is a condition for V, I, or R = infinity, ohm's law will fall apart (the voltage across the open will be undefined, which disagrees with what KVL tells you). There is no such thing as infinite resistance, infinite current flow, etc. in the real world anyway and most of the cases where you would approximate as open are actually nonlinear i-v relationships (e.g. dielectric in a capacitor will breakdown at a high enough DC voltage for example, even if we like to say the DC impedance is infinite).

Perhaps you could say Ohm's law is always true in LTI systems, but that is a very specific class of circuits, and so it is not universal. It would make sense why textbooks and professors stress the meaning of continuous, linear, time invariant, dynamic, and causality in circuits because they know that these conditions are necessary for our assumptions (V = IR) to be true.
 
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  • #33
psparky said:
Comon!
https://ece.uwaterloo.ca/~dwharder/N...wton/diode.png

KCL, KVL and V=IR are all true! You just proved my point.

Just because the voltage source is not strong enough to turn on the circuit...that does not mean the laws are cancelled!

It is a FACT you may not be able to solve all circuits with those three tools...but those three tools are at your disposal at any times...and there laws always hold true!

AHHHHHHHHHHHHHHHHHHHHH!

But for this circuit normally it is impossible to find analytical solution.
We a force to use numerical method or further simplify the circuit.
 
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  • #34
You made the point that the diode has a non IV relationship. Ok...so what.

At any point in time that voltage or current is based on the resistance of the diode.

In the circuit you showed...let's say there is a trickle current for the diode.

This trickle current will also trickle thru the resistor.

The current thru the diode multipled thru the resistance at that time...is the voltage drop.

The trickle current thru the resistor has a voltage drop based on same rules.

The sum of the two voltage drops equals the voltage source (KVL)Current into the source equals current out.(KCL) V=IR on all things described.

You have not convinced me...nor will you ever.
 
  • #35
psparky said:
You made the point that the diode has a non IV relationship. Ok...so what.

At any point in time that voltage or current is based on the resistance of the diode.

In the circuit you showed...let's say there is a trickle current for the diode.

This trickle current will also trickle thru the resistor.

The current thru the diode multipled thru the resistance at that time...is the voltage drop.

The trickle current thru the resistor has a voltage drop based on same rules.

The sum of the two voltage drops equals the voltage source (KVL)Current into the source equals current out.(KCL) V=IR on all things described.

You have not convinced me...nor will you ever.

So you say a diode still obey's ohm's law? In that case, you must be able to tell me the resistance or impedance of a diode. Can you show me a datasheet that tells what the R in ohm's law is for a diode? What your point is that incremental, small signal resistance exists, and I won't argue that. I will argue that a diode does not follow Ohm's law and you cannot apply Ohm's law to a diode.
 
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  • #36
Since I don't have much credibility, I will use a reference:

Read the 3rd paragraph here: http://www.phy.syr.edu/courses/PHY222.07Spring/manuals/ohm.pdf
 
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  • #37
After really thinking about it...I will have to agree that diodes do not follow ohm's law.

Why? Because if I have a 12 volt battery in series with a diode and resistor...I can change the value of the resistor and I will always have .7 volts dropped across the diode (assuming a .7 volt diode)...this clearly violates OHMS LAW..not to mention the couple sources you referenced.

Therefore, I clearly stand corrected. I'm so used to just dropping the .7 that I never stopped to think about it.

Are there any other devices that violate OHMS LAW?
 
  • #38
Linearity of indivdual components is not required for KVL and KCL to work. They work just fine on Diodes, FETs, whatever.
 
  • #39
Antiphon said:
Linearity of indivdual components is not required for KVL and KCL to work. They work just fine on Diodes, FETs, whatever.

Yep, the linearity discussion was only with regards to Ohm's law.
 
  • #40


gnurf said:
Lumped circuit-theory is already based on approximations of Maxwell's equations (e.g. capacitor current I = C*dV/dt is derived by ignoring the effect of the time-varying magnetic field in Faraday's Law). So within lumped circuit theory, I'd say KVL and KCL are always true.

pretty much agree.

What is this epsilon you mention?

any vanishingly small real and positive number. you know: "Given an \epsilon>0, find a \delta such that ..."

KVL and KCL ain't the perfect truth, but they are within \epsilon of it.

r b-j
 
  • #41
psparky said:
After really thinking about it...I will have to agree that diodes do not follow ohm's law.

Why? Because if I have a 12 volt battery in series with a diode and resistor...I can change the value of the resistor and I will always have .7 volts dropped across the diode (assuming a .7 volt diode)...this clearly violates OHMS LAW..not to mention the couple sources you referenced.

Therefore, I clearly stand corrected. I'm so used to just dropping the .7 that I never stopped to think about it.

Are there any other devices that violate OHMS LAW?

I think you are forgetting that Ohm's law is the law of a material property. If You say "a diode does not violate ohm's law" it would certainly be true for the material inside that diode given they conduct only by drift. But you know about the depletion layer and carrier injection and so and so. Do they violate ohm's law? Yes they do. As they conduct by diffusion as well as drift. However having solved this issue we can also say, diode's won't violate laws of electromagnetism and Poisson's equation. After all Ohm's law is just a special case. :smile:Are there other devices that violates ohm's law? Well roughly wherever depletion region is involved it will be violated.

EDIT: I just remembered you can roughly use ohm's law in linear portion of tunnel diode although it has depletion region. A very special case.
 
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  • #42
psparky said:
Are there any other devices that violate OHMS LAW?
Incandescent light globes are non-linear, and useful as a non-linear resistance in the region well below where the filament glows red. Carbon rods have a negative resistance with temperature, as I recall.
 
  • #44
Yes, the formula for voltage looks like that.
Myself, I would use a different representation of the same formula.

Do you have a method to find out what I, R, and A in that formula should be?
 
  • #45
That's exactly what I'm trying to figure out...no findings so far
 
  • #46
Typically, you would first simplify the circuit as much as possible.
The first step is to use what vk6kro showed in post #24.

The method I know (after the simplification), is to set up a differential equation with KVL and KCL and solve that (which I did).
But I'm pretty sure, that's not what you are expected to do, since you do not know (yet?) how to set up and solve a differential equation.

So you should have a different method in your textbook... but what is it?
 
  • #47
220px-RC_switch.svg.png


That formula seems unduly complex.

This one is more usual:

23bc4003a351f22399a2e89ecb70223a.png


where
7bbb7f3961b2ada8e476b2e5bbf16118.png
which is the time constant of the circuit. R is in ohms. C in Farads. (A convenient shortcut is to use Megohms and Microfarads here.)

For example,
if Vin =100 volts and the capacitor has no charge on it to start with.
C= 5 uF
R = 100 K (so RC = 0.5 seconds)
t = 0.6 second

V capacitor = 100 * (1 - 2.71828 ^ ( -0.6 / (5 * 0.1 )))
V capacitor = 100 * 0.6988 = 69.88 volts
 
  • #48
vk6kro said:
220px-RC_switch.svg.png


That formula seems unduly complex.

This one is more usual:

23bc4003a351f22399a2e89ecb70223a.png


where
7bbb7f3961b2ada8e476b2e5bbf16118.png
which is the time constant of the circuit. R is in ohms. C in Farads. (A convenient shortcut is to use Megohms and Microfarads here.)

i think if Femme blast out her equation, it will become equivalent to your more usual one.
 
  • #49
DragonPetter said:
You can use KCL and KVL on a capacitor. You represent it'd impedance with a phasor or in the laplace domain.

This is a snap shot at a certain time... you set the source to have a phase of zero and then each component has a different phase
 
  • #50
psparky said:
I agree. Kirchoff's laws and Ohm's law apply to 100% of the cases.

Adding a capacitor, inductor, diode...etc does not change that.

Accepting these laws seems to take time...but it shouldn't.

KVL, KCL and V=IR

Just accept it now.

only if Voltage = current * impedance. You need to translate into phasor form to analyze any non linear circuit.
 

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