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

  • #26
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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:
[itex] I = I_{s}e^{\frac{Vd}{nV_{T}}}[/itex]

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
psparky
Gold Member
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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
828
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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
psparky
Gold Member
884
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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
828
1
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
psparky
Gold Member
884
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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
828
1
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, [itex]V != IR[/itex], but rather [itex]V = nV_{T}ln\left(\frac{I}{I_{s}}\right)[/itex]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
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138
Comon!
https://ece.uwaterloo.ca/~dwharder/N...wton/diode.png [Broken]

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
psparky
Gold Member
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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
828
1
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
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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 [Broken]
 
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  • #37
psparky
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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
1,679
3
Linearity of indivdual components is not required for KVL and KCL to work. They work just fine on Diodes, FETs, whatever.
 
  • #39
828
1
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
rbj
2,226
9


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 [itex]\epsilon>0[/itex], find a [itex]\delta[/itex] such that ..."

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

r b-j
 
  • #41
376
1
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 wont 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
NascentOxygen
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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
I like Serena
Homework Helper
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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
Femme_physics
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That's exactly what I'm trying to figure out...no findings so far
 
  • #46
I like Serena
Homework Helper
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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 text book... but what is it?
 
  • #47
vk6kro
Science Advisor
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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
rbj
2,226
9
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
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
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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