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

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vk6kro
I.e.

http://img337.imageshack.us/img337/2701/15irk.jpg [Broken]

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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Gold Member
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. :)

vk6kro
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.

I like Serena
Homework Helper
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?
I though Kirchhoff's laws *always* apply.

I.e.

http://img337.imageshack.us/img337/2701/15irk.jpg [Broken]

Is this alright? Is this "legal"?
Is it legal and valid?
Yes.

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psparky
Gold Member
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.

vk6kro
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.

I like Serena
Homework Helper
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.

psparky
Gold Member
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"

I like Serena
Homework Helper
You know....I have to say.

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

--I like ILSe

Gold Member
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! :)

vk6kro
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?

rbj
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.

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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rbj
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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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?

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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psparky
Gold Member
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.

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?

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.

Ouabache
Homework Helper
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.

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?

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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vk6kro
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 [Broken]

which makes the calculations simpler.

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jim hardy
Gold Member
2019 Award
Dearly Missed
RE Lewin's video:

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