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

In summary: 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.
  • #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 [Broken]
 
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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 [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
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.
 
  • #51
I'd like to explain how the original circuit I posted works. I'll repost its image if I may:

http://img853.imageshack.us/img853/9302/diodescabal.jpg [Broken]

Once the switch is closed (and the push button remains unpressed) all the current flows towards the capacitor until it fully charges, then afterwards current flows through the resistor and LED's to the ground. When it passes the 5 mA for each diode they light up (in different times since the resistors don't have the same value). P is there to cause a shortcircuit thereby discharging the capacitor.

How did I do?
 
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  • #52
Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.

When you actually see the LEDs lighting up depends on the efficiency of the LED and the background lighting.
In a dark room, you may see the light at 1 or 2 milliamps.

Is that capacitor actually 500 MILLI Farads? ie half a Farad?
 
  • #53
Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.

Let me see if I have it straight. At the first moment the switch is closed, all the current flows towards the capacitor. The more time goes by, the more current starts flowing through the diodes (and resistors) until all the current flows through them. So the capacitor in fact just steals a bit from the current influx in the beginning until the diodes get it?

Is that capacitor actually 500 MILLI Farads? ie half a Farad?
Yes
 
  • #54
The capacitor will charge up to almost the supply voltage because the diodes draw very little current compared to the 3.75 amps that the capacitor initially draws.

The final voltage across the capacitor will be caused by the supply voltage (15 volts) minus the voltage drop due to the very small current drawn by the diodes. This will be something less than 32 mA so the drop will only be about 128 mV.

If you were to really build this, you would need to limit the current in the switch across the capacitor because this could be very large (depending on the internal resistance of the capacitor) and may damage the switch.

A good way to do this would be to put a small resistor between the capacitor and the 4 ohm resistor, then shorting that junction of the two resistors to ground. About 1 ohm should be adequate protection for the switch.

This way, the capacitor could fully discharge in a few seconds.
 
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  • #55
OK, thanks for this, but I'm trying to gauge this process based on time. What happens in the first moment it's closed, till the steady state. I was hoping my explanation above is correct?
 
  • #56
In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.
 
  • #57
vk6kro said:
In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.

How can you say that there is not enough voltage at the first moment if voltage is constant at 15 V?
 
  • #58
Femme_physics said:
How can you say that there is not enough voltage at the first moment if voltage is constant at 15 V?
Hi Femme_physics! vk6kro is saying there is not enough voltage on the capacitor plates to send current through the LEDs. It is the capacitor voltage which supplies the LEDs and their current-limiting resistors.
 
<h2>1. What is KLV and KCL?</h2><p>KLV (Kirchhoff's Voltage Law) and KCL (Kirchhoff's Current Law) are fundamental laws in circuit analysis that describe the conservation of energy and charge, respectively. These laws are used to analyze and solve complex circuits.</p><h2>2. Can KLV and KCL be used with a capacitor in the circuit?</h2><p>Yes, KLV and KCL can be applied to circuits that contain capacitors. However, special considerations must be made for capacitors, such as taking into account the charging and discharging of the capacitor.</p><h2>3. How do I apply KLV and KCL to a circuit with a capacitor?</h2><p>To apply KLV and KCL to a circuit with a capacitor, you must first identify all the branches and nodes in the circuit. Then, you can use KLV to analyze the voltage drops across the branches and KCL to analyze the current at each node, taking into account the behavior of the capacitor.</p><h2>4. Are there any limitations to using KLV and KCL with a capacitor?</h2><p>While KLV and KCL can be used with capacitors, there are some limitations. For example, these laws assume ideal conditions and may not accurately reflect the behavior of real-world circuits with non-ideal components. Additionally, KVL and KCL may not be applicable to circuits with non-linear elements, such as diodes.</p><h2>5. Why is it important to use KLV and KCL with a capacitor in a circuit?</h2><p>Using KLV and KCL with a capacitor in a circuit allows for accurate analysis and understanding of the circuit's behavior. It can help in designing and troubleshooting circuits, as well as predicting the behavior of the circuit under different conditions. Additionally, KLV and KCL are fundamental laws in circuit analysis and are essential for understanding and solving complex circuits.</p>

1. What is KLV and KCL?

KLV (Kirchhoff's Voltage Law) and KCL (Kirchhoff's Current Law) are fundamental laws in circuit analysis that describe the conservation of energy and charge, respectively. These laws are used to analyze and solve complex circuits.

2. Can KLV and KCL be used with a capacitor in the circuit?

Yes, KLV and KCL can be applied to circuits that contain capacitors. However, special considerations must be made for capacitors, such as taking into account the charging and discharging of the capacitor.

3. How do I apply KLV and KCL to a circuit with a capacitor?

To apply KLV and KCL to a circuit with a capacitor, you must first identify all the branches and nodes in the circuit. Then, you can use KLV to analyze the voltage drops across the branches and KCL to analyze the current at each node, taking into account the behavior of the capacitor.

4. Are there any limitations to using KLV and KCL with a capacitor?

While KLV and KCL can be used with capacitors, there are some limitations. For example, these laws assume ideal conditions and may not accurately reflect the behavior of real-world circuits with non-ideal components. Additionally, KVL and KCL may not be applicable to circuits with non-linear elements, such as diodes.

5. Why is it important to use KLV and KCL with a capacitor in a circuit?

Using KLV and KCL with a capacitor in a circuit allows for accurate analysis and understanding of the circuit's behavior. It can help in designing and troubleshooting circuits, as well as predicting the behavior of the circuit under different conditions. Additionally, KLV and KCL are fundamental laws in circuit analysis and are essential for understanding and solving complex circuits.

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