Thevenin's equivalent circuit for capacitor

In summary, the time constant of a circuit is Requivalent x C. Through calculations using Thevenin's equivalent circuit method, it has been determined that adding a resistor in parallel to the capacitor will decrease the time constant. However, this seems counterintuitive as increasing the resistance would typically increase the time constant. The explanation lies in the fact that the voltage divider in the circuit prevents the capacitor from charging to the source voltage, thus shortening the charging time and decreasing the time constant.
  • #1
Genji Shimada
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[Mod Note: Thread moved from technical forum hence no homework template.]

media%2F5cb%2F5cb51d17-b8f6-4316-850c-b59337c2474f%2Fphp40TRzE.png

Okay so the image i uploaded only because of the cirscuit never mind the text to it. I don't understand things mathematicaly verry well. I like to have a good physical grasp on what is going on with the circuit. That being said, i don't mean to insult any mathematicians, but for this one i need a good physical explanation on why is it that way.
So the time constant of this circuit is Requivalent x C. Through calculations using Thevenin's equivalent circuit method i have arrived at this outcome - If you add a resistor in parallel to the capacitor will decrease the Time constant. And now i ask Why and some maybe will say " Because with resistor in parallel the equivalent resistance is now less than it was before with only one resistor'' but that is not an answer atleast for me.
The way i think is that the current through the capacitor deposits the charge on it so if the current through it decreases it should take longer to charge up to a specific voltage. This is why increasing the resistance increases the time constant. However, from that Thevenin equivalent circuit method it doesn't seem that way. If you have two resistors in series and you add another branch in parallel with the bottom resistor, you would expect that now the current through the bottom resistor will decrease abit, maybe in half if the resistance of the new branch is equal. And that's pretty much what would happen in a circuit such as the one above in the picture after a time when the capacitor no longer behaves like a short circuit. So what I am saying is because of the 1k Ohm resistor the current through the capacitor should be less than if there was no 1kOhm therefore it would take more time for the capacitor to charge, but again according to the Thevenin equivalent circuit method that's not the way and it is infact the opposite. So what is happening, why adding resistors in parallel with the capacitor decreases the Time constant?

Thanks in advance
 

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  • #2
Genji Shimada said:
View attachment 219262
Okay so the image i uploaded only because of the cirscuit never mind the text to it. I don't understand things mathematicaly verry well. I like to have a good physical grasp on what is going on with the circuit. That being said, i don't mean to insult any mathematicians, but for this one i need a good physical explanation on why is it that way.
So the time constant of this circuit is Requivalent x C. Through calculations using Thevenin's equivalent circuit method i have arrived at this outcome - If you add a resistor in parallel to the capacitor will decrease the Time constant. And now i ask Why and some maybe will say " Because with resistor in parallel the equivalent resistance is now less than it was before with only one resistor'' but that is not an answer atleast for me.
The way i think is that the current through the capacitor deposits the charge on it so if the current through it decreases it should take longer to charge up to a specific voltage. This is why increasing the resistance increases the time constant. However, from that Thevenin equivalent circuit method it doesn't seem that way. If you have two resistors in series and you add another branch in parallel with the bottom resistor, you would expect that now the current through the bottom resistor will decrease abit, maybe in half if the resistance of the new branch is equal. And that's pretty much what would happen in a circuit such as the one above in the picture after a time when the capacitor no longer behaves like a short circuit. So what I am saying is because of the 1k Ohm resistor the current through the capacitor should be less than if there was no 1kOhm therefore it would take more time for the capacitor to charge, but again according to the Thevenin equivalent circuit method that's not the way and it is infact the opposite. So what is happening, why adding resistors in parallel with the capacitor decreases the Time constant?

Thanks in advance
I like your questioning style and emphasis on concepts. But in place of thinking what will happen if you please connect 1 K resistance afterwards just think what happens if you connect the capacitor afterwards after putting the switch on first. I think you will get the answer of your question.
 
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  • #3
" just think what happens if you connect the capacitor afterwards after putting the switch on first. I think you will get the answer of your question. "
Hmm all I can think of is that as you said if the voltage divider already existed and I connect the capacitor second, the first moment that I close the switch of the capacitor the 1k bottom resistor will be shortened by the capacitor and then as the capacitor gains voltage the potential difference across 1k resistor rises and current through it increases. Eventually as the capacitor reaches the voltage divider output voltage level, the voltage divider won't allow the voltage at the output to rise any further and so with the output and the capacitor at the same voltage there will be no more potential difference and therefore no reason for current to flow and charge the capacitor any further. As I think of it now it's kind of like the voltage divider is preventing the capacitor to charge to the source voltage and in a way is shortening the charging time. But is that why the time constant decreases?
 
  • #4
Genji Shimada said:
Hmm all I can think of is that as you said if the voltage divider already existed and I connect the capacitor second, the first moment that I close the switch of the capacitor the 1k bottom resistor will be shortened by the capacitor and then as the capacitor gains voltage the potential difference across 1k resistor rises and current through it increases. Eventually as the capacitor reaches the voltage divider output voltage level, the voltage divider won't allow the voltage at the output to rise any further and so with the output and the capacitor at the same voltage there will be no more potential difference and therefore no reason for current to flow and charge the capacitor any further. As I think of it now it's kind of like the voltage divider is preventing the capacitor to charge to the source voltage and in a way is shortening the charging time. But is that why the time constant decreases?
You will have to prove it theoretically I think. Conceptually the charging current passes through both the resistors connected in parallel, taking the source resistance is zero. It will not be proper to say that voltage is less so it takes less time because time constant is independent of the maximum voltage to which capacitor is charged.
 
  • #5
Okay then tell me the answer. This is not a homework I am not a student, I am just a man in his 20s who looks to learn electronics. So I need your help, tell me why is the time constant smaller with a voltage divider?
 
  • #6
Genji Shimada said:
Okay then tell me the answer. This is not a homework I am not a student, I am just a man in his 20s who looks to learn electronics. So I need your help, tell me why is the time constant smaller with a voltage divider?
Sorry, by forum rules we can't do that. Members are not allowed to provide solutions to other's problems posted in our homework forums. We can only provide hints, ask pertinent questions to guide you, spot errors in your work, etc.
 
  • #7
Have you studied Thevenin (and Norton) equivalents already? There's a fundamental property of these "equivalent" circuits that should lead directly to the answer you seek :wink:
 
  • #8
gneill said:
Have you studied Thevenin (and Norton) equivalents already? There's a fundamental property of these "equivalent" circuits that should lead directly to the answer you seek :wink:
Yes, I know a bit about Thevenin equivalent circuits. But then again, if you are not allowed to tell me the answer directly, can you at least tell me what is that "Fundamental property" I should go look for?
 
  • #9
Genji Shimada said:
Yes, I know a bit about Thevenin equivalent circuits. But then again, if you are not allowed to tell me the answer directly, can you at least tell me what is that "Fundamental property" I should go look for?
Sure, you should review what the "equivalent" actually implies in the term "Thevenin equivalent circuit".

In your text or course notes, where they introduce Thevenin and Norton equivalent circuits they should spell out what this equivalency means and implies.
 
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  • #10
Okay, Thanks!
 

Related to Thevenin's equivalent circuit for capacitor

1. What is Thevenin's equivalent circuit for capacitor?

Thevenin's equivalent circuit for capacitor is a simplified representation of a complex circuit that contains a capacitor. It consists of a voltage source in series with a single capacitor, with the value of the voltage source and capacitor chosen to produce the same behavior as the original circuit.

2. Why is Thevenin's equivalent circuit useful for capacitors?

Thevenin's equivalent circuit is useful for capacitors because it allows us to simplify complex circuits and analyze their behavior with just one capacitor and one voltage source. This makes calculations and analysis easier and more efficient.

3. How is Thevenin's equivalent circuit for capacitor calculated?

Thevenin's equivalent circuit for capacitor is calculated by first removing the capacitor from the original circuit and finding the open-circuit voltage across its terminals. This voltage becomes the value of the voltage source in the equivalent circuit. Next, the capacitor is replaced and the short-circuit current through it is found. This current is divided by the frequency of the circuit to determine the value of the equivalent capacitor.

4. Can Thevenin's equivalent circuit be used for circuits with multiple capacitors?

Yes, Thevenin's equivalent circuit can be used for circuits with multiple capacitors. Each capacitor can be treated separately and its equivalent circuit can be calculated using the same process as for a single capacitor. The overall equivalent circuit is then the combination of all the individual equivalent circuits for each capacitor.

5. What are the limitations of Thevenin's equivalent circuit for capacitor?

Thevenin's equivalent circuit for capacitor is only an approximation of the behavior of a complex circuit with a capacitor. It assumes that the behavior of the capacitor is linear and does not take into account non-ideal effects such as leakage current or temperature dependence. It is also only valid for circuits with sinusoidal inputs.

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