Thevenin's equivalent circuit for capacitor

[Genji Shimada]

[Mod Note: Thread moved from technical forum hence no homework template.]

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

 

[Let'sthink]

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.

 

[Genji Shimada]

" 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?

 

[Let'sthink]

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.

 

[Genji Shimada]

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?

 

[gneill]

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.

 

[gneill]

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

 

[Genji Shimada]

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

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?

 

[gneill]

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.

 

[Genji Shimada]

Okay, Thanks!