- #1

Genji Shimada

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[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

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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