Time constant in a circuit with many resistors. Please help me!

Thread Starter

subatomic particle

Joined May 8, 2018
35
Hello guys,
i just need to understand what R is in the equation of the time constant
Tao= RC
R=? In the example i took at school there was just one Resistor before the capacitor, so i made an example and i was wondering what is R in T=RC
in my example.
(Example is in the picture)

Thank you :)
 

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ronsimpson

Joined Oct 7, 2019
680
R2 is parallel with R3. Combine them into a single resistor. You should know how to do parallel resistors.
Now R1 adds to R2//R3. That is a simple add.
Again make Rs and Rc into one resistor.
It is not easy to see but Ry can add to the combination R total = Ry+(R1+(R2//R3)) Simple add.
Now redraw with only two resistors. One in parallel with C and one in series with C.
Now can you finish?
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
R2 is parallel with R3. Combine them into a single resistor. You should know how to do parallel resistors.
Now R1 adds to R2//R3. That is a simple add.
Again make Rs and Rc into one resistor.
It is not easy to see but Ry can add to the combination R total = Ry+(R1+(R2//R3)) Simple add.
Now redraw with only two resistors. One in parallel with C and one in series with C.
Now can you finish?
Thank you so much for your reply
i actually can do everything you said,this is not the problem, my problem is dont know which R should i take for T=RC
Do i take only the resistors before the capacitor? or do i take all the resistors?
In my opinion i shouldnt take the resistor that is parallel to C into consideration is that right? or should i also take it? My question here is for both cases (Charging the capacitor and discharging it
I hope you can help me again :D
 

WBahn

Joined Mar 31, 2012
25,760
What you want to do is look at the circuit from the perspective of the capacitor. So imagine turning off the source (setting it's output to 0 V, which makes it a short circuit) and then measuring the resistance that would be seen by removing the capacitor and putting the leads of an ohm meter in its place.
 

WBahn

Joined Mar 31, 2012
25,760
Thank you so much for your reply
i actually can do everything you said,this is not the problem, my problem is dont know which R should i take for T=RC
Do i take only the resistors before the capacitor? or do i take all the resistors?
In my opinion i shouldnt take the resistor that is parallel to C into consideration is that right? or should i also take it? My question here is for both cases (Charging the capacitor and discharging it
I hope you can help me again :D
You need to take everything into account. Also, getting the time constant isn't enough (unless that's specifically all you are tasked with determining) because the capacitor doesn't charge to the full source potential.

What you are really needing to do is to replace the rest of the circuit as seen by the capacitor with its Thevenin equivalent. If you haven't encountered that concept, now would be a good time to start exploring it as this kind of problem is usually one of the first problems given when studying Thevenin's (and Norton's) theorem.
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
What you want to do is look at the circuit from the perspective of the capacitor. So imagine turning off the source (setting it's output to 0 V, which makes it a short circuit) and then measuring the resistance that would be seen by removing the capacitor and putting the leads of an ohm meter in its place.
Hallo,
So its like finding out Rth (R Thévenin ) between the two ends of the capacitor?
In this case all the resistors on the left are parallel to all resistors on the right
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
You need to take everything into account. Also, getting the time constant isn't enough (unless that's specifically all you are tasked with determining) because the capacitor doesn't charge to the full source potential.

What you are really needing to do is to replace the rest of the circuit as seen by the capacitor with its Thevenin equivalent. If you haven't encountered that concept, now would be a good time to start exploring it as this kind of problem is usually one of the first problems given when studying Thevenin's (and Norton's) theorem.
Thank you so much! Now i understood.
Of course i know the Thevenin concept :)

Does it apply for both cases (charging and discharging)?
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
What's your thoughts on why it either would or would not apply for both cases?
I thought in charging case since i have the source then the voltage source should be my reference, because it is charging the capacitor and the voltage of it is being affected by the resistors, but now i undestood why it is false ;)
 

WBahn

Joined Mar 31, 2012
25,760
It depends on how the charging/discharging is performed. The circuit as shown is a purely DC circuit and so there is no charging or discharging. What was the circuit before the charging started? Was it the same except that the DC source was set to zero? Was there a switch that was either opened or closed? The same for discharging? What change is made to the circuit that starts the discharge? That is what will determine whether the time constant for charging is the same as the one for discharging, and that depends on details that haven't been provided.
 

ronsimpson

Joined Oct 7, 2019
680
The voltage source is considered a zero ohm source. (If you build a 10V source it will deliver 10V with out any resistance)

Lets say Vo=10Volts and has been at 10V for a long time and C has charged up to this 10V.
Now change Vo=0V. Now the capacitor will discharge down to 0V over time. Look at the current paths. Current flows through Rc, Rs, R2&R1, R3&R1 and through Ry and it will flow through the "0 ohm" in the voltage source.
1590695607522.png
I talked about resistance in post #2. Now we can talk current. It is easy to see that the current in Rs and current in Rc comes from the cap. (or into the cap) We can see that R3,R2 share the current. Current flows through R3//R2 & R1 & voltage source & Ry.
It does not matter if you find the resistor(s) or find the current flow(s) at the end the capacitor gets charged or discharged at a certain rate.
 

WBahn

Joined Mar 31, 2012
25,760
But you are assuming that the discharging is done by changing the output of Vs from 10 V to 0 V. That's only one possible way to effect the discharging. Another very common way is to simply disconnect Vs, either by removing it entirely or by opening a switch in series with it. The time constant for these two scenarios are different and there is nothing in the original problem to give preference to one over the other.
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
It depends on how the charging/discharging is performed. The circuit as shown is a purely DC circuit and so there is no charging or discharging. What was the circuit before the charging started? Was it the same except that the DC source was set to zero? Was there a switch that was either opened or closed? The same for discharging? What change is made to the circuit that starts the discharge? That is what will determine whether the time constant for charging is the same as the one for discharging, and that depends on details that haven't been provided.
I meant just by making the voltage source = 0 Volte like to short the circuit on the left where the source is( replace the source with just a wire)
I think in this case R will not change. I calculated it in the same way you told me.
 

Thread Starter

subatomic particle

Joined May 8, 2018
35
But you are assuming that the discharging is done by changing the output of Vs from 10 V to 0 V. That's only one possible way to effect the discharging. Another very common way is to simply disconnect Vs, either by removing it entirely or by opening a switch in series with it. The time constant for these two scenarios are different and there is nothing in the original problem to give preference to one over the other.
I know what you mean. If there was a switch the scenario would be different because R would change
But what i understood is: No matter what i change in the circuit if i use the Thevenin equivalent (look at the circuit from the perspective of the capacitor ) i will always get a right answer for R. Is that right?
 

ronsimpson

Joined Oct 7, 2019
680
But you are assuming
You are over thinking this. This is most likely a home work problem. It is simply all resistors "added up" .

In theory a voltage source does not care what voltage it is outputting. 10V or 0V it will drive that line to a voltage. It does not care if the current is going in or out. Does not care if positive or negative. It will hold that node at the voltage even at 1,000,000 amps. It looks like zero ohms to 10V or zero ohms to 0V.
I meant just by making the voltage source = 0 Volte like to short the circuit.
Yes it a short to some voltage. A short of 0 volts if you want.
 

WBahn

Joined Mar 31, 2012
25,760
I meant just by making the voltage source = 0 Volte like to short the circuit on the left where the source is( replace the source with just a wire)
I think in this case R will not change. I calculated it in the same way you told me.
Yes. Under that stipulation then the time constant will be the same. But hopefully you can now better appreciate how important it is to fully specify things because there are frequently more than one reasonable way to do something and sometimes the details make a big difference. Know what does, and does not, need specifying takes experience.
 

MrAl

Joined Jun 17, 2014
7,592
Hi,

It's a little interesting that the voltage source has zero impedance always so the Thevenin Rth is zero always if the voltage source is shorted out or not. Since there are actual resistors in series with it, the only resistance then are those that are in series with it. So the Rth does not change.
This analysis is of course valid in the static case.

If the voltage source ever goes open circuit, then that means none of those series resistors are in the circuit anymore so the Rth would have to change probably very drastically.

Another side problem would be to think about what happens if the voltage source goes from max to zero as a 50 percent duty cycle. That means it could be 10v for 50 percent of the time and 0v for the other 50 percent.
Then try to figure out the Rth for the average capacitor voltage.
A second problem would be to make the voltage source go from 10v to open circuit then back again at a 50 percent duty cycle and figure out the Rth again for the average cap value.
 
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