Calculate RC at steady state

Thread Starter

Josh1233

Joined Aug 25, 2022
9
Hi,
at first switch open , then close . Then after t=0 switch close. I want to calculate time constant when circuit reach steady state. T=RC . R will be 6//1+12 right?

Mod:lightened image.


image.jpg
 
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WBahn

Joined Mar 31, 2012
27,389
Hi,
at first switch open , then close . Then after t=0 switch close. I want to calculate time constant when circuit reach steady state. T=RC . R will be 6//1+12 right?

Mod:lightened image.

View attachment 275594
There are a number of things that you can do to help improve your ability to have good conversations about circuits. You might think that some of these are nit picky, but I can't begin to emphasize enough the importance and value of having good attention to detail.

So, in no particular order, your diagrams show a capacitor but indicate that it is 20 microhenries, which is an inductance. So is this and RC circuit or an LR circuit.

Then, you talk about "the time constant when the circuit reach steady state". The response is an exponential, so in theory it will never reach steady state. In practice, exponential responses are generally considered to have reached their final state after five time constants. This is an arbitrary convention, but after five time constants the system has settled to within nearly half a percent of the final value and that is "good enough" for nearly all purposes. After one time constant it has only gotten a bit under 2/3 of the way to the final value. A better way to phrase this is simply, "the time constant for this circuit."

Your diagrams show that the switch is open at t=0 and closed at t=∞. That really doesn't define when the switch actually closes, since those diagrams would apply equally well to a switch that closed at , say, t=10 s. Your text description does an adequate job of making it clear that the switch closes at t=0. To make it so that the diagrams convey this say notion, show the switch open at T=0- in the first diagram and closed at t=0+ in the second (the - and + are usually written as superscripts).

Finally, get in the habit of tracking your units properly. Resistance is not a pure number -- the units are a part of nearly any physical quantity and should not be left off. So use "R = (6Ω||1Ω)+12Ω". Furthermore, you really should include the parentheses because a slightly different circuit would have "R = 6Ω||(1Ω+12Ω)" and lots of people would sloppily write "R = 6Ω||1Ω+12Ω" in either case without thinking that there is no defined (or widely accepted, even) order of operations between combining things in series and combining them in parallel. Instead, they would see the grouping they intended and assume that their readers would, too, when in reality they are forcing their readers to either become mind readers or guess -- engineering is not about either of those.

NOTE: You might notice that I didn't include units when saying t=0 or t=∞. This is because 0 ms, 0s, 0 hr, 0 years, and 0 centuries are all the same thing. The same when using infinity. The exception is when we are talking about measurements made on a relative scale, such as Celsius or Fahrenheit temperatures.
 
Last edited:

Thread Starter

Josh1233

Joined Aug 25, 2022
9
There are a number of things that you can do to help improve your ability to have good conversations about circuits. You might think that some of these are nit picky, but I can't begin to emphasize enough the importance and value of having good attention to detail.

So, in no particular order, your diagrams show a capacitor but indicate that it is 20 microhenries, which is an inductance. So is this and RC circuit or an LR circuit.

Then, you talk about "the time constant when the circuit reach steady state". The response is an exponential, so in theory it will never reach steady state. In practice, exponential responses are generally considered to have reached their final state after five time constants. This is an arbitrary convention, but after five time constants the system has settled to within nearly half a percent of the final value and that is "good enough" for nearly all purposes. After one time constant it has only gotten a bit under 2/3 of the way to the final value. A better way to phrase this is simply, "the time constant for this circuit."

Your diagrams show that the switch is open at t=0 and closed at t=∞. That really doesn't define when the switch actually closes, since those diagrams would apply equally well to a switch that closed at , say, t=10 s. Your text description does an adequate job of making it clear that the switch closes at t=0. To make it so that the diagrams convey this say notion, show the switch open at T=0- in the first diagram and closed at t=0+ in the second (the - and + are usually written as superscripts).

Finally, get in the habit of tracking your units properly. Resistance is not a pure number -- the units are a part of nearly any physical quantity and should not be left off. So use "R = (6Ω||1Ω)+12Ω". Furthermore, you really should include the parentheses because a slightly different circuit would have "R = 6Ω||(1Ω+12Ω)" and lots of people would sloppily write "R = 6Ω||1Ω+12Ω" in either case without thinking that there is no defined (or widely accepted, even) order of operations between combining things in series and combining them in parallel. Instead, they would see the grouping they intended and assume that their readers would, too, when in reality they are forcing their readers to either become mind readers or guess -- engineering is not about either of those.

NOTE: You might notice that I didn't include units when saying t=0 or t=∞. This is because 0 ms, 0s, 0 hr, 0 years, and 0 centuries are all the same thing. The same when using infinity. The exception is when we are talking about measurements made on a relative scale, such as Celsius or Fahrenheit temperatures.
Thank you . I will fix my error
 

MrChips

Joined Oct 2, 2009
27,112
At t = 0, what is the voltage across the capacitor?
What is the current through the 12Ω resistor?

At steady state, what is the voltage across the capacitor?
What is the current through the 12Ω resistor?
 

dcbingaman

Joined Jun 30, 2021
720
Please correct the circuit and repost your question, as the question is not logical. The Time Constant of the circuit has nothing to do with steady state. Are you interested in the time constant or what the steady state DC values of the circuit are? Are we dealing with an Inductor or a Capacitor, we have the symbol for a capacitor and have the units of inductance?
 

MrAl

Joined Jun 17, 2014
9,533
Hi,
at first switch open , then close . Then after t=0 switch close. I want to calculate time constant when circuit reach steady state. T=RC . R will be 6//1+12 right?

Mod:lightened image.

View attachment 275594

Hello there,

I have to say these two circuits are a little confusing to me too because of a couple reasons.
The first, already somebody addressed, is the 'capacitor' is labeled as having a value of "20uH" which is not the right units of a capacitor. A capacitor is specified in Farads or microfarads or similar, and in some circuits as 'imaginary' Ohms, but never in Henries or similar. So that is the first thing to clear up.

Then there is the question of which circuit you want to find the time constant for because there are actually two time constants here, one for the top drawing and one for the bottom drawing, and they will be slightly different.

Just to maybe help a little, if that really is a capacitor and the switch is closed, then your calculation of the resistance to be used in the time constant looks right with 6 Ohms in parallel with 1 Ohm, then that result in series with 12 Ohms. Once you calculate that you can then use it to calculate the time constant T=R*C.
With the switch open, the time constant is slightly different and i bet you can calculate that too.

It is also not too clear why you have drawn two circuits, and why you mention t=inf. From the drawing it looks like you want to calculate the response after the switch is closed, at t approaching infinity, because as t approaches infinity the exponential part of the response dies to zero, something that is nearly zero long before that but not exactly zero until t=inf.
So maybe you can explain why you have the two circuits drawn and why you included that note about t=inf.
At t=inf that would take an infinite number of time constants to elapse, so i dont think it would help to understand the time constant very much.

So if you can clear a few things up i am sure you will get some good help here.
 
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