# L-R circuit and time constants.

#### Brandons95

Joined Dec 16, 2015
6
Okay so Im quite new to this. I've been given this question as part of my HNC electrical engineering course. I'm really stuck and all I know is that for part a I need to start by dividing 160 by 5. Help would be much appreciated.

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#### kyka

Joined Jun 7, 2015
24
Apply KCL. Solve the differential equation. Define what do you consider as approximately and you're good to go.

#### MrAl

Joined Jun 17, 2014
7,965
Hi,

You can usually use a value of 5 time constants for approximate calculations. So it takes 5 time constants to reach roughly the max value.
Write the equation for this and figure out the resistance. Then halve it and see how long it takes to get to the maximum value with the new resistor value in the time domain equation. You could also think about the new max current value when the new value resistor is used.

#### Brandons95

Joined Dec 16, 2015
6
Okay so the equation I've been working with is t = L/R but I'm told I need to re arrange it. Is there any chance someone could give me an example? I missed the lesson on this and all my tutor is pretty much saying is I need to re arrange the equation. Also dividing 160 by 5 what does that give me like I know it's 32 but what units.

#### WBahn

Joined Mar 31, 2012
26,398
Okay so the equation I've been working with is t = L/R but I'm told I need to re arrange it. Is there any chance someone could give me an example? I missed the lesson on this and all my tutor is pretty much saying is I need to re arrange the equation. Also dividing 160 by 5 what does that give me like I know it's 32 but what units.
Well, if you track your units they will tell you what units.

The 160 has units of milliseconds.
The 5 is just a number (it is how many time constants is generally considered sufficient to reach the final value for most practical purposes).

So you have (160 ms)/(5). What are the units of the result?

#### Brandons95

Joined Dec 16, 2015
6
Well, if you track your units they will tell you what units.

The 160 has units of milliseconds.
The 5 is just a number (it is how many time constants is generally considered sufficient to reach the final value for most practical purposes).

So you have (160 ms)/(5). What are the units of the result?
Well, if you track your units they will tell you what units.

The 160 has units of milliseconds.
The 5 is just a number (it is how many time constants is generally considered sufficient to reach the final value for most practical purposes).

So you have (160 ms)/(5). What are the units of the result?
Still milliseconds?

#### Brandons95

Joined Dec 16, 2015
6
So if I've got tau (32) and 470mH I can make R the subject and do 32 x 470 then I have my resistance? Which would equal to 15,040 milliohms or 15.04 ohms?

#### WBahn

Joined Mar 31, 2012
26,398
Still milliseconds?
Yep.

You treat units just like any other factor. A physical quantity (usually) has two parts that are "multiplied" together to yield a physically meaningful value. The numeric part and the unit part.

It is meaningless to say that I am 72 tall or that I am 6 tall -- and certainly 72 = 6 is incorrect. But both of those are just numbers, not a length measurement. However, 72 inches is a length measure, as is 6 feet, and indeed

72 inches = 6 feet

is a perfectly correct equation.

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#### WBahn

Joined Mar 31, 2012
26,398
So if I've got tau (32) and 470mH I can make R the subject and do 32 x 470 then I have my resistance? Which would equal to 15,040 milliohms or 15.04 ohms?
32 is not a time constant, it is a number. You need to multiply it by the proper units, namely milliseconds.

What are the units of your "32 x 470"? Let's find out.

τ = 32 ms
L = 470 uH

By multiplying them together to get a resistance, you are claiming that

R = τ·L

$$R \; = \; \tau \; L \; = \; \(32 \; ms$$(470 \; \mu H\) $$\frac{\(1 \; \frac{V \cdot s}{A}$$ }{1 \; H}\) $$\frac{1 \; \Omega}{\(1 \; \frac{V}{A}$$}\) \; = \; 15040 \; n\Omega \cdot s^2 \; = \; 15.04 \; \mu \Omega \cdot s^2
\)

The units work out to be micro-ohm · seconds-squared.

Are those units of resistance?

No. So you KNOW that this answer is WRONG.

Now, you will probably contend that you don't have time to "waste" on all this units tracking. But consider that, by not tracking your units, you have guaranteed that anything you do beyond this point is wrong and just wasted effort.

Most mistakes you make will mess up the units and let you catch the mistake almost immediately. But that is only true IF you properly track the units so that they are there to get messed up by your mistakes (and you WILL make mistakes -- we all do). Tracking units is very probably the single most effective error-detectioin tool available to the engineer. Instead, what you did was through a bunch of numbers at a calculator and then tacked on the units that you WANTED the result to have.

#### Brandons95

Joined Dec 16, 2015
6
Ugh I thought I was getting somewhere there. This will teach me not to miss lessons.

#### WBahn

Joined Mar 31, 2012
26,398
Ugh I thought I was getting somewhere there. This will teach me not to miss lessons.
Considering that you (or someone on your behalf) is spending good money for you to be able to attend those lessons, hopefully you will do so whenever possible.

Having said that, it is a sad reality that most texts and professors of engineering fail miserably to give the issue of the proper use of units the attention they deserve. Personally, I think this is unforgivable given the extreme value they have in detecting errors and the extreme consequences that uncaught errors can (and have) had. But most text authors and college professors have little to no "real world" experience and only see units as a formality that deserves a bit of lip-service now and then.

#### Papabravo

Joined Feb 24, 2006
15,068
Considering that you (or someone on your behalf) is spending good money for you to be able to attend those lessons, hopefully you will do so whenever possible.

Having said that, it is a sad reality that most texts and professors of engineering fail miserably to give the issue of the proper use of units the attention they deserve. Personally, I think this is unforgivable given the extreme value they have in detecting errors and the extreme consequences that uncaught errors can (and have) had. But most text authors and college professors have little to no "real world" experience and only see units as a formality that deserves a bit of lip-service now and then.
They also fail to convey the importance of time management skills which, it seems, are completely lacking in the student population. Waiting until the last minute to do a homework set or a research paper is an all to common occurrence. If you know at the outset that you have a problem, at least you have time to do something about it. Wait till the last minute and you are out of options.

#### WBahn

Joined Mar 31, 2012
26,398
They also fail to convey the importance of time management skills which, it seems, are completely lacking in the student population. Waiting until the last minute to do a homework set or a research paper is an all to common occurrence. If you know at the outset that you have a problem, at least you have time to do something about it. Wait till the last minute and you are out of options.
True, but I certainly can't claim that I was guilty of it before, during, and after my college days and right up to today. But while I can relate and sympathize with excessive procrastination, that doesn't mean I should absolve it's practitioners from the consequences of their choices -- quite the opposite.

And in all fairness to faculty, many of us try very hard to stress the importance of not waiting until the last minute. I conduct weekly Help Sessions and I set homework due dates such that you have to be thinking about the homework several days before it's due in order to ask questions about it during the help session. The end result, sadly predictable, is that people don't come to the help sessions because they aren't prepared to do so because they haven't started working on it. I have an occasional assignments that is complex enough that it is particularly important that they not put it off until the last minute and I will emphasize that in class every single day and yet a significant portion of the class will not have even looked at it the class before it is due. There's only so much that can be done.

The frustrating part for me, personally, is that when I returned to college after getting out of the service I was NOT a procrastinator -- I read and studied assigned chapters before class, I started work on assigned homework the day it was assigned (and usually finished it that same day) and stayed up as late as it took on Friday to have ALL of my assignments completely done. The result was remarkable -- I had the entire weekend free to do whatever I wanted to do (and without a moment's guilt over what I should be doing), I didn't study for any finals except one (diffy-Q), and I walked to an easy 4.0 despite taking 23 semester hours. It was, by far, the most enjoyable semester I had in my entire college career -- with the following semester running a close second.

But even knowing this, I still find myself putting things off until the last minute. Something that I hope to train out of my daughter!

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#### JoeJester

Joined Apr 26, 2005
4,390
WBahn,

Your story should be told .... Completely with the results you obtained, and have your successful students do the same. Maybe it might sink in a little. I know its an uphill battle.

I run across a similar situation twice a year on another forum when people compete in the service wide competition. Also see it annually with people who haven't got their CEUs completed.

Now that the last of the training is published, I can finish my stuff up that is due prior to the 29th.

#### WBahn

Joined Mar 31, 2012
26,398
WBahn,

Your story should be told .... Completely with the results you obtained, and have your successful students do the same. Maybe it might sink in a little. I know its an uphill battle.
I have. I usually tell it in every class (not always, but usually) every semester. Doesn't seem to make a bit of difference (there's been a very small handful of exceptions over the years).

#### JoeJester

Joined Apr 26, 2005
4,390
I agree. I've written and talked for over a decade on the competition concerning successful preparation.

Apparently it does work for few. I told one woman it was time for her to put her big girl panties on and take care of business. I also said that was my only talk she was getting on that subject of completing her CEUs.

#### MrAl

Joined Jun 17, 2014
7,965
Okay so the equation I've been working with is t = L/R but I'm told I need to re arrange it. Is there any chance someone could give me an example? I missed the lesson on this and all my tutor is pretty much saying is I need to re arrange the equation. Also dividing 160 by 5 what does that give me like I know it's 32 but what units.
So if I've got tau (32) and 470mH I can make R the subject and do 32 x 470 then I have my resistance? Which would equal to 15,040 milliohms or 15.04 ohms?
Hi,

You got close but the algebraic manipulation was not exactly done right...that was the main error. Since you missed the lesson i'll provide some more information here. The only assumption is that 5 time constants provides enough time to reach a close enough value to the max because that gets the approximation error down to less than 1 percent and it's a whole number. This assumption could vary however depending on what your professor considers adequate for the course.

You started with 160ms and you reasoned that if that was 5 time constants then one time constant would be 32ms, and that was correct. But now we have the algebra with the formula:
t=L/R {t here is the time constant in units of seconds}

You need the value for R, so you need to first solve for R. Multiply both sides by R and we get:
t*R=L

divide by t and we get:
R=L/t

Now we have an equation for directly calculating R.

Checking the units as Wbahn suggests, we get:
Ohms=(V*s/A)/s=V/A=Ohms

so the units check out ok.

You would next use R=L/t to find the right value of the resistance. Surprisingly, it comes out close to what you got before which shows how you can get almost the right answer by doing it wrong sometimes.

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#### Sinus23

Joined Sep 7, 2013
246
The frustrating part for me, personally, is that when I returned to college after getting out of the service I was NOT a procrastinator -- I read and studied assigned chapters before class, I started work on assigned homework the day it was assigned (and usually finished it that same day) and stayed up as late as it took on Friday to have ALL of my assignments completely done. The result was remarkable -- I had the entire weekend free to do whatever I wanted to do (and without a moment's guilt over what I should be doing), I didn't study for any finals except one (diffy-Q), and I walked to an easy 4.0 despite taking 23 semester hours. It was, by far, the most enjoyable semester I had in my entire college career -- with the following semester running a close second.

But even knowing this, I still find myself putting things off until the last minute. Something that I hope to train out of my daughter!
I'm have a similar story with the exception of even though I've attended all the classes, read all the notes and done all the problems...I still cram like there is no tomorrow for a test. I can't get a goodnight sleep until all the tests are over...

but that's just me

#### Brandons95

Joined Dec 16, 2015
6
Hi,

You got close but the algebraic manipulation was not exactly done right...that was the main error. Since you missed the lesson i'll provide some more information here. The only assumption is that 5 time constants provides enough time to reach a close enough value to the max because that gets the approximation error down to less than 1 percent and it's a whole number. This assumption could vary however depending on what your professor considers adequate for the course.

You started with 160ms and you reasoned that if that was 5 time constants then one time constant would be 32ms, and that was correct. But now we have the algebra with the formula:
t=L/R {t here is the time constant in units of seconds}

You need the value for R, so you need to first solve for R. Multiply both sides by R and we get:
t*R=L

divide by t and we get:
R=L/t

Now we have an equation for directly calculating R.

Checking the units as Wbahn suggests, we get:
Ohms=(V*s/A)/s=V/A=Ohms

so the units check out ok.

You would next use R=L/t to find the right value of the resistance. Surprisingly, it comes out close to what you got before which shows how you can get almost the right answer by doing it wrong sometimes.
Hello there, thank you for clearing this up for me. We don't really work on re arranging equations much at college to be fair so I'm pretty poor at it if I'm honest.