# RC Circuit Revision

#### RysterHD

Joined Dec 15, 2015
8
Hello, i'm new here and am struggling with RC circuits, i am trying to do some practice questions for my exam on Thursday and am struggling with two which i do not believe i have learnt but think something along those lines will be on the exam.

I have changed the number up so that any help i do recieve i will still have to work out on my own.

V1 - 10V
R1 - 200 ohms
L1 - 400 mH

I have worked out the time constant, which i believe is 200/0.4 = 500

And now comes my problem, it says that i should work out the "Voltage across the inductor after it has been charging for 300 µs"

and then the "current flowing in the circuit after it has been discharging for 700 µs"

I know the rules state to write what i've already done, but i've gone through 5~ pieces of paper so far and got absolutely nowhere, i just need a nudge in the right direction so im more confident for my exam on thursday please, thank you

- Ryan

#### WBahn

Joined Mar 31, 2012
26,398
Hello, i'm new here and am struggling with RC circuits, i am trying to do some practice questions for my exam on Thursday and am struggling with two which i do not believe i have learnt but think something along those lines will be on the exam.

I have changed the number up so that any help i do recieve i will still have to work out on my own.

V1 - 10V
R1 - 200 ohms
L1 - 400 mH

I have worked out the time constant, which i believe is 200/0.4 = 500
Welcome to AAC.

If you stick around here long enough, you will quickly learn that I am a self-confessed units Nazi. The reason is simple -- tracking units is perhaps the single most-effective error detection tool available to the engineer. If you track your units, then most (not all) mistakes you make will mess up the units allowing you to catch them early on.

Let's see how this works in practice.

You are claiming that the time constant of the LR circuit (despite the thread title and your leading sentence which both mention RC circuits???) is

$$\tau \; = \; \frac{200}{0.4} \; = \; 500$$

500 what? Seconds? Minutes? Milliseconds? Goats?

Well, let's see what the units of your computation actually work out to be:

$$\tau \; = \; \frac{200 \; \Omega}{0.4 \; H} \; = \; 500 ??$$

Ohm's Law tells us that

$$V \; = \; I \cdot R R \; = \; \frac{V}{I}$$

So resistance has units of volts/ampere

The constitutive relation for an inductor will give us the units for the Henry

$$V \; = \; L\frac{di}{dt} L \; = \; V\frac{dt}{di}$$

So inductance has units of volt-seconds/ampere

So what would the units of resistance/inductance be?

$$\tau \; = \; \frac{200 \; \Omega}{0.4 \; H} \( \frac{\(\frac{V}{A}$$}{\Omega}\)$$\frac{H}{\(\frac{V\cdot s}{A}$$}\)\; = \; 500 s^{-1}
\)

So your result has units of recipicol time, which is not what you want and, therefore, you KNOW the answer is wrong and can stop at this point and find and fix the error before wasting a bunch of time doing a bunch of work that is only guaranteed to produce a result that is wrong.

In addition to always, always, always tracking your units, you also want to always ask if the answer makes sense. In this case, does it make sense that the time constant for components like these would be the next best thing of ten minutes? You may or may not have a good enough feel for things to be able to answer that question right now, but this is the kind of thing that you want to start developing a feel for -- what is reasonable and what is unreasonable to expect in the results.

And now comes my problem, it says that i should work out the "Voltage across the inductor after it has been charging for 300 µs"

and then the "current flowing in the circuit after it has been discharging for 700 µs"

I know the rules state to write what i've already done, but i've gone through 5~ pieces of paper so far and got absolutely nowhere, i just need a nudge in the right direction so im more confident for my exam on thursday please, thank you

- Ryan
And now consider that everything on those five pieces of paper was wasted time that could have been avoided right at the start.

Now that you have the right time constant, show us some work (doesn't have to be a lot) to show how you are trying to answer the first question. We can't tell you what you are doing wrong until you show us what you are doing.

#### dtekumse

Joined Dec 15, 2015
9
Hello, this is my first message also.

I would like to stress how important WBahn's advice is, as I am a professional lecturer at a Faculty of Electrical Engineering and I meet such errors on a daily basis. This shows ignorance of basic principles of physics. Even though engineering is quite different from physics, we should never forget it actually came out of it and it is limited by basic laws governing everything in this universe.

The advice I give is: one row of calculation, three rows of sense.

#### RysterHD

Joined Dec 15, 2015
8
Hi, thanks for the reply, im just doing electrical at college and so haven't done as advanced into it as you have, thanks for helping me out with the first one, here's some working out i've just tried:

I = 10V/0.4H = 25

It = I (1-e^-Rt/L)
Which i have worked out as
It = 25 ( 1-e^-500t)
which would mean It = 25 and this is where i got confused, because i thought it would then be
25/25 = 1-e^-500t
and then -1 to both sides, making it
-1 = -e^-500t
then i thought i would have to do the log, but i can't do the log of -1 so i'm stuck here, i know it's probably all wrong but im really struggling to grasp what's happening

Sorry if im annoying you by how little i know, i feel like i've been thrown in the deep end, thanks for wanting to help me out

#### MrAl

Joined Jun 17, 2014
8,243
Multiply both sides by -1 first.

#### RysterHD

Joined Dec 15, 2015
8
Multiply both sides by -1 first.
At which point?
Won't it be after -1 = -e^-500t to make it 1=e^-500t?

#### MrAl

Joined Jun 17, 2014
8,243
Hi,

You're welcome.

Yes, then you can take the log as you desired.

Note i did not check anything before this so correctness is up to you.

#### JoeJester

Joined Apr 26, 2005
4,390

#### WBahn

Joined Mar 31, 2012
26,398
Hi, thanks for the reply, im just doing electrical at college and so haven't done as advanced into it as you have, thanks for helping me out with the first one, here's some working out i've just tried:

I = 10V/0.4H = 25
And, once again, because you refuse to track your units, you start off on your very first line guaranteeing that everything you do is going to be wrong.

$$I \; = \; \frac{10\; V}{0.4\;H} \cdot \( \frac{H}{\(\frac{Vs}{A}$$}\) \; = \; 25 \; \frac{A}{s}
\)

Yet, because you refuse to track your units, you forfeit the opportunity to see that the above result is completely wrong and, instead, proceed to blindly waste your time and effort chasing a rabbit down a hole.

Sorry if im annoying you by how little i know, i feel like i've been thrown in the deep end, thanks for wanting to help me out
I'm more than willing to help you out, but you don't seem willing to try to learn.

#### Bill B

Joined Nov 29, 2009
61
The internet is a wonderful place full of all kinds of information. It is quite apparent that you don't understand how the circuit works. First, do some research on how the circuit works. There are many sites (including the textbook link given in this thread) that explain exactly how things work. An example: http://www.electronics-tutorials.ws/inductor/lr-circuits.html

Just knowing how to run the calculations without understanding what the circuit is doing and how it reacts given different conditions really doesn't do you any good. WBahn is giving you good advice, but even if you track your units your mistakes probably won't be apparent to you if you don't understand how the circuit works. Trust me, I found this out the hard way.

#### Bordodynov

Joined May 20, 2015
2,867
Tau=L1/R1=0.4/200=0.02 sec.
I=10v/R1*(1-exp(-Time/Tau))=10/200*(1-exp(-Time/Tau))=0.05*(1-exp(-Time*R1/L1))=0.05*(1-exp(-Time*500))

#### JoeJester

Joined Apr 26, 2005
4,390
@Bordodynov

You might want to revisit your tau calculation unless you posted exactly what you wanted to post.