# NCEES prob. 105 - Electronics



## schmidty99 (Mar 5, 2010)

Hey all. I'm working this problem again, and I see how they do it (sort of), but I'm hoping for some clarification. Basically, they want you to solve for a RC ckt at a given time, but the equation for the voltage is v(t)= A+Be^(-t/RC). Voltage v(t) is a charged capacitor. What do the values of A and B stand for? Is A = v(0) and B = v(infinity)? I see similar eqautions in the EERM and some other study books but not the exact one above.

Any help or clarification would be appreciated. Thanks!


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## rshankle PE (Mar 6, 2010)

schmidty99 said:


> Hey all. I'm working this problem again, and I see how they do it (sort of), but I'm hoping for some clarification. Basically, they want you to solve for a RC ckt at a given time, but the equation for the voltage is v(t)= A+Be^(-t/RC). Voltage v(t) is a charged capacitor. What do the values of A and B stand for? Is A = v(0) and B = v(infinity)? I see similar eqautions in the EERM and some other study books but not the exact one above.
> Any help or clarification would be appreciated. Thanks!


Hi,

I don't have the book in front of me (and I'm too lazy to go looking for the last time this was posted) but I believe this is the one where you have a 6 volt battery connected to 2 resistors in series, with a capacitor in parallel to the second resistor. Then the switch connecting the battery to the circuit is open at t=0, find Vc(t).

For me, I broke the problem into two circuits. One for before the switch is open (t&lt;0) and one after it is open (t=&gt;0). For the first circuit (t&lt;0) the capacitor has been there a long time and is charged up to whatever the voltage is across the resistor it is in parallel with. (I think using the voltage divider rule, the voltage across that resistor was 4 volts.)

That's all you need from the first circuit is Vc(t=0) = 4volts.

The thing to remember with capacitors is their voltage can not change instantaneously so V(t=0-) = V(t=0+).

Now for the second circuit all you have is a charged capacitor in parallel with a resistor. Look at page 30-2 of the EERM and it takes you through all the math to the final equation you are looking for Vc(t) = Vo e ^ (-t / RC), where Vo is the charge at Vc(0) or 4V.

It might help if you copied example 30.1 onto some scratch paper just so you get use to working through the math, and then you will recognize the correct equation to use next time from Table 30.1.

So to answer your question, A = 0 (since the battery is removed at t=0, no forcing function), B= Vo since it is the coefficient in front of the exponent and tau = RC.

Hope this helps.


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## schmidty99 (Mar 6, 2010)

Yup, that's the one. Ok, I'll try to work through it as you suggest. You da man! Thanks!


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## rshankle PE (Mar 6, 2010)

schmidty99 said:


> Yup, that's the one. Ok, I'll try to work through it as you suggest. You da man! Thanks!


Thanks!


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## schmidty99 (Mar 6, 2010)

Sparrow:

In looking at this again, I understand the equation and the math in the EERM, but what is A and what is B? I don't see that exact form in the book? I looked over your post again, and you say "A=0 since the battery is removed." So all that equation means is that A is the battery voltage and B (in this case) is the capacitor voltage? BJS


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## rshankle PE (Mar 7, 2010)

schmidty99 said:


> Sparrow:
> In looking at this again, I understand the equation and the math in the EERM, but what is A and what is B? I don't see that exact form in the book? I looked over your post again, and you say "A=0 since the battery is removed." So all that equation means is that A is the battery voltage and B (in this case) is the capacitor voltage? BJS


Yes B is Vo also known as the initial voltage of the capacitor for circuit 2.


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## benbo (Mar 7, 2010)

Sort of a general comment. As Sparrow was pointing out, it is usually best for these DC RC or RL charging/discharging circuits to try to understand what is happening rather than to look for an equation to plug into. Of course, if you can't figure it out exactly, plugging into something is a last resort and will give you a better chance.

I was always confused by these things, until I really sat down and started thinking about what it meant by "connected for a really long time" etc and what was happening over time. I actually had to plug in some numbers for time to get it in a couple cases. Of course, even these generalizations can be decieving. A capacitor can be connected across a battery for a really long time and never charge to the full value of the battery if it doesn't have sufficient capacity. But they don't usually throw curve balls like that at you without making it really clear.

Anyway, it may help to google around for a site that you can understand to help you out if you have trouble -

For example -

http://www.ac.wwu.edu/~vawter/PhysicsNet/T...t/RCSeries.html

In my case, for these DC transient circuits, the mathematics (DEs) was no help in understanding them conceptually. First I understood the concepts, then I looked at the mathematics behind them. YMMV.


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## rshankle PE (Mar 7, 2010)

benbo said:


> Sort of a general comment. As Sparrow was pointing out, it is usually best for these DC RC or RL charging/discharging circuits to try to understand what is happening rather than to look for an equation to plug into. Of course, if you can't figure it out exactly, plugging into something is a last resort and will give you a better chance.
> I was always confused by these things, until I really sat down and started thinking about what it meant by "connected for a really long time" etc and what was happening over time. I actually had to plug in some numbers for time to get it in a couple cases. Of course, even these generalizations can be decieving. A capacitor can be connected across a battery for a really long time and never charge to the full value of the battery if it doesn't have sufficient capacity. But they don't usually throw curve balls like that at you without making it really clear.
> 
> Anyway, it may help to google around for a site that you can understand to help you out if you have trouble -
> ...


Benbo thanks for the link.


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## schmidty99 (Mar 7, 2010)

Thanks for the help guys. I think I got it. BJS


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## benbo (Mar 8, 2010)

Sparrow said:


> Benbo thanks for the link.


No problem. I'm not sure that is even the best one. You may even be able to figure out "what is happening over time" in the circuit just from the EERM. But sometimes I just had to sit down and think about it for a while, which was sometimes unpleasant.


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## rshankle PE (Mar 9, 2010)

benbo said:


> Sparrow said:
> 
> 
> > Benbo thanks for the link.
> ...



Hey Schmitdy,

I like Benbo's idea to try to think about what's going on before grabbing for a mathematical equation.

In this example problem you know the capacitor is being charged when the battery is connected, and you know it will discharge when the battery is taken away. You also know the value to which the capacitor is charged is based upon the voltage it sees (which is the voltage across the resistor.)

My question for you would be, what would happen if we replaced the capacitor with an inductor? Don't grab for an equation, but think about how inductors work....


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## benbo (Mar 9, 2010)

This is not to say you have to re-invent the wheel on every prob on the exam. If a circuit configuration exactly matches a known commoon configuration, of course you can just plug in. But now during study time I really think it helped me to take the time to try and understand the mechanism and source of the equation, as study time permits.

A perfect example is op-amps. I found and printed out a few pages with all the typical open and closed loop op-amp circuits. Integrators, filters, differentiators, amps, comparators, etc. I brought them into thee exam just in case. But I never looked at it. I was able to figure out every op amp on the test using the "Golden Rules for Op Amps", and KCL. But I was fully prepared to use the notes if I ran out of time or I had to.


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## schmidty99 (Mar 9, 2010)

Sparrow:

I've always been pretty good with capacitors and inductors, so I think I'm OK with understanding the "how it works" part. If they wouldn't have thrown that form of the equation at me, I probably would have got it. The answer you are looking for (I think) is that a capacitor cannot discharge voltage instantly and a inductor cannot discharge current instantly...and go from there.

Benbo:

Gotchya. For the record, I try to understand every problem, but sometimes its hard to find the steps necessary. I mean, I've taken a whole night (2+ hours) to try and find an answer. But its getting better. I know that I'm going through a few problems in the NCEES review again, and they make a lot more sense now. So I think its going well. If you have a sheet with the "Golden Rules for Op-Amps," I'd love to see it. That's another thing that can get confusing awful fast.

Thanks again for all your help you guys. BJS


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## benbo (Mar 9, 2010)

schmidty99 said:


> If you have a sheet with the "Golden Rules for Op-Amps," I'd love to see it. That's another thing that can get confusing awful fast.


http://hyperphysics.phy-astr.gsu.edu/hbase...nic/opampi.html

Just a fancy way of saying you can always assume V(inverting) = V(noninverting) and there is not any current into the inputs. Then just do the node analysis.

Only works for negative feedback, but that's at least 80% of the circuits.


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## rshankle PE (Mar 9, 2010)

Schmidty,

Very good, that's what I was thinking about inductors.

Looking forward to the next problem if you have any. (I just can't answer any from Electrical Engineering Sample Examinations 3rd Edition since I'm saving that for a 8 hour timed test at the end of the month.  )

Take care


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## schmidty99 (Mar 10, 2010)

Thanks benbo!

Sparrow:

Wow, you're a glutton for punishment. There's a problem in the NCEES practice exam on linear regression. I'm trying to figure that one out, but going nowhere fast. I don't remember the problem number, but its in the first 20 of the morning section. Thanks! schmidty


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## rshankle PE (Mar 10, 2010)

schmidty99 said:


> Thanks benbo!
> Sparrow:
> 
> Wow, you're a glutton for punishment. There's a problem in the NCEES practice exam on linear regression. I'm trying to figure that one out, but going nowhere fast. I don't remember the problem number, but its in the first 20 of the morning section. Thanks! schmidty


It's problem 112 in the same book of problems.

The key here is to realize that linear regression is based upon the coefficients being linear, not the data. In the problem X1, X2, .... represents the data. I had one of our statistics guys come and explain the answer to me. (Even if X1 was logrythmic data, you could let some other variable say X4 = exp(X1) and now follows a straight line.)

The coefficients are bo, b1, b2 ... , so the correct answer is D, because b2 is in the exponent. (That's basically the extent of me understanding that problem.)

This in not my strong suit so hopefully Benbo or someone else can chime in if you have additional questions.


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## schmidty99 (Mar 10, 2010)

Yikes! I'm not sure I get it. I'll try to re-visit that one later tonight or tomorrow. In the meantime, I'll do a little more research. Thanks! Schmidty


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