Thevenin's Theorem Question - Page 1
Forum section: electronics / beginners

author: tec5c
author_rank: Frequent Contributor
timestamp: April 16, 2012, 09:42:06 am
content: Hi there,First I understand this may be a very basic question though I just want to understand entirely so I'm not overlooking anything with this topic.The subject I am studying at the moment is called "Circuit Analysis" and of course, the Thevenin method falls under this module.Below is the circuit I'm dealing with..The question asks "..sketch the Thevenin equivalent circuit at the terminals AB."I get the feeling I'm trying to make life harder for myself than I need to, as all of the examples I have done while studying have all been done with terminals across a component, in which my notes say to remove the component and find ETH using an already understood method. ie Mesh, Node...So to find RTH, is it simply a matter of adding the two resistors as they're in series, so RTH= 50 Ohms? And ETH is then just plainly 20v?I'll also add that I am currently doing my studies by correspondence and I don't have much help to turn to when I get stuck, hence this post.Any help is greatly appreciated.Thanks.

author: cybergibbons
author_rank: Frequent Contributor
timestamp: April 16, 2012, 10:21:47 am
content: Hi,For the VT, take open circuit voltage, this is just a voltage divider, so 10/(40+10), 5V.For the RT, set voltage sources to 0. It's a 10 and 40 Ohm resistor in parallel - 10*40/(10+40) = 8 Ohms.To add, it could be the layout of the circuit is throwing you. Just set it up as a normal voltage divider and it might be clearer?

author: ejeffrey
author_rank: Super Contributor
timestamp: April 16, 2012, 10:27:23 am
content: cybergibbons is correct except that he did the voltage divider wrong. The Vth is 16 volts and Rth is 8 ohm. The tricky part is just learning to see R1 and R2 in parallelrelative to AB.

author: cybergibbons
author_rank: Frequent Contributor
timestamp: April 16, 2012, 10:42:53 am
content: Whoops, indeed. Sorry. Couldn't see the picture and reply box on my phone and got them the wrong way round.

author: tec5c
author_rank: Frequent Contributor
timestamp: April 16, 2012, 11:20:19 am
content: Thanks for the replies, I have been a little stressed with trying to complete this course from home (full-time job, mixed with a full-time course from home will do that! Be forewarned!!) but it is quite refreshing to know that there's useful help out there.Quote from: cybergibbons on April 16, 2012, 10:21:47 amYou're quite right, by placing the resistors vertically it was definitely a "D'OH!" moment.Thanks again, and I'm sure I'll be back with more questions if that's alright with the crowd!

author: tec5c
author_rank: Frequent Contributor
timestamp: April 17, 2012, 01:03:47 am
content: Back already!There's a second part to this question and much like the first, I'm not certain of the way about solving this one.It involves the same circuit..Though the question asks to 'Determine the value of resistance which will draw 0.5A when connected to AB.'I'm sure there's a right way to go about this besides just plugging in different values of resistors until you get the 500mA flowing through it, though I have not done any questions like this one and I want to be sure I know the correct way to do it.Could anyone be of any assistance to lead me in the right direction please?Thanking you.

author: Andy
author_rank: Contributor
timestamp: April 17, 2012, 01:39:00 am
content: In the thevenin equivalent circuit, you have a voltage source, Vth, in series with Rth, and your unknown resistor between A and B, Rab. Now using ohms law:Vth = I*(Rth+Rab)Now substitute in Vth = 16, Rth = 8, I = 0.516 = 0.5*(8+Rab)32 = 8+RabRab = 24 ohms

author: IanB
author_rank: Super Contributor
timestamp: April 17, 2012, 03:38:41 am
content: Another way to solve this problem:If the current between A and B is 0.5 A, then,Vab = Vth - Iab Rth = 16 - 0.5 * 8 = 12 VTherefore,Rab = Vab / Iab = 12 / 0.5 = 24 ohmsAll solutions just involve the application of Ohm's law to the known circuit elements.

author: tec5c
author_rank: Frequent Contributor
timestamp: April 17, 2012, 03:46:17 am
content: Thanks Andy and IanB for the replies. I am starting to regret choosing to complete this course by correspondence as I seem to stress myself into a state where I can't think straight!I have another question. This one is to do with Norton's theorem. Below shows the original circuit and my attempt at its Norton equivalent circuit. One thing that is throwing me off is that the procedure in my notes says to remove the load resistor and replace it with a short. Now these questions I am doing all involve open terminals (AB), so does one just skip the step of removing a component and just put a short between the open terminals and proceed from there?I'm not confident in my answer and would like to clarify the procedure.Thanking you.

author: IanB
author_rank: Super Contributor
timestamp: April 17, 2012, 04:07:22 am
content: You can check your solution, which may give some insight into the method. If the Norton circuit is to be equivalent to the original circuit then the results should match in two particular cases: the open circuit voltage between terminals A & B should match, and the short circuit current between terminals A & B should also match.

author: IanB
author_rank: Super Contributor
timestamp: April 17, 2012, 04:21:51 am
content: (By the way, I'm not saying you got the answer wrong, just in case you are worried. I'm just mentioning a standard check you can perform to confirm you got the answer right.)

author: tec5c
author_rank: Frequent Contributor
timestamp: April 17, 2012, 05:29:07 am
content: Thanks Ian,Playing around with those circuits in LTSPICE and another circuit simulator program, I find it a little tricky to cross-check the equivalent circuit with the original. I believe this may just be the way I am using the programs though I wanted to ask how would you yourself go about checking it?

author: slateraptor
author_rank: Frequent Contributor
timestamp: April 17, 2012, 05:49:59 am
content: Quote from: tec5c on April 17, 2012, 05:29:07 amCalculate VABfor both circuits and compare. Are they equal?For the first circuit:VAB= [ R3/ (R1+ R3) ] * V1= 57.6VNote that R2is not part of a closed loop, therefore zero current passes through it and thus we can pretend that it doesn't exist for the sake of calculating VAB. If a load were connected across AB, that's a different story and we cannot neglect R2, however, if the circuit were transformed into its Thevenin equivalent, then R2will be accounted for.For the Norton equivalent:VAB= I1* R4= 57.6V

author: IanB
author_rank: Super Contributor
timestamp: April 17, 2012, 06:38:07 am
content: Quote from: tec5c on April 17, 2012, 05:29:07 amTo be honest, you absolutely should not be using a circuit simulator program to work out such simple circuits. You will simply fail to get any understanding of what you are trying to learn. It is vital to be able to solve those problems with a pencil and paper, or even solve them in your head by inspection after you have a bit of practice.For instance, to find the short circuit current in the Norton circuit example, you connect terminals A & B together. Then proceed as follows:R2||R3 = (20 x 80) / (20 + 80) = 16 ohmsR1 + (R2||R3) = 20 + 16 = 36 ohmsThe total current in the circuit is then:I = 72 V / 36 ohms = 2 ATo find the current in the R2 branch, use the current divider theorem:Iab = 80 / (80+20) x 2 A = 1.6 AThat gives you your Norton current source.As slateraptor showed you, the open circuit voltage is:Vab = 80 / (80 + 20) x 72 = 57.6 V (voltage divider theorem)It follows that your Norton equivalent resistance must be:R4 = 57.6 V / 1.6 A = 36 ohmsAs you can see, checking the solution is also solving the problem. If you work out the problem two different ways and get the same result each time, you have a check of your answer.

author: Bored@Work
author_rank: Super Contributor
timestamp: April 17, 2012, 08:27:28 am
content: Quote from: IanB on April 17, 2012, 06:38:07 amThat, and the point of homework is that you, not we, do the homework.

author: tec5c
author_rank: Frequent Contributor
timestamp: April 17, 2012, 09:33:16 am
content: Thanks Ian,I agree that using a simulator will not help me learn, though I was not using it as a learning tool. It was a way to check the answers I got on pencil and paper. As I said I am doing this work by correspondence so I do not have a teacher I can turn to when I get stumped on something. That's where I brought in the simulator, so I can check my working. All the simulator will give is values at certain points in the circuit, I am only interested in those values once I have worked through a question. It helps to confirm that I am on the right path and that I am studying the correct things.Quote from: BoredAtWork on April 17, 2012, 08:27:28 amWow, never knew that one. Thanks!

author: Bored@Work
author_rank: Super Contributor
timestamp: April 17, 2012, 11:08:56 am
content: Quote from: tec5c on April 17, 2012, 09:33:16 amGlad I could help and show you the way to enlightenment. No go and sin no more.

author: cybergibbons
author_rank: Frequent Contributor
timestamp: April 17, 2012, 12:07:09 pm
content: However, I do find once you move onto analysing transistors and other semiconductor devices, it can be interesting to model them in SPICE alongside your simplified models. You learn the limitations quicker like that. Better still, actually set up a demo as well.

author: steve_w
author_rank: Regular Contributor
timestamp: April 18, 2012, 11:13:56 am
content: A good way to get this into your head is to do lots of examples, it doesn't matter if you get some wrong as long as you understand why you got them wrong.and as boredatwork says you have to do the work