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BigJohnny 
Posted: October 12, 2011 12:47 am

Member Group: Trusted Members Posts: 200 Member No.: 25,390 Joined: July 27, 2009 
I'm starting to get really frustrated with this because I know it's quite simple, but the way we are being taught isn't helping.
the instructor is very quite, and very Asian which only adds to the confusion as she rushes through lessons and seems to over complicate things. So we have been learning DC circuits, series, parallel and combination. We are also getting into the equivalent circuits and talking about kirchoffs voltage law, voltage divider law, thevenin equivalents, and rxry etc and having tons of formulas thrown around. So now we are faced with this and I keep getting these damn things wrong, cause I'm trying to remember when and where to use stuff. I just today had it click what the hell R1R2 means, and that I'm supposed to to use (RxRy/Rx+Ry)Vs=Rx,y So can anyone make this lesson easier to understand and how to solve these friggin diagrams, and when and where to use what formulas. Determine the thevenin equivalent as seen from terminals A and B. (Again, forgive the diagram quality. Smartdraw 2010 sucks) 
draget 
Posted: October 12, 2011 03:01 pm

Forum Addict ++ Group: Cleanup Taskforce Posts: 6,323 Member No.: 1,770 Joined: December 31, 2004 
A Thevenin equivalent has only a voltage source and series resistor, so for your case here we need to know two things:
1. The unloaded voltage. 2. The resistance looking into AB. So, the first one can be worked out by realising that the current across R4 when the circuit is unloaded is zero, and then you can have a simple series loop, which you can work out the current in (its 2.5V over R1+R2+R3). Multiply this current by the value of R2 and you'll get the voltage across it, which, when unloaded is the voltage across AB. Next, replace the voltage source with a short and establish the total resistance looking into AB. You will have R4 + (R3+R1)R2 (work graphically in your head or start off drawing lots of diagrams until you can identify which components are in series or parallel with each other). Then you can create the Thevenin equivalent. You'll get the hang of doing this sort of thing  I always avoid first principles approaches with writing out a whole heap of KVL/KCL equations if at all possible, you'll get it done much quicker once you learn a few tricks. You will be taught this stuff, mesh analysis, nodetodatum and superposition theorems, but you'll probably find that you will never use those methods, however, the intuition you gain practising will be useful in later circuit analysis. 
millwood 
Posted: October 12, 2011 05:20 pm

Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
it is fairly simple.
TE is looking to represent any (linear) circuit with an ideal voltage source + output / serial resistance. open A and B: figure out what the voltage is between A and B > that's the voltage of your ideal voltage source; short A and B: what's the current going through A and B. that's your shortcircuit current. Your resistance = the voltage of your ideal voltage source / the shortcircuit current. done! in this particular case, if you open up A and B, the voltage there is (78) / (100+78+48) * 2.5 = 0.87v; if you short A and B, the shortcircuit current is 2.5v/(100+22+47)=15ma. so the circuit is equivalent to an ideal voltage source of 0.87v in serial with a resistor of 0.87v/15ma=58ohm. 
millwood 
Posted: October 12, 2011 05:21 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
the currentequivalent of it is an ideal current source in parallel with a resistor. 

millwood 
Posted: October 12, 2011 06:04 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
that's wrong. when you short A and B, R4/R2 becomes parallel. so total current going through R4//R2 = 2.5v / (100+R4//R2 + 47) = 2.5v / (100+17.16+47) = 15ma. etc. the key really is to make a V+R or I//R circuit behave identically to your circuit when A/B is open AND shortcircuited. btw, all of that from 2 minutes of googling + ohm's law that I learned before dropping out of highschool. 

BigJohnny 
Posted: October 12, 2011 11:20 pm

Member Group: Trusted Members Posts: 200 Member No.: 25,390 Joined: July 27, 2009 
Ok, Im still lost.
millwood, i do search google before I post anything here. Sometimes I get the answer, sometimes it seems like they are talking neurosurgery and I end up more lost on a simple problem. Let me take a crack at this one. Vab = 100K+56K/(100K+56K)1.5V = (no this wont work) ..... so i've been sitting here for twenty minutes looking over my notes and I can't even begin to figure out where to start this problem?! I have to find the voltage across A and B, but the above formula results in 1.5 which I dont think is actually possible given resistance, so I must be figuring out resistance wrong. 
draget 
Posted: October 13, 2011 06:27 am

Forum Addict ++ Group: Cleanup Taskforce Posts: 6,323 Member No.: 1,770 Joined: December 31, 2004 
You want 56k/(100k+56k) 1.5 V
Voltage divider laws are nice, but it makes more sense if you work if from Ohms Law. Work out the current through R1+R2 due to the 1.5V then use V=IR to work out the voltage across R2. It will come to the same. 
millwood 
Posted: October 13, 2011 12:05 pm

Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
again, it is incredibly simple  you just need to remember the principles behind what you do.
1) open circuit A/B, and find the voltage Vab = R2 / (R1 + R2) * V = 0.5v; 2) short circuit A/B, and find the current Iab = V / R1 = 0.015ma. so the TE has a voltage of Vab and (serial) resistance of Vab / Iab = 30k+ (that number will be precisely R1//R2). 
millwood 
Posted: October 13, 2011 12:09 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
you may find this offensive but if that's really reflective of your knowledge on math / ohm's law, you can get to TE a lot faster by going back and retaking those classes so you have a solid understanding of the basics that are prerequisite of more advanced classes. 

CWB 
Posted: October 13, 2011 12:09 pm

Forum Addict ++ Group: Spamminator Taskforce Posts: 21,945 Member No.: 15,154 Joined: May 15, 2008 
sometimes , "simple" is a relative term .
 "Know how to solve every problem that has been solved"
R. Feynman '88 
millwood 
Posted: October 13, 2011 01:02 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
conceptually, here is how you approach it. all TE (the voltage source version of it) is trying to do is to simplify any (linear) circuit into a voltage source (V) in serial with a resistor ®. so the TE circuit on the right should behave like the original circuit on the left under ANY CIRCUMSTANCE. in this particular case, we need two circumstances so solve for V and R. the two cases we used is when Rl = infinity (open circuit) and when Rl=0 (short circuit). for the open circuit case, you calculate the voltage across Vab, and your TE circuit should produce that voltage across A/B. in this case, because it is open circuit, Vab = V. for the short circuit case, you calculate the current going through A/B (Iab), and your TE circuit should produce that same current, which should be V/R. problem solved and all you need is some simple math + ohm's law.
i guess so. but if a highschool dropout like me can do it, anyone can do it. 

BigJohnny 
Posted: October 13, 2011 01:07 pm


Member Group: Trusted Members Posts: 200 Member No.: 25,390 Joined: July 27, 2009 
This is coming as quite a shock to me in all honesty. I've never had this much trouble with anything, I finished my previous college prep math course with an A, and blew through it like it was 3rd grade math. There is a lot more going on in life in general now which I think is distracting my attentions, but that's no excuse. At this point I just feel like I look at this stuff and it can't get into my head. Look how much trouble I'm having with one simple thing. I just cant seem to figure out what to do for some reason and it causes a great deal of anxiety. Anyway, that's no excuse, and I agree with all of you, it is extremely simple stuff. I feel like the worlds biggest idiot. I'm seeking help for this, not just forum help I mean the real kind. back to the question at hand. millwood: I know what an open and short circuit are, but relating to the diagrams what exactly does it mean? draget: could you explain why I want to use 56k/(100k+56k) 1.5 V, or more precisely, why those numbers in that way? I think that's what I'm not getting here. I know the formula, but not clear on what it's doing, or how to determine what values to insert into the formula. 

BigJohnny 
Posted: October 13, 2011 11:51 pm


Member Group: Trusted Members Posts: 200 Member No.: 25,390 Joined: July 27, 2009 
I understand this much of it, and seeing the difference between open and short helps. Where I'm lost here is how to sum the resistances in which order, to achieve the correct Rth or Vth. 

draget 
Posted: October 14, 2011 03:15 am


Forum Addict ++ Group: Cleanup Taskforce Posts: 6,323 Member No.: 1,770 Joined: December 31, 2004 
Its a ratio thing, the same current flows in both resistors, which is determined by ohms law. So, I = V/R = 1.5/(100k+56k) Now you want the voltage across the lower 56k, so Ohms law says V=IR = I*56, so then you sub the I we've found above and you get the nice divider ratio 

millwood 
Posted: October 15, 2011 01:15 pm


Forum Addict ++ Group: Trusted Members Posts: 1,823 Member No.: 25,377 Joined: July 26, 2009 
the 2nd half of the sentence suggests that the 1st half of the sentence isn't true. many times, there is a huge difference between what you think you know and what you know. one way to tell is to close your book and try to articulate to yourself in your own words what you just learned. 

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