Thévenin's theorem, open circuit voltage calculation? - Page 1
Forum section: electronics / beginners

author: Sudo_apt-get_install_yum
author_rank: Regular Contributor
timestamp: September 18, 2018, 01:35:23 pm
content: Hi everyone!So my friend showed me a question he got on a test (see attached image) and I realized that I don’t know how to solve this.Now I’ve googled around and skimmed through one of my books (Analog electronics) and I am pretty sure that I have to use "Thévenin's theorem" to solve this. Reason why is because it’s a linear circuit and fits the Wikipedia explanation of the theoremhttps://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theoremSo I fired up LTSpice and simulated the circuit, the resulting voltage between AB is ~12V witch confuses me even more since this voltage is higher than both sources.Could someone please explain to me what’s going on?I’ve never encountered a real life example of this, where is this used and in what circumstance is it used, my friend didn’t solve it correctly so I’m also indirectly asking on his behalf.Thanks for any help!

author: Mr. Scram
author_rank: Super Contributor
timestamp: September 18, 2018, 01:37:36 pm
content: Following for fun.

author: kulky64
author_rank: Regular Contributor
timestamp: September 18, 2018, 02:00:41 pm
content: I calculated this equivalent model:V = 12.073170731707 VR = 1.9512195121 kOhm

author: Mario87
author_rank: Frequent Contributor
timestamp: September 18, 2018, 02:02:48 pm
content: Ok, I THINK I know how to do this, but bear with me as I'm only just starting to go through the AoE textbook and lab books which funnily enough cover voltage dividers & theveins theorem in the first chapter. Then again I could be TOTALLY wrong in what I'm about to explain, so hopefully someone else can clarify.The way I looked at it was to take the first part of the circuit as a voltage divider with 1k & 2.2k10v over those resistors gives you 6.875v across the 2.2k resistorFrom there we can calculate the resistance of the current source. So using ohms law, V=IR which goes onto 6.875=0.003R, therefore R=2291So now our 2.2k resistor is in parallel with a 2291 ohm resistor which is no different than a single resistor of 1122 ohmsSo now we have a voltage divider with 3 resistors. First one is 1k, second is made up of the parallel devices which we have just calculated have a resistance of 1122 ohms and the final resistor is 5kSo then, if we have 10v across all 3 resistors, that means the voltage across the final 5k resistors is 7.02vTake that 7.02v and add on the 5v supply you have in series with it and you get an output of12.02vLike I said, I could be TOTALLY wrong and may have just stumbled upon the answer by sheer chance, but that's how I have understood it.If someone with more knowledge can confirm it would be great, and if I am correct it would be good to know that I am actually understanding what I am reading in these books.

author: oPossum
author_rank: Super Contributor
timestamp: September 18, 2018, 02:04:46 pm
content: 2k2 * 3mA = 6.6V10V + 6.6V - 5V = 11.6V11.6V / (1k + 2k2 +5k) = 1.4mA5k * 1.4mA = 7.07V5V + 7.07V = 12.07V(1k + 2k2) * 1.4mA = 4.53V10V + 6.6V - 4.53V = 12.07

author: rstofer
author_rank: Super Contributor
timestamp: September 18, 2018, 02:05:29 pm
content: Nodal analysis will get it done but it's too early in the morning...Assign a node N1 where R1, R2 and I1 intersect and call the voltage at that node VxThere is a current coming into the node through R1 and it is (V1 - Vx) / R1There are two currents leaving the node I1 + (Vx - Va) / R2The sum of these 3 currents is 0There are three currents entering the node at terminal A: I1 + (Vx-Va) / R2 + (V2-Va) / R3The sum of these 3 currents is 0There are two unknowns Vx and Va. Fortunately, we have two equations so it ought to work out.Something like that...Now that we know the output voltage Va, we need the equivalent series resistance. Replace the current source by an open circuit and both voltage sources by short circuits and solve the series/parallel resistor problem using Norton's Theorem.

author: Sudo_apt-get_install_yum
author_rank: Regular Contributor
timestamp: September 18, 2018, 02:22:11 pm
content: Quote from: kulky64 on September 18, 2018, 02:00:41 pmQuote from: oPossum on September 18, 2018, 02:04:46 pmThanks for the help!It seems like the answer is 12.07 just like the simulation said. Could you please explain a little bit more, like how you add the voltages together in step two? I have a hard time imagining it. I’m definitively not used to the kinds of circuits.

author: rmacintosh
author_rank: Regular Contributor
timestamp: September 18, 2018, 02:29:26 pm
content: Looks like a simple current source /voltage source conversion problem.Convert current source to voltage source.Combine any voltage sources and resistances.Convert back to current sources combine.Back to voltage source and you have your answer

author: oPossum
author_rank: Super Contributor
timestamp: September 18, 2018, 02:44:48 pm
content: It can be solved by knowing two things....1) Ohms law2) A current source in parallel with a resistor is equivalent to a voltage source in series with a resistor. The diagram in the post above shows that.

author: iMo
author_rank: Super Contributor
timestamp: September 18, 2018, 03:28:01 pm
content: See below

author: rstofer
author_rank: Super Contributor
timestamp: September 18, 2018, 03:30:40 pm
content: I stuffed the equations I gave above into wxMaxima. The voltage VA is indeed 12.07V. Attached is the program and the output. The output starts where the equations are rewritten at "(eq1)"The last 5 lines of code sort the variables. The program itself was written long ago for another example. I just changed the equations a bit.ETA:You can see in eq1 the one current going into node Vx and the two currents leaving. Similarly, at eq2, you can see the 3 currents entering node Va. Positive currents are entering and negative currents are leaving. The assumed direction is irrelevant and you can just as well have all of the current entering and none leaving. Or you can have them all leaving and none entering. How can that be? In the solution, some of the entering or leaving currents will turn out to be negative. If this happens, your assumed direction was incorrect. No big deal, just realize what the negative result means. But it doesn't affect the solution!There are ways to simplify the circuit by various manipulations (like converting the current source to a voltage source with a series resistor. These shortcuts can be quite handy. But the thing about Kirchhoff's Laws is that they take a methodical approach to solving these circuits. You make some assumptions about direction but even that doesn't have to be right.Two simple rules: The currents entering and leaving a node must add up to zero. Current can not spill out on the floor. Second, the sum of the voltage drops around a mesh must add up to zero. Voltage can not pile up.ETA (again): Only the first two equations have anything to do with the solution. There are two equations (eq1 and eq2) and two unknowns (Vx and Va). The other equations just allow me to name the components and assign values. That keeps numerics out of the first two equations and that allows the equation to look a lot like the circuit. What if I used numerics and all the resistors had the same value. It would be confusing!

author: Zero999
author_rank: Super Contributor
timestamp: September 18, 2018, 04:03:26 pm
content: The voltage is higher because the current source and resistance adds to V1.I solved the first two steps in my head, with no calculator. It should be fairly self-explanatory, just by looking at the schematics.Step 1, a current source in parallel with a resistor, is equivalent to a voltage source in series with a resistor.Step 2, the voltages and resistances of the two sources can be added together to form one voltage source and series resistance, as both sources and resistances are in series.The final step was a bit more tricky.Calculate the current through R2 and R3, which is the same, as they're in series.I = V/RV = V1 - V2 = 11.6VR = R1 + R2 = 5k + 3.2k = 8.2kI = 11.6/8200 = 0.0014146 = 1.4146mACalculate the voltage across R3:V = I*RV = 0.0014146*5 = 0.007073 = 7.073VCalculate the voltage at node A, which is simply the voltage across R3 added to V2.VA= 5 + 7.073 = 12.073VWhat's more, the whole circuit can be simplified to a simple voltage source and a series resistor. The voltage is equal to the voltage at node A and the resistance equal to R2 and R3 in parallel.R1 = (3.2k*5k)/(3.2k+5k) = 16k/8.2k = 1.951k

author: rstofer
author_rank: Super Contributor
timestamp: September 18, 2018, 04:29:06 pm
content: Quote from: Sudo_apt-get_install_yum on September 18, 2018, 02:22:11 pmAs I understand the solution,Step 2: The current source and parallel resistor are now treated as a voltage source of 6.6V. Think about 2 batteries in series, they add. So, 10V + 6.6V => 16.6V equivalent. Then the 5v source is subtracted when a mesh equation is written around the 3 sources and 3 resistors (step 3). The next two steps use the mesh current across the 5k resistor to get the final output voltage. The last 2 steps cross-check that the same voltage is contributed by the 16.6V source as is contributed by the 5V source and the 5k resistor. It balances so the output is indeed 12.07V.

author: rstofer
author_rank: Super Contributor
timestamp: September 18, 2018, 04:50:43 pm
content: Strictly speaking, my solution uses Kirchhoff's Laws, not Thevenin's Theorem. Here is a better description of the process.https://www.electronics-tutorials.ws/dccircuits/dcp_7.html

author: Vtile
author_rank: Super Contributor
timestamp: September 18, 2018, 05:04:17 pm
content: There is many ways to skin this cat, which is still really simple mixed source problem and like said above can be solved in many ways.I just want to point out that remember to simplify the networks. You can even go as far as the attachment 2 (network2) and ask yourself what is the current in "ground leg".Edit2. Better drawn network2.