Solve this circuit - Page 1
Forum section: electronics / beginners

author: tom66
author_rank: Super Contributor
timestamp: November 16, 2012, 10:19:05 pm
content: I was given this circuit on a homework sheet:Code:Va ____ Vb+---|____|---+------------+| 10 ohm | || | || | | | | || | 1 ohm | | 2 ohm | | 2 ohm| | | | | || | || | |/ \ / \ ---12V 10V -\ / \ /| |--- ---- -I'm supposed to find Va and Vb. I tried using superposition rules but I got the wrong answers. How can I solve it?

author: w2aew
author_rank: Super Contributor
timestamp: November 16, 2012, 10:34:00 pm
content: Simple if you do it in two steps. Combine the 1 and 10 ohm resistors, solve for vb, the only unknown. Use the fact that the sum of the currents entering/leaving any given node is always equal to zero. Once vb is solved, then solving for va is easy.

author: sorin
author_rank: Frequent Contributor
timestamp: November 17, 2012, 12:30:07 am
content: Kirchhoff's laws

author: tom66
author_rank: Super Contributor
timestamp: November 17, 2012, 11:33:27 am
content: I can easily solve when the voltages are the same, or when I'm dealing with current sources, but I can't easily see how to solve this using KCL or KVL. Someone told me to solve it using simultaneous equations and loop analysis, two of my weak points. But if I draw a loop of current from the 12V to 10V... what happens to the other loop going from 10V through the 2 ohm resistors? How is the current forced back into the source; does the source behave here as an open circuit?

author: Bored@Work
author_rank: Super Contributor
timestamp: November 17, 2012, 12:06:51 pm
content: A loop of current? What should that be?Kirchhoff's laws have voltage loops:The sum of all the voltages around a loop is zero.and current nodes:The sum of all currents in a network node is zero.Of the three apparent voltage loops (you can construct even more), you can take two, otherwise your system of equations will be overdetermined. So you add a current sum to your system of equations and then solve the system. That is really absolutely basic stuff, first week of electrical engineering 101.

author: tom66
author_rank: Super Contributor
timestamp: November 17, 2012, 12:08:54 pm
content: Sorry, mixed my terminology; I just meant "current", not "loop of current".I get that there should be 2V between the three resistors, but I'm not sure how to cope with the 2 ohm to ground. If that were deleted I could work it out very easily (it's just a voltage divider.)I feel like I missed something. I have no problem with KVL when applied to simple voltage sources or resistor networks where voltages are defined, but this problem just doesn't "click", because the two are combined. I just can't "see" it in my head like I can for most problems. I'm on my 8th week of ELEC1510 first year... not quite 1st week.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: November 17, 2012, 12:13:12 pm
content: Use LTspice.or Mesh analysis and KVLcall current flowing up through R1 as Ir1call current flowing up through R2 as Ir2Ir4 = Ir1Ir3 = Ir1 + Ir2Kvl sum of voltages in a circuit = 02 circuits (loops where the sum of the voltages = 0)v1+Vr1 + Vr4 -Vr2 -v2 = 0v2+ Vr2 -Vr3 = 0express in terms of Ir1 and Ir2.so two equations and two unknowns should be solvable.

author: Unknown
author_rank: Unknown
timestamp: November 17, 2012, 12:14:04 pm
content: The sum of all the voltages in a loop is zero, a voltage source has a fixed voltage, so the voltage across the other elements in the loop should sum up to negative this voltage (mind your sign conventions!). Treat the voltage source as an element between the ground node and the voltage node with a fixed voltage. This element completes the loop.Superposition should also work fine here, but you should really have a good grasp on Kirchoff's circuit laws if you study anything related to electronics. I would recommend solving the circuit via both methods, the results should be identical.

author: tom66
author_rank: Super Contributor
timestamp: November 17, 2012, 12:18:51 pm
content: I have the solutions from my teacher and I -could- just write them down but that wouldn't help in an exam. I already used spice to solve the problem and got the same answers he gave (I think 6 2/3V for Va, 9 1/3V for Vb, from memory...) However, when I replaced each voltage source with a short (superposition), I never could get the right answer. Perhaps I was doing it wrong, I will have another look at it tonight. I need to brush up on simultaneous equations... it seems very unusual for such a simple thing, I always am fine after I remember how to do them, but I keep forgetting!

author: Unknown
author_rank: Unknown
timestamp: November 17, 2012, 02:09:11 pm
content: Superposition gives me the same answers as LTspice (5.58V and 11.42V). This is the most elegant way to solve this problem, but you really should be able to solve the system of linear equations that you can derive from KVL/KCL.

author: w2aew
author_rank: Super Contributor
timestamp: November 17, 2012, 02:17:50 pm
content: You can write a simple equation for the currents in each leg from node Vb. Vb/2, (Vb-10)/2, etc. the sum of these three MUST equal zero. Solve for Vb. once you have Vb, solving for Va is easy since you'll know the current in that leg.

author: tom66
author_rank: Super Contributor
timestamp: November 17, 2012, 02:27:52 pm
content: Quote from: alm on November 17, 2012, 02:09:11 pmI got those answers from superposition. I thought I was doing it wrong! How did I do it wrong in spice? Now I am reallyIt's only really this question that's tripped me up, I've been able to solve most of the other kvl/kcl questions.

author: kg4arn
author_rank: Unknown
timestamp: November 17, 2012, 02:30:22 pm
content: tom66,Since you say that you have the teacher's solution already, I am posting a solution here.The first key to this problem is writing the correct loop equations. Look at the image titled "Loop Equations" first without looking at the "Loop Analysis Solution." See if you can on your own, produce the same equations. The key here is that each loop shares a common branch with the 10V source and the 2 ohm resistor. Therefore, each equation will have both of these terms. Also, watch out for the proper polarity of the terms.Once you have the right loop equations, it's a matter of algebra to solve the 2 equations with 2 unknowns (the loop currents) and then you can calculate Va and Vb.I included the superposition solution. Notice the polarity dot on the 10 ohm resistor. The sources force current in opposite directions through this resistor.In practice, you would use spice for complicated loops because it is so easy to make algebra errors.BUT!!!! You must understand how to do this simple example by hand so you can understand what happens in circuits sharing a common branch.I hope this helps.

author: tom66
author_rank: Super Contributor
timestamp: November 17, 2012, 02:42:27 pm
content: I appreciate your help!I am trying to understand this. My intuition fails me when looking at circuits like this. Cheers.

author: sorin
author_rank: Frequent Contributor
timestamp: November 17, 2012, 02:46:24 pm
content: now is just mathematic

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2012, 07:09:13 pm
content: You have been given a lot of responses in this thread, but I'm drawn to the fact that w2aew gave you the answer in the very first reply.Always consider if you can simplify the problem from the one given you before you try to solve it.So you were given this:Code:Va ____ Vb+---|____|---+------------+| 10 ohm | || | || | | | | || | 1 ohm | | 2 ohm | | 2 ohm| | | | | || | || | |/ \ / \ ---12V 10V -\ / \ /| |--- ---- -But by inspection it is the same as this:Code:Vb+------------+------------+| | || | || | | | | || | 11 ohm | | 2 ohm | | 2 ohm| | | | | || | || | |/ \ / \ ---12V 10V -\ / \ /| |--- ---- -Now you can simply use Ohm's law combined with the knowledge that the currents must add up. So:(12 - Vb)/11 + (10-Vb)/2 = (Vb - 0)/2=> (1/2 + 1/2 +1/11) Vb = 12/11 + 10/2=> Vb = (12/11 + 10/2) / (1/2 + 1/2 + 1/11)=> Vb = 5.58 VNow you have Vb, you can go back and use the lever rule for Va:(12 - Va) / 1 = (Va - Vb) / 10=> (1/10 + 1) Va = 12 + Vb / 10=> Va = (12 + Vb / 10) / (1/10 + 1)=> Va = 11.42 V

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2012, 07:13:37 pm
content: Quote from: alm on November 17, 2012, 02:09:11 pmI seem to have a mental blank on how superposition works. Could someone give me the quick two minute summary on how it is applied to this example?

author: Unknown
author_rank: Unknown
timestamp: November 17, 2012, 07:47:15 pm
content: The idea of superposition (which only works for linear circuits) is to short each of the voltage sources and open all of the current sources, then turn on the sources one by one, calculate the voltages of interest, and sum up the voltage resulting from each of the sources.Short the 12V source:Code:Va ____ Vb+---|____|---+------------+| 10 ohm | || | || | | | | || | 1 ohm | | 2 ohm | | 2 ohm| | | | | || | || | || / \ ---| 10V -| \ /| |--- ---- -This is a resistive divider with 2 ohm on top and a parallel combination of 2 and 1+10 ohm at the bottom, so it's easy to calculate Va and Vb. I estimated the parallel resistors to be about 2 ohm (2 ohm parallel with much larger value), so Vb would be a little under 5V in this case.Now short the 10V source:Code:Va ____ Vb+---|____|---+------------+| 10 ohm | || | || | | | | || | 1 ohm | | 2 ohm | | 2 ohm| | | | | || | || | |/ \ | ---12V | -\ / || |--- ---- -To calculate Vb, you can treat this as a voltage divider with 11 ohm on top and 2 // 2 ohm at the bottom, or 1/12 * 12 V. So Vb is conveniently 1 V. The sum of the two values of Vb is 6 V, which is pretty close to the real value because I was too lazy to calculate the parallel resistance. You can also derive Va by superposition, but I treated it just as a voltage divider of 1 and 10 ohm between 12 V and 6V, so Va = Vb + (12 V - Vb) * 10/11, or close to 12V.

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2012, 08:02:04 pm
content: Quote from: alm on November 17, 2012, 07:47:15 pmOK, thanks for the explanation. I'll have to get pencil and paper and prove to myself how that method comes up with the right answer.

author: Bored@Work
author_rank: Super Contributor
timestamp: November 17, 2012, 08:05:39 pm
content: Quote from: IanB on November 17, 2012, 07:09:13 pmThat was just Kirchhoff, expressed in a convoluted way. One current node, two voltage loops.An EE student should be capable of doing that in his sleep. So stop shitting on us who pointed out the correct way to do it in a lot of responses. Stick it to the student who didn't do it.

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2012, 08:13:04 pm
content: Quote from: Bored@Work on November 17, 2012, 08:05:39 pmAnd who curdled your milk this morning? I don't recall directing any responses in your direction. What are you getting so upset about?QuoteIsn't that exactly what "You (tom66) have been given a lot of responses in this thread..." is actually saying? Or do you have reading comprehension problems along with a sour personality?

author: tom66
author_rank: Super Contributor
timestamp: November 18, 2012, 01:16:31 am
content: Thank you for all the responses and help, and a special thank you to Bored@Work for his well-crafted thank-you in his own unique way.IanB; I did of course try simplifying the 10+1 ohm into a single resistor, that was the first thing I tried. I fall down when you have parallel voltage sources of different values. I am not very good there, as it's not something seen very often in electronics (though I'm sure I'll come across it at one point or another.) However, I did not realise with superposition that you could work out the voltages and add them together. I always did it the awkward way -- by adding the currents and calculating V=IR.

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2012, 01:51:40 am
content: Quote from: tom66 on November 18, 2012, 01:16:31 amOn the contrary, this kind of problem is at the very core of all circuit analysis. It's the kind of problem you should develop the ability to look at and see the solution almost without thinking. The general principles of current balances and voltage sums used to solve this problem are the same principles used to solve all circuits in electronics, however complex. (But not superposition: as alm mentioned, this only works for linear circuits.)Did you at least follow the working I set out? You don't actually have parallel voltage sources (if this has any meaning), what you have are three known voltages (12 V, 10 V and 0 V) and two unknown voltages (Va and Vb). Since you have two unknown voltages, you need to write two independent equations to solve for those voltages. Or you could simplify as w2aew indicated and eliminate one of the voltages. Then you only need to write one equation to solve for one unknown voltage at a time.I'm not quite sure what level you are at in your studies: sixth form, technical college or university? If you are still in the sixth form or at technical college your difficulty with this could be forgiven. If you happen to be at university studying for an EE degree you are in trouble. This sort of problem should be second nature to you.

author: tom66
author_rank: Super Contributor
timestamp: November 18, 2012, 02:05:06 am
content: I am first year in university, in my ninth week as of tomorrow, yes, I am studying for a MEng in Electrical and Electronics Engineering. Forgive me. I will make sure to understand this problem. I did reduce it to simultaneous equations. However, for some reason, I went blank trying to solve them...It seems silly, perhaps even stupid, but I couldn't think how they are solved.

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2012, 02:10:42 am
content: You seem to be suffering from the problem with education described in this video:For physics you could easily substitute engineering. I'm afraid you will need to concentrate a lot on developing your mathematical skills to get through an engineering degree. Engineering is all maths.

author: tom66
author_rank: Super Contributor
timestamp: November 18, 2012, 02:21:31 am
content: I agree. It is something I will work on. The university has moved the maths exam from a paper to online now though, it is taken in 11 individual homeworks, which is easier than doing an exam (I tend to get exam stress, great eh?) I achieve an average of about 80% on those, which involve integral and differential calculus, complex numbers, and linear algebra. However, I do sometimes find I need to refer to a textbook to look up how to do a particular problem. I am getting better, but it is definitely my weakest area.The grade for a 1st is 70% average overall, which I am aiming for (2:1 is 60%, which would still be good, but 1st is better), though the first year is not accounted for in the final result (you must still pass though!)

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2012, 02:32:27 am
content: Quote from: tom66 on November 18, 2012, 02:21:31 amLucky youIt was the same for me when I did my degree, many years ago (Bath, Chem Eng, '84). I also aspired for a First, but I flunked and had to settle for a 2(I). Good luck with your studies!

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2012, 02:38:33 am
content: Quote from: tom66 on November 18, 2012, 02:21:31 amBy the way, there is no shame in looking things up in textbooks. Doing research to improve your knowledge is the sign of a good engineer. The key thing is to recognize what kind of problem it is and where to look to find the necessary information.

author: w2aew
author_rank: Super Contributor
timestamp: November 18, 2012, 02:51:33 am
content: Also, remember that things like Kirchoff's, Norton's, loop analysis, superposition, etc. are all tools. You don't have to use strictly one tool to solve a given problem. Learn how to apply the principles, along with a common sense look at the problem, and use whatever combination of tools and techniques that help you to solve the problem. That was my approach with my suggestion to help you with this problem.Your studies will help put the tools in your toolbox. It's up to you to learn how best to use these tools to solve the problems at hand.