Find the voltage at the 2 junctions of a 3 element voltage divider across a supply voltage of 8.4 volts. The two junctions of the divider both supply external current of 5mA.
A more basic problem can be solved using Thevenin's idea with only 2 resistors of say 570 and 240 ohms and a 6.9 volt supply, and 5mA of current from the single junction.
The output impedance at the junction is the 2 resistors in parallel, or about 169 ohms. The open circuit voltage ignoring the 5mA is about 2.044 volts, and so the voltage drop on R1 (240) is 2.044 - (169 * .005) = 1.2 volts.
But I don't see an easy way to apply Thevenin to the other case where there are 3 or more resistors and junctions with known currents from the junctions.
I've labeled a few things above. Vc=8.4, Vg=0, and you are interested in Va and Vb, which are the voltages at node a and node b, respectively. I assume the two 5mA currents are "outward" due to the arrows showing. It's easier to use these conductances:
G1 = 1/R1 G2 = 1/R2 G3 = 1/R3
Now, examine node a. It's equation looks like:
(1) Va*(G1+G2) + 5mA = Vb*G2 + Vc*G1
Before you say you may not be able to remember how to do this, or even understand how I got the above equation, just stop thinking about voltage _differences_ and think in terms of superposition (currents flow two ways, at once) instead.
Current flows both inward and outward. The currents outward flow via the only three paths that exist: R1, R2, and the 5mA outgoing flow. The currents flowing inward flow via just two paths (we've already accounted for the outward current, by definition): R1 and R2.
Equation 1 then must first start with this realization: "Current outward equals current inward." Which you must know, since it isn't possible for electrons to pile up in the node. Current in and current out must indeed match up. So what is left is to put the currents IN and the currents OUT onto the two sides.
Equation 1's left side reads like this: "The current that spills outward from node a is simply Va times the conductance outward via G1 and G2 (since it flows outward via both, you add them), plus of course the 5mA that is also flowing outward." Think of Va, here, as any arbitrary voltage that mysteriously (without needing to know the surrounding voltages) _forces_ current outward via R1 and R2. Since the
5mA is _known_ by definition, Va doesn't interact with it. It just _is_. So just add that current since it is an outward going one and belongs on this half of the equation.
Equation 1's right side reads like this: "The current that spills inward into node a is simply Vc times G1 plus Vb times G2." We've already accounted for the 5ma and it doesn't spill inward. So it doesn't appear on this side. However, the surrounding voltages _do_ force a reverse current via whatever conductances they have available to them. So, of course, you must include them on the right side.
That explains the first equation you need. Now, examine node b. It's equation looks like:
(2a) Vb*(G2+G3) + 5mA = Va*G2 + Vg*G3
However, since we know out of hand that Vg=0, we can drop that term:
(2b) Vb*(G2+G3) + 5mA = Va*G2
Okay, now you have two equations and two unknowns. Solve them. From equation 2b we have:
Having solved for Va, let's compute it. For the values you gave, I get slightly over Va=5.169V. Plugging that into equation 3, I get just over Vb=346mV.
Holy cow, Jon, that's a lot of work you did there... :-)
For the "working man," I'm thinking John's approach or just superposition is rather easier. E.g., for superposition:
Vb due to 8.4V source: 8.4*100/(100+570+240) = 923mV Vb due to 5mA flowing out of Vb node: -100*.005*(570+240)/(100+570+240) = -445mV Vb due to 5mA flowing out of Va node: -100*.005*240/(100+570+240) = -132mV
Add'em all up, and you get 346mV
For Va, working back up the ladder you get Vb+(.005+Vb/100)*570=5.17V. For double-checking, you can work from the top down...
8.4-240*(.005+.005+Vb/100)=5.17V as well.
John's approach perhaps has the best combination of "fast" and "lowest chance of error."
I suppose another approach is to just write a single node voltage equation and solve for the current in R3: 8.4 = 100*I + 570*(I+.005) +
240)*(I+.005+.005) --> 8.4 = (100+570+240)*I + (570*.005+240*(.005+.005)) =
910*I+5.25 --> I=3.46mA. Hmm, this approach has a lot to recommend it -- it's easier to see how it would be extended to 4 or 5 or more resistors and (potentially unequal) currents, and then you just work your way back up after getting the bottom-most current.
This could be a fun interview question, just to see what approach someone uses.
I think you could convert the current sources to Thevenin sources, and then use the Millman theorem to get a system of 2 equations for the two voltages.
I used superposition. The concept depends upon it. Just not the same setup.
The method I illustrated is also how spice sets up its own solutions. An approach I've come to appreciate. It is something very hard to confuse and allows a narrow focus upon each node without consideration of the rest. I happen to like narrowing down the field of view. Also, it sets up your matrix beautifully, should you be dealing with something a little more like a Y or delta or nasty combinations of that plus not just resistors but currents and voltages all ascatter. Not having to _think_ about too much at once is really very nice.
Um. You are _working_, too!!
I prefer the view I laid out. I've found it bullet-proof to missing some detail since all you need to look at are the paths right in front of your face, one node at a time. I've made many fewer mistakes that way.
I suppose so! ;)
Just by way of closing this, once I mastered the "spill" viewpoint, I've found I can mechanically solve quite a wide variety of problems. It turns out that the exact same approach works in solving a variety of electric charge and field questions of rather complex structures (not just spheres, but those one might find in an electric gun with the Wehnelt cap and hole, etc. It's a thinking tool that solves many _fundamental_ problems one may encounter. Of course, it also works with the infinite resistor problem, too, and allows an brain-dead easy setup of a solution for it that you simply cannot fail to get right.
V1 is known (8.40V), so this is 2 equations in two unknowns and easily solved for V2 = 5.169230769 V, V3=0.346153846 V.
Best regards, Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
Hey there, we wouldn't want some question of the correct value of the NINETH digit on the 'ol Agilent 3458A, now would we.. ;-)
Best regards, Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
Well, eight and a half, so not directly applicable in this particular case, but it's not totally ridiculous calculating DC voltages to 10 or
11 places when such tools are at hand. Certainly to 7 places.
Best regards, Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
Everyone is making it too difficult. Just write it down in sequence, the answer falls right out...
formatting link
At MIT, I was spared (*) from Guillemin's obtuseness, I had Harry B. Lee for passive circuit analysis ;-)
(*) I was awarded honors at the end of freshman year and went into course VI-B (just six of us, made for nice class sizes :-) ...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...
That's the same as what I ended up doing (in the later part of my post), except that you use "V3/100" rather than just "I" for the current in the lowest resistor. :-)
I had a guy who was a pretty talented teacher (he'd won a couple awards for it, and I found him quite understandable), although he had little if any real-world design experience.
I'd read some of Guillemin's book, and while I think the guy was pretty darned sharp, I disagree with his notion that you have to have an incredibly thorough understanding of network analysis down pat before you can get useful circuit design or analysis done. Something like his "Introductory Circuit Theory" strikes me as a good graduate-level course! (Be weary of any book with the term "introductory" in its title...)
My opinion here is that in the real world your approach or mine (or superposition if you insist) tends to be rather more productive and insightful than Jon's somewhat pedantic approach. As soon as you go down the path of, "well, this is how SPICE sets up the equations," I think you're largely giving up on gleaning insight directly from the equations themselves and have to run sweeps or Monte Carlo simulations to get some back.
I had a fields & waves professor who thought I was good enough he let me sign up for a graduate level EM course as a sophomore. Let's just say I was a bit lost through much of it... :-)
Did you get to do the problems where, with a linear network, you'd transform it to a "dual" network and then be able to directly compute 1st-order sensitivities (e.g., dV_output_of_a_small_signal_amp/dC_be or similar -- the first homework problem is usually dV_output_of_a_simple_voltage_divider/dR_esistor_in_one_leg since it's easy to compute and compare directly) by solving for the various node voltages and loop currents? I thought that was amazingly cool when they taught it to us; I rather doubt it would have ever occurred to me that you could get such powerful results with such simple circuit machinations.
I actually take a very different view. And I'll tell you why. When the infinite resistor grid problem showed up here for the first time _I_ saw it (and it has, more than once, as I participated in at least two such instances), the method for setting up the closed solution integral as well as forumlating a trivial numerical solution was obvious to me _because_ of that "pedantic approach." It provides a very broadly applicable method that applies across many fields and provides a useful thinking tool that will serve well no matter where you find yourself.
None of this takes away from what you are saying, either. I just don't think you should belittle such a powerful tool.
Yabbut... few practicing engineers encounter the inifnite resistor problem in real-world problems, whereas the "three resistors with current draws" is something much closer to a real-world problem and, in fact, it used to be a
100% real-world problem back when tube radios would have a long, multi-tapped wirewound resistor that supplied different voltages to different parts of the circuit (...long before just giving everything its own regulated rail was viable).
Oh, I think SPICE is great; I expect that in a given day for, e.g., Jim, well over 90% of it is spent in SPICE with less than 10% spent doing algebra long-hand to try to analyze or design circuits. So I certainly wouldn't suggest colleges drop coverage of the standard node voltage/loop current matrix approach that's been taught for many decades, rather I'd add emphasis on not-so-commonly-taught-but-useful concepts such as inverted poles and zeroes that are quite helpful when resorting to long-hand algebra. (And I'd find the time to teach this in that these days in a circuirt analysis class you're not going to have to sit there and eliminate variables in a set of simultaneous equations one by one when any decent calculator today will solve for all the unknowns in well under a second.)
Not having the benefit of a formal training in electronics, I'd solve the problem by this method or some other approach using the same method of logical derivation.
I1 = I2 + 5mA = I3 + 10mA
I3 = (8.4-I1*R1-I2*R2)/R3
= (8.4 - (I3+10mA)*R1 - (I3+5mA)*R2)/R3
I3*R3 = 8.4 - (I3+10mA)*R1 - (I3+5mA)*R2
= 8.4 - I3*R1 - 10mA*R1 - I3*R2 - 5mA*R2
I3*(R3+R1+R2) = 8.4 - 10mA*R1 - 5mA*R2
I3*910 = 8.4 - 2.4 - 2.85 = 3.15
I3 = 3.15/910 = 3.46153846mA
I1 = 13.46153846mA
I2 = 8.46153846mA
The voltages at the two tap points come by subtracting the voltage drops across R1 and R2.
Upper tap = 5.1692307696V
Lower tap = 0.3461538474V, also = I3*R3
Variations to this method, but using the same logic, would be to derive one of the unknown voltages first instead of I3.
A general formula for any values of "external curents" and any resistance could be composed by using variables instead of numerical values.
This approach may be cumbersome, plodding and mechanical, but it does the job without relying on any established theorem except Ohm's law, simple arithmetic and logical thinking.
It can also be expanded for any number of resistors and tapped currents.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.