Ohm's Law Problem
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From: Joel Koltner on 15 Feb 2010 12:36 "Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote in message news:11sin51vdubl64ouhcgn3b2omf36q111in(a)4ax.com... > Everyone is making it too difficult. Just write it down in sequence, > the answer falls right out... > http://analog-innovations.com/SED/OhmsLaw_SED_JustWriteItDown.pdf 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. :-) > At MIT, I was spared (*) from Guillemin's obtuseness, I had Harry B. > Lee for passive circuit analysis ;-) 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 was awarded honors at the end of freshman year and went into > course VI-B (just six of us, made for nice class sizes :-) 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. ---Joel
From: Paul Hovnanian P.E. on 15 Feb 2010 13:00 Rich Webb wrote: > On Mon, 15 Feb 2010 00:48:46 -0500, Spehro Pefhany > <speffSNIP(a)interlogDOTyou.knowwhat> wrote: > >>On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned Bill Bowden >><wrongaddress(a)att.net> wrote: >> >>>Ohm's Law Problem: >>> >>> 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. >>> >>> V1 +8.4 >>> | >>> R1 = 240 >>> | >>> V2 .---------> 5 mA >>> | >>> R2 = 570 >>> | >>> V3 .---------> 5 mA >>> | >>> R3 = 100 >>> | >>> | >>> GND >> >> >> >>(V1-V2)/R1 = 5mA + (V2-V3/R2 >>(V2-V3)/R2 = 5mA + V3/R3 >> >>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. > > The Curse of the Calculator strikes again! > > I3 = 3.5 mA, V3 = 0.35 V, and V2 = 5.2 V. Hard to justify more than two > significant figures here. > This calls for a slide rule. -- Paul Hovnanian paul(a)hovnanian.com ---------------------------------------------------------------------- Have gnu, will travel.
From: Jon Kirwan on 15 Feb 2010 14:25 On Mon, 15 Feb 2010 09:36:26 -0800, "Joel Koltner" <zapwireDASHgroups(a)yahoo.com> wrote: >"Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote in >message news:11sin51vdubl64ouhcgn3b2omf36q111in(a)4ax.com... >> Everyone is making it too difficult. Just write it down in sequence, >> the answer falls right out... >> http://analog-innovations.com/SED/OhmsLaw_SED_JustWriteItDown.pdf > >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. :-) > >> At MIT, I was spared (*) from Guillemin's obtuseness, I had Harry B. >> Lee for passive circuit analysis ;-) > >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. ><snip> 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. Jon
From: Joel Koltner on 15 Feb 2010 14:43 Hi Jon, "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message news:pe7jn5pnjc9sq679eck53hsd2kbbhjned1(a)4ax.com... [infinite resistor problem] > 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. 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). > None of this takes away from what you are saying, either. I > just don't think you should belittle such a powerful tool. 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.) ---Joel
From: pimpom on 15 Feb 2010 15:17
Bill Bowden wrote: > Ohm's Law Problem: > > 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. > > +8.4 > | > R1 = 240 > | > .---------> 5 mA > | > R2 = 570 > | > .---------> 5 mA > | > R3 = 100 > | > | > GND > > > 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. > > +6.9 > | > R1 =240 > | > .----------> 5 mA > | > R2 =570 > | > GND > > 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. > > Any ideas? > > -Bill 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. |