Ohm's Law Problem
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From: Rich Webb on 15 Feb 2010 16:25 On Tue, 16 Feb 2010 01:47:20 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >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. Guys, it's not that hard: 8.4 = 100 * I + 570 * (I + 0.005) + 240 * (I + 0.010) and solve for I. One equation, one unknown. Bada-bing. >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 If I was grading this, I'd give credit for the approach but I'd probably subtract points for carrying out the result to an unreasonable number of significant figures. I3 here is *not* 3.46153846mA, it's 3.5 mA. That's as far as the precision is known; farther than that is just making stuff up. Not picking on you in particular, pims. It's an all too common practice nowadays, as evidenced by most of the other, similar, answers in the thread. -- Rich Webb Norfolk, VA
From: Jim Thompson on 15 Feb 2010 16:40 On Mon, 15 Feb 2010 13:29:30 -0800, Joerg <invalid(a)invalid.invalid> wrote: >Joel Koltner 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 >> > >Ohm's law? I thought that had been repealed ... :-) > > >> 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. ... > > >Sadly, that's the kind of notion that drives potential EE candidates >away, at least from analog. And now we have a serious shortage of those. >They think they have to be a rocket scientists to be able to thrive and >make money in analog. Which is wrong. I learned the majority of my >skills by "winging it". IOW I built RF stuff before I knew squat about >any of that. And it actually worked, some still does. > >Note to potential candidates: If an author or professor says something >like what must have been stipulated in this book, that you must be a top >notch network analyst, do not listen. It's much more important to >experiment, experiment, experiment, get a "feel" for what works, _then_ >dive into the theory. Not the other way around. Just my 2 cents. > >Whew. Now I feel better ... I always vote for too few analog engineers ;-) Virtually everything I attack, I have no initial clue about how to make it work. That's what makes it fun ;-) ...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.
From: Jim Thompson on 15 Feb 2010 17:04 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...) Indeed! Be "weary". As a member of the MIT Honors (electrical engineering) VI-B curriculum I had the joy of choosing _extra_ electives, so be _wary_ of courses with elementary in the title, such as "Elementary Number Theory" , I had an "F" at mid-term :-(, "B" at the end... what a ride :-( > >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. Yep. I always prefer to reach in there and find the obvious, and then work outward from there. > >> (*) 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 I am quite familiar with duality, but I don't find it all that useful in actual practice. I probably do more "Wye" to "Delta" (and vice versa) transformations if I'm working a difficult network. I'm a bit of an Algebra nut, so none of this fazes me very much ;-) ...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.
From: pimpom on 15 Feb 2010 17:04 Rich Webb wrote: > On Tue, 16 Feb 2010 01:47:20 +0530, "pimpom" > <pimpom(a)invalid.invalid> > wrote: > >> 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. > > Guys, it's not that hard: > > 8.4 = 100 * I + 570 * (I + 0.005) + 240 * (I + 0.010) and solve > for I. > One equation, one unknown. Bada-bing. Yep. That's the kind of general formula I said later on (in the part you snipped) that can be put together and expanded for general use. I just presented it as quoted to illustrate the logic of 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 > > If I was grading this, I'd give credit for the approach but I'd > probably subtract points for carrying out the result to an > unreasonable number of significant figures. > > I3 here is *not* 3.46153846mA, it's 3.5 mA. That's as far as > the > precision is known; farther than that is just making stuff up. > > Not picking on you in particular, pims. It's an all too common > practice nowadays, as evidenced by most of the other, similar, > answers in the thread. That's OK. The 9-digit precision is just for illustration. A practice encouraged, in part, by easy access to electronic aids. Still, for a practical design, I'd probably use somewhat more precision than two significant figures. Just how many digits would depend on how the results are to be used. To determine the operating point of a BJT linear amplifier for example, I'd probably use 3.46mA. If the result is to be used in a few more steps of calculation where errors can become cumulative, I may use 3.4615mA. Where relatively small differences can be important, I may use even more digits.
From: life imitates life on 15 Feb 2010 17:06
On Mon, 15 Feb 2010 16:38:42 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Mon, 15 Feb 2010 12:24:59 -0800, the renowned life imitates life ><pasticcio(a)thebarattheendoftheuniverse.org> wrote: > >>On Mon, 15 Feb 2010 10:49:56 -0500, Spehro Pefhany >><speffSNIP(a)interlogDOTyou.knowwhat> wrote: >> >>>On Mon, 15 Feb 2010 07:22:37 -0800, the renowned life imitates life >>><pasticcio(a)thebarattheendoftheuniverse.org> wrote: >>> >>>>On Mon, 15 Feb 2010 09:46:23 -0500, Spehro Pefhany >>>><speffSNIP(a)interlogDOTyou.knowwhat> wrote: >>>> >>>>>On Mon, 15 Feb 2010 06:35:19 -0500, the renowned Rich Webb >>>>><bbew.ar(a)mapson.nozirev.ten> 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. >>>>> >>>>>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 >>>> >>>> >>>> Nineth? >>>> >>>> Doth thee haveth a nineth digit? >>> >>>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 >> >> >> The word is Ninth. And no, that meter doesn't have that many either. > >http://www.eefocus.com/data/myspace/3/17997//blog/2a65afe9.jpg > It is an 8 and a half digit meter. |