Ohm's Law Problem
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From: Spehro Pefhany on 15 Feb 2010 18:40 On Mon, 15 Feb 2010 15:02:09 -0800, the renowned life imitates life <pasticcio(a)thebarattheendoftheuniverse.org> wrote: > > The only two values the first digit can have with the others filled up >is zero or one. That is only half a digit in meter speak. I'll leave it >to you, this time around, to figure out the simple reason why. So... if an Engineer makes $175,000 per year, does she have a 6-digit income or a 5-1/2 digit income? Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff(a)interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
From: Jim Thompson on 15 Feb 2010 18:52 On Mon, 15 Feb 2010 09:11:40 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote: >On Sun, 14 Feb 2010 17:25:29 -0800, Jon Kirwan ><jonk(a)infinitefactors.org> wrote: > >>On Sun, 14 Feb 2010 16:12:04 -0800 (PST), 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. >>> >>> Vc=+8.4 >>> | >>> R1 = 240 >>> | >>> Va a---------> 5 mA >>> | >>> R2 = 570 >>> | >>> Vb b---------> 5 mA >>> | >>> R3 = 100 >>> | >>> | >>> Vg GND >> >>><snip> >> >>> Any ideas? >> >>Use nodal analysis. >> >>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: >> >>(3) Vb = (Va*G2 - 5mA) / (G2+G3) >> >>We can now substitute equation 3 into equation 1: >> >>(4a) Va*(G1+G2)+5mA=G2*(Va*G2-5mA)/(G2+G3)+Vc*G1 >>(4b) Va*(G1+G2)=Va*G2^2/(G2+G3)-G2*5mA/(G2+G3)+Vc*G1-5mA >>(4c) Va*(G1+G2)-Va*G2^2/(G2+G3)=Vc*G1-5mA-G2*5mA/(G2+G3) >>(4d) Va*(G1+G2-G2^2/(G2+G3))=Vc*G1-5mA*(1+G2/(G2+G3)) >>(4e) Va*(G1+G2-G2^2/(G2+G3))=Vc*G1-5mA*(1+G2/(G2+G3)) >>(4f) Va=(Vc*G1-5mA*(1+G2/(G2+G3)))/(G1+G2-G2^2/(G2+G3)) >> >>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. >> >>Jon > >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 > >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 Reminiscing over Harry B. Lee I found this... http://www.eecs.mit.edu/facts.html He was my instructor in passive analysis (1959-1960) when he received the Goodwin Medal. ...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: Joel Koltner on 15 Feb 2010 19:01 "Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote in message news:ebnjn5l6ca4nbonndu6dmncqtgqkthhfhv(a)4ax.com... > http://www.eecs.mit.edu/facts.html --> First Master of Engineering degree awarded (1994) My undergraduate alma mater (University of Wisconsin - Madison) is always sending me mail, trying to get me to sign up for this (http://mepp.engr.wisc.edu/). Sounds like a soft-core business degree designed for those who got tired of actual engineering work and wanted to get into management. :-) Of course, tuition is much higher than regular old undergraduates pay... I'm not planning on taking it... ever.
From: life imitates life on 15 Feb 2010 20:23 On Mon, 15 Feb 2010 18:40:58 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Mon, 15 Feb 2010 15:02:09 -0800, the renowned life imitates life ><pasticcio(a)thebarattheendoftheuniverse.org> wrote: > >> >> The only two values the first digit can have with the others filled up >>is zero or one. That is only half a digit in meter speak. I'll leave it >>to you, this time around, to figure out the simple reason why. > >So... if an Engineer makes $175,000 per year, does she have a 6-digit >income or a 5-1/2 digit income? > > >Best regards, >Spehro Pefhany She has a one and three quarter tenths of a million dollar a year salary.
From: Bitrex on 15 Feb 2010 20:36
Joerg 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 ... > Here's a question I'd like an honest answer to: I've studied analog design for a long time as a hobby, but I was never able to attend a university program of study for it during my 20s due to a series of illnesses, and having to work hard when I was well enough to stay afloat. I'm now 30, and I'm fortunate to currently be in a stable enough situation to attempt a degree. I love EE, and would like to study it formally, but if the future is truly as bleak for US engineers (particularly ones who who will be at least 35 before they land their first job) I should probably really let that sink in before starting. |