Ohm's Law Problem
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From: Przemek Klosowski on 16 Feb 2010 01:41 On Sun, 14 Feb 2010 16:12:04 -0800, Bill Bowden wrote: > +8.4 > | > R1 = 240 > | > .---------> 5 mA > | > R2 = 570 > | > .---------> 5 mA > | > R3 = 100 > | > | > GND > Warning: brute force method. The advantage here is that it always works, pretty much regardless of the circuit. The problem has five unknowns: currents through each resistor, and voltages on the node below R1 and R2. Let's denote them i1, i2, i3 and V1 and V2. From quick inspection we can write five Kirchhoff equations: 240*i1 + V1 = 8.4 (voltage drop on R1 plus V1 is the supply voltage) and of course the currents have to add up in each node, i.e.: i1=.005+i2 and i2=.005+i3. Finally, the remaining voltages must be: V1=V2+i2*570 V2=i3*100 So, five equations with five unknowns--can be done by hand, but it's easiest to use a linear algebra program like Octave (or Matlab if you have extra cash and don't have Octave). With little practice it's trivial to just write out the matrices, but for completeness here's how to do it systematically. We gather coefficients for each unknown in the respective column ordered as [i1 i2 i3 V1 V2], and move the free terms to the RHS: 240*i1 + 0 + 0 + V1 +0 = 8.4 1*i1 - 1*i2 + 0 + 0 +0 = .005 0 + 1*i2 -1*i3 + 0 +0 = .005 0 + 0 +100*i3 + 0 -V2 = 0 0 + 570*i2 + 0 - V1 +V2 = 0 From this, it's clear that it is the linear system of equations A*x=B, where x is the vector of unknowns [i1 i2 i3 V1 V2], and A=[240 0 0 1 0 1 -1 0 0 0 0 1 -1 0 0 0 0 100 0 -1 0 570 0 -1 1 ] B=[ 8.40000 .005 .005 0 0]' and solve simply as a set of linear equations, using the command A\B resulting in: 0.0134615 0.0084615 0.0034615 5.1692308 0.3461538 again, these are the three current i1, i2 and i3, and two voltages, V1 and V2. The result is of course the same as everyone else, V1=5.169V and V2=0.346V.
From: Jon Kirwan on 16 Feb 2010 02:15 On Tue, 16 Feb 2010 17:02:19 +1100, "David L. Jones" <altzone(a)gmail.com> wrote: ><snip> >It's much more fun to just built it and measure it: >http://www.eevblog.com/2009/08/15/eevblog-25-the-infinite-resistor-puzzle/ I saw it, Dave! :) And enjoyed very much the time you spend there. Of course, you are a very special person. Not many would just jump in like that and I very much respect what you did and what you offered to others in preparing the video, as well. It was (and is) a generous gesture, well above par. Jon
From: Jon Kirwan on 16 Feb 2010 02:32 On Mon, 15 Feb 2010 13:29:30 -0800, Joerg <invalid(a)invalid.invalid> wrote: ><snip> >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. It's hard for most of us to get a feel for what works without first having some idea of what to expect. Theory is primary to interpreting and understanding experimental result. What isn't known through theory defines the word 'random.' Jon
From: Bill Bowden on 16 Feb 2010 02:39 On Feb 14, 9:48 pm, Spehro Pefhany <speffS...(a)interlogDOTyou.knowwhat> wrote: > On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned Bill Bowden > > > > <wrongaddr...(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. > > Best regards, > Spehro Pefhany > -- > "it's the network..." "The Journey is the reward" > sp...(a)interlog.com Info for manufacturers:http://www.trexon.com > Embedded software/hardware/analog Info for designers: http://www.speff.com Yes, that makes sense, thanks. But the trick is finding the most useful equations. I tend to start with less useful stuff like VR1+VR2+VR3 = 8.4 and then substitute currents for voltages like (IR1 * 240) +(IR2 * 570) +(IR3 * 100) = 8.4 which ends up with 3 unknowns and needs a couple sheets of paper to figure out. But, I misstated the current flow direction which should have been the other way. I get confused with source and sink currents and think of current going from the negative point to the positive, while others think the other way. The circuit has 15mA in the 100 ohm resistor for a voltage of 1.5 at V3 and 10mA in the 570 resistor for a voltage of 5.7+1.5 = 7.2 volts at V2 and the last 5mA goes into the 240 resistor for a voltage drop of 1.2. All equals 8.4 So, using your idea with the current going the other way, (V1-V2)/R1 = (V2-V3/R2) - .005 (V2-V3)/R2 = V3/R3 - .005 V1-V2 = R1*((V2-V3/R2) -.005) V2-V3 = R2*(V3/R3 - .005) V2 = R2 * (V3/R3 - .005) + V3 ( R2 * (V3/R3 - .005) + V3) = R2*(V3/R3 - .005) +V3 R2*V3 / R3 - 2.85 +V3 = R2*(V3/R3 - .005) +V3 (570V3/ 100) - 2.85 = 570* (V3/100 -.005) 5.7V3 - 2.85 = 5.7V3 - 2.85 0V3 -2.85 = -285 And so... 0 = 0 , which doesn't say much. It should be V3 = 1.5 What did I do wrong? I think I forgot the 8.4 volt known supply voltage which should be in here somewhere? I'll look at it again tomorrow. -Bill
From: Joel Koltner on 16 Feb 2010 03:27
"Bill Bowden" <wrongaddress(a)att.net> wrote in message news:b13fe59a-8f59-4458-8b6f-a49f1e118e7f(a)p13g2000pre.googlegroups.com... > What did I do wrong? You never substituted your equation for V2 into the preceeding equation (V1-V2=R1*...) -- you just took your equation for V2 and shuffled both sides around until you got 0=0! (Effectively you had one equation but two unknowns still -- V2 and V3 -- so the "0=0" tells you that there are infinitely many values of V2 and V3 that satisfy the equation. You need the additional constraint of the other equation to obtain a unique solution.) ---Joel |