Ohm's Law Problem
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From: Jon Kirwan on 14 Feb 2010 20:40 Here are the commands I used and the results: g1=1/240 g2=1/570 g3=1/100 vc=8.4 gz=g2+g3 va=(vc*g1-5e-3*(1+g2/gz))/(g1+g2-g2^2/gz) vb=(va*g2-5e-3)/gz va 5.169230769 vb 0.3461538462 Jon
From: Joel Koltner on 14 Feb 2010 21:22 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. ---Joel
From: Bitrex on 14 Feb 2010 23:02 Joel Koltner wrote: > 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. > > ---Joel 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.
From: Jon Kirwan on 14 Feb 2010 23:11 On Sun, 14 Feb 2010 18:22:51 -0800, "Joel Koltner" <zapwireDASHgroups(a)yahoo.com> wrote: >Holy cow, Jon, that's a lot of work you did there... :-) Nah. It just flows. Done it too many times. >For the "working man," I'm thinking John's approach or just superposition is >rather easier. E.g., for superposition: 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. >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 Um. You are _working_, too!! >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 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 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. > >---Joel 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. Jon
From: Spehro Pefhany on 15 Feb 2010 00:48
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. 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 |