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Quote from: Martin.au on January 14, 2017, 05:00:03 PMX+y = 10How many unknowns? oh Jesus. Now Martin is going to school me on matrix algebra. Go ahead, Martin, come up with 10 independent equations. School me.
X+y = 10How many unknowns?
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Quote from: Dave Hawkins on January 14, 2017, 05:07:19 PMQuote from: Martin.au on January 14, 2017, 05:00:03 PMX+y = 10How many unknowns? oh Jesus. Now Martin is going to school me on matrix algebra. Go ahead, Martin, come up with 10 independent equations. School me. That's not matrix algebra. That's a simple linear equation.With how many unknowns?
Quote from: Martin.au on January 14, 2017, 05:17:47 PMQuote from: Dave Hawkins on January 14, 2017, 05:07:19 PMQuote from: Martin.au on January 14, 2017, 05:00:03 PMX+y = 10How many unknowns? oh Jesus. Now Martin is going to school me on matrix algebra. Go ahead, Martin, come up with 10 independent equations. School me. That's not matrix algebra. That's a simple linear equation.With how many unknowns?4 But feel free to use 10 if you like. I am anxious to see how you will come up with 10 independent equations to solve it.
Here's the answer to some of the questions I just posed, Davie-face-planter:Spoiler (click to show/hide)Six nodes, one Kirchoff's Current Law for each node is... six linear equations.Four closed loops, one Kirchoff's Voltage Law per loop is ... four linear equations.Ten linear equations total.With a unique solution in terms of R1, R2, ... R6.I'll leave it for you to expound on why this is.
Quote from: Dave Hawkins on January 14, 2017, 05:25:51 PMQuote from: Martin.au on January 14, 2017, 05:17:47 PMQuote from: Dave Hawkins on January 14, 2017, 05:07:19 PMQuote from: Martin.au on January 14, 2017, 05:00:03 PMX+y = 10How many unknowns? oh Jesus. Now Martin is going to school me on matrix algebra. Go ahead, Martin, come up with 10 independent equations. School me. That's not matrix algebra. That's a simple linear equation.With how many unknowns?4 But feel free to use 10 if you like. I am anxious to see how you will come up with 10 independent equations to solve it. Lol. Now there's a big problem with Hawkinsing other people's posts.According to Dave, x + y = 10 has 4 unknowns.Read things carefully Dave.
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Quote from: Martin.au on January 14, 2017, 05:31:01 PMQuote from: Dave Hawkins on January 14, 2017, 05:25:51 PMQuote from: Martin.au on January 14, 2017, 05:17:47 PMQuote from: Dave Hawkins on January 14, 2017, 05:07:19 PMQuote from: Martin.au on January 14, 2017, 05:00:03 PMX+y = 10How many unknowns? oh Jesus. Now Martin is going to school me on matrix algebra. Go ahead, Martin, come up with 10 independent equations. School me. That's not matrix algebra. That's a simple linear equation.With how many unknowns?4 But feel free to use 10 if you like. I am anxious to see how you will come up with 10 independent equations to solve it. Lol. Now there's a big problem with Hawkinsing other people's posts.According to Dave, x + y = 10 has 4 unknowns.Read things carefully Dave. i'm reading just fine, my friend. And I'm waiting on you to show me the 10 independent equations that you are going to use to solve for your 10 unknowns.Lol
Hey suit yourself. Solve it anyway you like. But solve it.
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Quote from: VoxRat on January 14, 2017, 04:35:16 PMQuote from: Dave Hawkins on January 14, 2017, 04:28:58 PMQuote from: uncool on January 14, 2017, 04:07:39 PMFun fact, Dave: solving this circuit is an application of linear algebra.True. And I'm waiting on Mr. Smart Guy to explain why he thought there were 10 unknowns. Quote from: JonF on January 14, 2017, 01:41:44 PMThere are six nodes with unknown currents and four nodes with unknown voltages. Ten unknowns.Well, dave...ARE there six nodes with unknown currents and four nodes with unknown voltages?Or are there not?No there are only four true unknowns - the voltages of those four nodes. If you know those four voltages then everything else can be determined. Yeah you could pick four DIFFERENT unknowns if you like but you are still only going to have four true unknowns because everything else can be calculated once you find these plus the resistor values given.
Quote from: Dave Hawkins on January 14, 2017, 04:28:58 PMQuote from: uncool on January 14, 2017, 04:07:39 PMFun fact, Dave: solving this circuit is an application of linear algebra.True. And I'm waiting on Mr. Smart Guy to explain why he thought there were 10 unknowns. Quote from: JonF on January 14, 2017, 01:41:44 PMThere are six nodes with unknown currents and four nodes with unknown voltages. Ten unknowns.Well, dave...ARE there six nodes with unknown currents and four nodes with unknown voltages?Or are there not?
Quote from: uncool on January 14, 2017, 04:07:39 PMFun fact, Dave: solving this circuit is an application of linear algebra.True. And I'm waiting on Mr. Smart Guy to explain why he thought there were 10 unknowns.
Fun fact, Dave: solving this circuit is an application of linear algebra.
There are six nodes with unknown currents and four nodes with unknown voltages. Ten unknowns.
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Quote from: Pingu on January 14, 2017, 04:56:57 PMNice to see you getting the rust off those cognitive circuits Dave!You'll be able to work out that nested hierarchy before long!You can't ever give an honest compliment without a spin for your own agenda can you? That's pretty sad.
Nice to see you getting the rust off those cognitive circuits Dave!You'll be able to work out that nested hierarchy before long!
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Quote from: Dave Hawkins on January 14, 2017, 05:03:36 PMQuote from: Pingu on January 14, 2017, 04:56:57 PMNice to see you getting the rust off those cognitive circuits Dave!You'll be able to work out that nested hierarchy before long!You can't ever give an honest compliment without a spin for your own agenda can you? That's pretty sad. actually this is not even a compliment it all because you're telling me that my cognitive circuits have neverworked very well. I think you probably need to get some rust remover for yourself.
Quote from: JonF on January 14, 2017, 01:49:20 PMHere's the answer to some of the questions I just posed, Davie-face-planter:Spoiler (click to show/hide)Six nodes, one Kirchoff's Current Law for each node is... six linear equations.Four closed loops, one Kirchoff's Voltage Law per loop is ... four linear equations.Ten linear equations total.With a unique solution in terms of R1, R2, ... R6.I'll leave it for you to expound on why this is.Uh.There are 2 loops, not 4, and Kirchhoff's law only says something at 4 (maybe 5, depending on how you count the "grounding") of the nodes. 7 equations (5 if you use the 2 obvious substitutions, as I did), 7 variables (or 5 equations, 5 variables).Or, if you work in terms of the voltages, 2 (for my simplified circuit looking only at inner nodes), 4 (in two ways), or 6 (looking at all nodes, with V1 = 1 as an equation).
Quote from: Dave Hawkins on January 14, 2017, 02:04:49 PMSimple circuit my ass ... Reaching way back into my memory banks, I think this is how I would set it up Use KCL for the first three equations and then KVL applied to both loops. There are four unknowns - The unknown voltages of the four nodes - so this should be enough to solve it. That looks right. Since it took you so long to figure out I doubt it came from your memory."Simple circuit" is a technical term and refers to the lack of mathematical complexity, not how much calculation is required. See the definitions I posted. Especially there's no time dependence, only one source, and no nonlinearities. It's just linear equations with an answer guaranteed. Simple.
Simple circuit my ass ... Reaching way back into my memory banks, I think this is how I would set it up Use KCL for the first three equations and then KVL applied to both loops. There are four unknowns - The unknown voltages of the four nodes - so this should be enough to solve it.
Topics are like music for me. Some days I'm "in the mood" for Pavarotti and some days I'm "in the mood" for Willie Nelson. And I have no control over when a particular mood will strike. One of these days, I'll get around to reading Oard's book in the dropbox. And one of these days, I'll return to NH's. Yesterday I was in the mood for dusting off the cobwebs from Kirchoff's Laws.