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I gotta say, this doesn't seem to be all that simple of a problem.
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Quote from: [Serious] on January 16, 2017, 02:01:33 AMQuote from: Lugubert on January 15, 2017, 09:44:51 PMQuote from: Dave Hawkins on January 15, 2017, 06:20:02 PMAll the R values are knowns. That is, the goal was to write the voltages in terms of R's only which would be given.I'm with Dave. That it could be interpreted in another way never occurred to me.Really doesn't work. Defining an unknown in terms of other unknowns doesn't change anything to a known
Quote from: Lugubert on January 15, 2017, 09:44:51 PMQuote from: Dave Hawkins on January 15, 2017, 06:20:02 PMAll the R values are knowns. That is, the goal was to write the voltages in terms of R's only which would be given.I'm with Dave. That it could be interpreted in another way never occurred to me.Really doesn't work. Defining an unknown in terms of other unknowns doesn't change anything to a known
Quote from: Dave Hawkins on January 15, 2017, 06:20:02 PMAll the R values are knowns. That is, the goal was to write the voltages in terms of R's only which would be given.I'm with Dave. That it could be interpreted in another way never occurred to me.
All the R values are knowns. That is, the goal was to write the voltages in terms of R's only which would be given.
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I was just a lowly radio tech, but isn't that just a DC version of a Wien Bridge?
Quote from: Testy Calibrate on January 16, 2017, 09:54:15 PMI gotta say, this doesn't seem to be all that simple of a problem. Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know?
And he wrote 10 equations with 10 unknowns
... blindfolded with his hands tied behind his back! Honest. Just ask him.
And his simple solution is comin' right up.
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Jesus Christ.
Quote from: Dean W on January 16, 2017, 09:41:54 PMThis is your KCL for nodes 2 and 5, with current directions per my diagram. I don't think they are independent because of the shared (V_{5} - V_{2})/R_{3} term.1: i_{0} + i_{3} = i_{2,4} (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: i_{1,5} = i_{3} + i_{6} (1 - V_{5})/R_{1,5} = (V_{5} - V_{2})/R_{3} + V_{5}/R_{6}Rearrange equation 2.1: (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: (1 - V_{5})/R_{1,5} - (V_{5} - V_{2})/R_{3} = V_{5}/R_{6}Add them and the (V_{5} - V_{2})/R_{3} term drops out. Then you are left with:(1 - V_{2})/R_{0} + (1 - V_{5})/R_{1,5} = V_{2}/R_{2,4} + V_{5}/R_{6}That's one equation with two unknowns. At best you can get a proportional relationship between V_{2} and V_{5}, but I don't think you can get absolute values for them without another independent equation. Again, circuit shit is not my forte and I acknowledge that I could be missing something here.The two equations you have are independent. The fact that both have a (V2 - V5) term grouped together is irrelevant; all that matters is that they don't fully cancel for any nontrivial manipulation. Try simply writing the equations you have using the resistances to make coefficients for V2 and V5, using distribution to separate the V2 - V5 term.
This is your KCL for nodes 2 and 5, with current directions per my diagram. I don't think they are independent because of the shared (V_{5} - V_{2})/R_{3} term.1: i_{0} + i_{3} = i_{2,4} (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: i_{1,5} = i_{3} + i_{6} (1 - V_{5})/R_{1,5} = (V_{5} - V_{2})/R_{3} + V_{5}/R_{6}Rearrange equation 2.1: (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: (1 - V_{5})/R_{1,5} - (V_{5} - V_{2})/R_{3} = V_{5}/R_{6}Add them and the (V_{5} - V_{2})/R_{3} term drops out. Then you are left with:(1 - V_{2})/R_{0} + (1 - V_{5})/R_{1,5} = V_{2}/R_{2,4} + V_{5}/R_{6}That's one equation with two unknowns. At best you can get a proportional relationship between V_{2} and V_{5}, but I don't think you can get absolute values for them without another independent equation. Again, circuit shit is not my forte and I acknowledge that I could be missing something here.
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Quote from: Dean W on January 16, 2017, 09:41:54 PMSpoiler (click to show/hide)This is your KCL for nodes 2 and 5, with current directions per my diagram. I don't think they are independent because of the shared (V_{5} - V_{2})/R_{3} term.1: i_{0} + i_{3} = i_{2,4} (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: i_{1,5} = i_{3} + i_{6} (1 - V_{5})/R_{1,5} = (V_{5} - V_{2})/R_{3} + V_{5}/R_{6}Rearrange equation 2.1: (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: (1 - V_{5})/R_{1,5} - (V_{5} - V_{2})/R_{3} = V_{5}/R_{6}Add them and the (V_{5} - V_{2})/R_{3} term drops out. Then you are left with:(1 - V_{2})/R_{0} + (1 - V_{5})/R_{1,5} = V_{2}/R_{2,4} + V_{5}/R_{6}That's one equation with two unknowns. At best you can get a proportional relationship between V_{2} and V_{5}, but I don't think you can get absolute values for them without another independent equation. Again, circuit shit is not my forte and I acknowledge that I could be missing something here.The two equations you have are independent. The fact that both have a (V2 - V5) term grouped together is irrelevant; all that matters is that they don't fully cancel for any nontrivial manipulation. Try simply writing the equations you have using the resistances to make coefficients for V2 and V5, using distribution to separate the V2 - V5 term.
Spoiler (click to show/hide)This is your KCL for nodes 2 and 5, with current directions per my diagram. I don't think they are independent because of the shared (V_{5} - V_{2})/R_{3} term.1: i_{0} + i_{3} = i_{2,4} (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: i_{1,5} = i_{3} + i_{6} (1 - V_{5})/R_{1,5} = (V_{5} - V_{2})/R_{3} + V_{5}/R_{6}Rearrange equation 2.1: (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: (1 - V_{5})/R_{1,5} - (V_{5} - V_{2})/R_{3} = V_{5}/R_{6}Add them and the (V_{5} - V_{2})/R_{3} term drops out. Then you are left with:(1 - V_{2})/R_{0} + (1 - V_{5})/R_{1,5} = V_{2}/R_{2,4} + V_{5}/R_{6}That's one equation with two unknowns. At best you can get a proportional relationship between V_{2} and V_{5}, but I don't think you can get absolute values for them without another independent equation. Again, circuit shit is not my forte and I acknowledge that I could be missing something here.
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Spoiler (click to show/hide)Quote from: uncool on January 16, 2017, 11:44:14 PMQuote from: Dean W on January 16, 2017, 09:41:54 PMThis is your KCL for nodes 2 and 5, with current directions per my diagram. I don't think they are independent because of the shared (V_{5} - V_{2})/R_{3} term.1: i_{0} + i_{3} = i_{2,4} (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: i_{1,5} = i_{3} + i_{6} (1 - V_{5})/R_{1,5} = (V_{5} - V_{2})/R_{3} + V_{5}/R_{6}Rearrange equation 2.1: (1 - V_{2})/R_{0} + (V_{5} - V_{2})/R_{3} = V_{2}/R_{2,4}2: (1 - V_{5})/R_{1,5} - (V_{5} - V_{2})/R_{3} = V_{5}/R_{6}Add them and the (V_{5} - V_{2})/R_{3} term drops out. Then you are left with:(1 - V_{2})/R_{0} + (1 - V_{5})/R_{1,5} = V_{2}/R_{2,4} + V_{5}/R_{6}That's one equation with two unknowns. At best you can get a proportional relationship between V_{2} and V_{5}, but I don't think you can get absolute values for them without another independent equation. Again, circuit shit is not my forte and I acknowledge that I could be missing something here.The two equations you have are independent. The fact that both have a (V2 - V5) term grouped together is irrelevant; all that matters is that they don't fully cancel for any nontrivial manipulation. Try simply writing the equations you have using the resistances to make coefficients for V2 and V5, using distribution to separate the V2 - V5 term.Please go ahead and write out the solution.
Spoiler (click to show/hide)Solving it:1) Let's make a simplified circuit that's equivalent, by "combining" the edges denoted with 2 and 4, and the edges 1 and 5. This allows us to break it down to dealing with only 5 currents. The circuit will have 5 wires.The vertices will be denoted 0, 1, 2, and 3, with 0 as the "high voltage" end, 3 as the "low voltage" end, and 1 and 2 connected in the middle (corresponding to vertices 2 and 5 in the original diagram). The edges will be 1 from 0 to 1, 2 from 0 to 2, 3 from 1 to 2, 4 from 1 to 3, and 5 from 2 to 3. The equations we get are then:I1 - I3 - I4 = 0 (Total current through 1 is 0)I2 + I3 - I5 = 0 (Total current through 2 is 0)R1 I2 - R2 I2 + R3 I3 = 0 (Voltage around first loop is 0)R3 I3 - R4 I4 + R5 I5 = 0 (Voltage around second loop is 0)R1 I1 + R4 I4 = 1 (Voltage from 0 to 3 is 1)The problem challenges us to find I5. We use Cramer's rule, and get (according to Wolfram Alpha)(R1 R3 + R1 R4 + R2 R4 + R3 R4 + R1 R2 R5 + R1 R3 R5 + R1 R4 R5 + R2 R4 R5 + R3 R4 R5)/(R1 R2 R3 + R1 R2 R4 + R1 R2 R5 + R1 R3 R5 + R1 R4 R5 + R2 R3 R4 + R2 R4 R5 + R3 R4 R5)If you want to rephrase it in terms of the original problem: my R1 is the original R0, my R2 is the original R1 + R5, my R3 is the original R3, my R4 is the original R2 + R4, and my R5 is the original R6.
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Quote from: Testy Calibrate on January 16, 2017, 09:54:15 PMI gotta say, this doesn't seem to be all that simple of a problem. Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know? And he wrote 10 equations with 10 unknowns ... blindfolded with his hands tied behind his back! Honest. Just ask him. And his simple solution is comin' right up.
Quote from: Dave Hawkins on January 17, 2017, 04:28:47 AMQuote from: Testy Calibrate on January 16, 2017, 09:54:15 PMI gotta say, this doesn't seem to be all that simple of a problem. Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know? And he wrote 10 equations with 10 unknowns ... blindfolded with his hands tied behind his back! Honest. Just ask him. And his simple solution is comin' right up.Dave, even I know that a simple phenomenon isn't necessarily a "simple problem" to solve.I mean, the four colour map problem hasn't been solved yet. Or has it?