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| How can this be? (picture) |
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| ebastler:
--- Quote from: billbyrd1945 on February 17, 2019, 02:00:50 pm ---Okay. I'm thinking that a rule would be: When using conventional flow, the series resistor's output cannot be divided on its way back to the power source (or the next component). And such a resistor (or component having resistance) would be calculated in the analysis as a whole (non-inverted) value. In this case, the value would be 1000, not 1/1000. --- End quote --- No, sorry, you still got this wrong. It does not matter at all what comes after the series resistor's output. There could be another complex network following the output of R5 in your example (including a division into multiple paths); and R5 would still be in series with both, the complex network in front of its input, and the complex network behind its output. The only thing that matters is that there is no "bypass" in parallel with R5, where the current could flow without going through R5. Sometimes we develop a mental block, and you seem to be in this situation -- maybe overthinking things. Perhaps you should ask that 12-year-old granddaughter of yours to explain it in her words. ;) |
| Brumby:
I think the OP needs to stick to resistance - forget conductance altogether. (Yes, it can make the math easier, but it is confusing the hell out of understanding the concept!) Resistors in series: Rtot = Ra + Rb Resistors in parallel: Rtot = 1/( 1/Ra + 1/Rb) Please - just stick to these equations. |
| Brumby:
Process: (A) Solve for R1 and R3 in series (B) Solve for R2 and R4 in series (C) Solve for (A) and (B) in parallel (D) Solve for (C) and R5 in series The value for (D) is the equivalent resistance for the circuit - and you can get the current flowing, using Ohm's Law. If you then want to find out what current is flowing within the network: Using the current calculated above, you can find the voltage across R5 (Ohm's Law again) You can then work out the voltage that is across (A) and (B) and, once more, using Ohm's Law, work out the current passing through each. Note: These 2 current values should add up to the total current. You can then work out the voltages across each resistor: R1, R2, R3 and R4. |
| billbyrd1945:
Okay. Believe it or not, you provided the rule with this "The only thing that matters is that there is no "bypass" in parallel with R5, where the current could flow without going through R5. Thank you. I'm done. Thanks to everyone. Sorry if I seem a little hard headed. :D |
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