Author Topic: How can this be? (picture)  (Read 4190 times)

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Offline billbyrd1945Topic starter

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Re: How can this be? (picture)
« Reply #25 on: February 17, 2019, 12:52:59 pm »
To ebastler: No, I understand that R1 and R2 are in series within a parallel branch and therefore added together and then inverted (2 steps). Ditto for R3 and R4. And I intuitively know that R5 would be a series resistor outside the parallel branches. But how would I explain to my 12 year old granddaughter how to rule out R5 as a parallel circuit resistor? How would she know to use it's value rather than its inverse? Thank you
 

Offline ebastler

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Re: How can this be? (picture)
« Reply #26 on: February 17, 2019, 12:59:13 pm »
To ebastler: No, I understand that R1 and R2 are in series within a parallel branch and therefore added together and then inverted (2 steps). Ditto for R3 and R4. And I intuitively know that R5 would be a series resistor outside the parallel branches. But how would I explain to my 12 year old granddaughter how to rule out R5 as a parallel circuit resistor? How would she know to use it's value rather than its inverse? Thank you

Well, all of the current that wants to flow from one battery terminal to the other needs to flow through R5. There is no alternative (parallel) path to R5 which the current could take. The current first has to find its way through the resistor network made of R1 to R4, and then (afterwards, sequentially, serially) flows though R5. So R5 is in series with the other resistor network.
 

Offline Brumby

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Re: How can this be? (picture)
« Reply #27 on: February 17, 2019, 01:06:35 pm »
But how would I explain to my 12 year old granddaughter how to rule out R5 as a parallel circuit resistor? How would she know to use it's value rather than its inverse? Thank you

RE-DRAW THE CIRCUIT - in the 'conventional' format: Power rails at top and bottom - circuit in between, working left to right

eblaster has done this for you:
 

Offline ebastler

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Re: How can this be? (picture)
« Reply #28 on: February 17, 2019, 01:08:22 pm »
To add to my prior post:

I still like the "plumbing" analogy to explain such simple circuits. Think of your circuit as a system of water pipes. "Current" is the flow rate of the water passing through a pipe (gallons per minute, liters per second or whatever). "Voltage" is the water pressure you apply (e.g. via an elevated reservoir and a pump which brings the water back up to the reservoir).  "Resistors" are narrow sections of tubing which constrain the flow.

If your force the water to flow through two narrow sections in series, their resistances will add up -- for a given applied pressure, less water will flow through that circuit. If you provide two parallel paths to the water, the water can chose either way. You will get more water flow through those two paths combined than through each individual path. In other words, each parallel path can "conduct" water -- their conductivities (inverse resistances) add up.
 

Offline billbyrd1945Topic starter

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Re: How can this be? (picture)
« Reply #29 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.
 

Offline ebastler

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Re: How can this be? (picture)
« Reply #30 on: February 17, 2019, 02:09:52 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.

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.  ;)
 

Offline Brumby

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Re: How can this be? (picture)
« Reply #31 on: February 17, 2019, 02:28:59 pm »
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.
 

Offline Brumby

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Re: How can this be? (picture)
« Reply #32 on: February 17, 2019, 02:39:30 pm »
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.
« Last Edit: February 17, 2019, 02:42:38 pm by Brumby »
 

Offline billbyrd1945Topic starter

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Re: How can this be? (picture)
« Reply #33 on: February 17, 2019, 04:45:17 pm »
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
 
The following users thanked this post: Brumby, ebastler, tooki


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