simpler voltage divider idea

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Alec_t

Joined Sep 17, 2013
12,890
If the resistor has a short across it then the voltage across the resistor will be zero, regardless of the current through the short.
 

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Capernicus

Joined Jun 24, 2022
87
Thankyou for the garnered interest.

You can see another way, is if you have two equally powered resistors on the ends as well.

Yes it is a short over the resistor, but if there was a component on that short, it would have a very low amount of volts on it, (the voltage drop of the resistor.) because its meeting at a double junction either side of the resistor, and there is very little "voltspace" between the 2 ends of the resistor, so you only get the potential difference provided by the length of the resistance wire.

Theres more to this... I need to continue on or ill never get my conduction model done. and falstad will never have a decent competitor.
 

MrChips

Joined Oct 2, 2009
26,481
What you claim is correct if the voltage source has internal resistance.
This becomes equivalent to a 2-resistor voltage divider.
 

MrAl

Joined Jun 17, 2014
9,157

Can u guys concur this for me?
Hi,

In a completely linear circuit that is generally true, but there are a few exceptions.
The first is if the voltage across the resistor is regulated. That would usually mean that the voltage will stay constant unless you overload the voltage regulator, and then it will fall abruptly.

If the circuit is not linear then it's anybody's guess what will happen. The voltage could go up, go down, or even stay the same.
I guess for this question though it would be reasonable to assume that you are talking about a linear circuit as that is how what a lot of circuit theory starts out with.

Oh on more case, if you make the resistance too low in some circuits you will blow the protective fuse and so the voltage will fall to zero :)
That would actually be a case included in the non linear class. A fuse is linear up to the point where it blows, then it is incredibly non linear.

What you say about the short doesnt seem to make any sense though. If you place a short across a resistor the voltage will drop to zero and in power circuits it is most likely going to blow the fuse or circuit breaker. It could easily cause other parts to burn up too.
 

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Capernicus

Joined Jun 24, 2022
87
Ok... I think im onto something, and its just there is more than 1 way to reduce voltage on a circuit with resistors than just that single 2 resistor voltage divider.

There is other topologies which also reduce the volts, and if u haven't handled them, your simulator isn't correct is it?
 

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Capernicus

Joined Jun 24, 2022
87
voltage dividers happen when u have junctions. (split offs and recombinations) when the wires come apart and come back together.

Ohms law designates the current due to the voltage, but voltage dividers and further voltage loss which actually reduces the power at certain points in the circuit, I think its diff!!!
 

BobTPH

Joined Jun 5, 2013
5,227
Just as I thought. None of us has any idea what he is talking about. Now I get the idea that he has simulated something if Falstad and does not think the simulation is correct.
 

Ya’akov

Joined Jan 27, 2019
5,998
voltage dividers happen when u have junctions. (split offs and recombinations) when the wires come apart and come back together.

Ohms law designates the current due to the voltage, but voltage dividers and further voltage loss which actually reduces the power at certain points in the circuit, I think its diff!!!
This is simply wrong. It would probably do some good to look into Thevenin's Theorem and network analysis.
 

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Capernicus

Joined Jun 24, 2022
87
Thevenin's theorem, which I wasnt aware of till now, is when ur discovering the global resistance for a network?
Thats only part of the analytical process, you have to take the junctions into account specifically as well. At each junction, you can thevenin reduce it as well per junction segment, to get its total resistance - which produces the draw from the battery.
 
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