Voltage divider with and without a common point

Thread Starter

Yami

Joined Jan 18, 2016
352
I have done a simulation of a voltage divider question from the worksheet. My question is why is the voltage different once I remove the common connecting wire. I know it something to do with the reference point not being the same, but how/where does the 71.14V comes from?
I have attached a screenshot of the simulation.
 

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Jony130

Joined Feb 17, 2009
5,314
My question is why is the voltage different once I remove the common connecting wire
Ask the software design team why is that. In the real world, the voltmeter will show 0V because in reality, the voltmeter needs "closed loop" to work as a voltmeter (voltmeter need a very small current to flow). So, the voltage in the second case is undetermined.
 

WBahn

Joined Mar 31, 2012
26,398
In order for the simulator to work correctly, every node needs a DC path to ground in order for the simulator to calculate the DC operating point. Without it, there are an infinite number of solutions. Some simulators will simply reject the input file if this isn't the case. Others will automatically install a large resistance to ground for troublesome nodes. Yet others will just do the best they can and come up with one of the infinite number of solutions. That's what your's is doing.

Even your top circuit lacks a ground reference. Try asking it what the voltage on one of the nodes is and you'll probably get a nearly random value.
 

MrAl

Joined Jun 17, 2014
8,864
I have done a simulation of a voltage divider question from the worksheet. My question is why is the voltage different once I remove the common connecting wire. I know it something to do with the reference point not being the same, but how/where does the 71.14V comes from?
I have attached a screenshot of the simulation.
Hello there,

As others have alluded to, the software has a lot to do with it, and the methods used to solve circuits.

The real answer would depend on how you want to view the voltmeter and the electric fields. If you allow even a tiny conductance for the volt meter (a large impedance but not infinite) then you would actually see zero, as long as the fields did not influence the readings much, and that is hard to describe.

The simulator answer is what the simulator wants to do with floating nodes and floating power sources. A common method is to use the concept of a ground "at infinity" and then solve the network as usual. This has various implications though, some of which will be very strange, so it's best just to avoid this situation because you might see things that make no sense as it tries to work out the ground.
For example, try replacing the left source with TWO voltage sources in series, which effectively does NOT change the TERMINAL voltage of the 100v source in real life, yet it may change the overall results in the simulator (that 71v). The voltage 'across' the two nodes then might change drastically. It still depends on the solution methods though what happens here, as if it knows how to lump sources then it may turn out the same result, but if it does not then it may turn up an entirely different voltage. Kind of funny i guess.

If you were given this exercise as some homework, then you should have been told how your particular simulator handles these situations. If not, then you will have to do some experimenting to find out how it handles various unusual cases.

If you were to rotate the left side network by 90 degrees counter clockwise and the right side network by 90 degrees clockwise, you could then estimate the effect the fields might have on the reading with an infinite impedance volt meter, but in reality you'd have to know the construction of every piece of hardware in the circuit including the length of wire and how it is positioned. The 90 and 90 degree rotations would give you the simplest case if the two networks were also very close together after the rotation.
 
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MrAl

Joined Jun 17, 2014
8,864
I was going through some worksheet question from all about circuits
https://www.allaboutcircuits.com/worksheets/voltage-divider-circuits/
Hi,

That's a very good idea for your practice, but i do not see them doing that circuit, the one without the lower connection. Please check again. Maybe you just got curious when you did not use that connecting line at the bottom.

It's probably not worth mentioning, but i tried this with Micro Cap and it seems clear that it is using the ground at infinity concept, but it also uses a ground seeking method that tries to find the zero voltage level in a DC circuit without a ground based on impedances.
For example, in the left circuit if i make the 47k resistor a 50k resistor (for more clear results) the divider ratio is then Vs*2/3 which is two thirds of the supply voltage. However, the voltage does not appear as 66.6666 it appears as:
100*2/3=66.6666
66.6666-50=16.6666
16.6666*2/3=11.1111 volts, final result.
That makes the ground found at:
11.1111/2=5.5555 volts.
That makes the upper voltage 44.4444v and the lower voltage -55.5555v (and of course the difference is 100v).
This could change though with the addition of more parts unless they are arranged in the same impedance ratio.
Probably not worth mentioning though because this doesnt seem to serve any useful purpose.
 

Thread Starter

Yami

Joined Jan 18, 2016
352
Hi,

That's a very good idea for your practice, but i do not see them doing that circuit, the one without the lower connection. Please check again. Maybe you just got curious when you did not use that connecting line at the bottom.
The circuit is shown when you press the "Reveal answer" button
 

JoeJester

Joined Apr 26, 2005
4,390
@MrAl

An artificial connection between the sources would mean both circuits results are the same as both were active during the simulation. Here is the circuit, drawn again, but the picture is not of just the circuit but the whole TINA window. It was the DC analysis. I just add the blue box, using the available tools in TINA, around it to highlight just the area I would normally copy and paste to a file for a forum or elsewhere on the internet.

AAC-HW-180310-1.png
 

WBahn

Joined Mar 31, 2012
26,398
Hi,

That might be artificially connecting the two minus terminals of the two sources.
Try adding some stuff, changing some stuff, see what happens.
No simulator worth anything at all is going to do that and it's for sure that this simulator isn't because, if it were, then the bottom circuit would give the same reading as the top circuit.

Most likely, the voltmeter model includes a large parallel resistance of perhaps 1 GΩ.
 

MrAl

Joined Jun 17, 2014
8,864
Hi,

That might be artificially connecting the two minus terminals of the two sources.
Try adding some stuff, changing some stuff, see what happens.

Hello again,

I am not sure what i was referring to there (first line above) as it is not what i meant to say that should be obvious :)

However, adding some stuff, changing some stuff could reveal more about how the simulator handles things and that is true for any simulator.
 

WBahn

Joined Mar 31, 2012
26,398
Hello again,

I am not sure what i was referring to there (first line above) as it is not what i meant to say that should be obvious :)

However, adding some stuff, changing some stuff could reveal more about how the simulator handles things and that is true for any simulator.
While I have no problem with experimenting around to tease out how things work, the best place to start, particularly if it's a simulator used for serious design work, is in the documentation. The behavior under these conditions is often well described. For instance, in some simulators that don't have a ground node defined, the first net encountered in the top level file is redefined as node 0.

But that aside, the more important thing for people to do is to learn how to use the simulator correctly, which includes properly defining the common node in every simulation schematic set, that every node needs a DC path to ground for the DC and OP analyses, and how to provide these using IC commands if needed.
 

MrAl

Joined Jun 17, 2014
8,864
While I have no problem with experimenting around to tease out how things work, the best place to start, particularly if it's a simulator used for serious design work, is in the documentation. The behavior under these conditions is often well described. For instance, in some simulators that don't have a ground node defined, the first net encountered in the top level file is redefined as node 0.

But that aside, the more important thing for people to do is to learn how to use the simulator correctly, which includes properly defining the common node in every simulation schematic set, that every node needs a DC path to ground for the DC and OP analyses, and how to provide these using IC commands if needed.
Hi,

Sure, but then again some people want to understand the reason behind things they see, even if not "proper". That's part of what makes us human.
 
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