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Parallel EMF.

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lord loh.

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What happens if two different EMFs are connected parallel.

My teachers told me that the lower of the two EMFs appear at the output of the setup.

However, SPICE simulation got me unexpected results. The output was a value between the two EMFs. A series resistance is must for simulations.

I used the internal series resistance of a 1nano ohm.

When I varied the series resistance, the results changed drastically.

emf1=5v
emf2=20v

1. I kept both internal resistance at 1n ohm the output was ~12v
2. When the internal resistance of EMF1 was made zero, the output was 5v
3. When the internal resistance of EMF2 was made zero, the output was 20v

both cannot be made zero for simulation.

What ought to happens in the ideal case? and what shall I get in a practical case?

Attached is a swCAD file.
 

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  • draft3.zip
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Parallel EMFs

lord loh, I think you are seeing the limitation of reliance on simulations to answer your questions when you get into regions that are not adequately described in models or are undfinable. I think it is not to your advantage to be plugging things into a Spice simulation instead of understanding the fundamental concepts. (This is not intended to be insulting to you or to deride your education, but rather to call your attention to the ABSOLUTE NEED to understand the basic concepts and only use the Spice simulations where appropriate.)

Your question, "what happens if two different EMFs are connected in parallel," can only be answered by consideration of the internal resistances of the two EMF sources and the ability of the sources to sustain whatever current drain results. I'm not sure simulation software can give you a correct answer unless it contains models or is programmed for abnormal conditions. (This is where I must say I have never used spice simulation software.)

The first consideration might be that you should never connect two different EMF sources in parallel unless you have precise knowledge of their characteristics, current capabilities, damage thresholds, and internal resistances. Your Spice software requires that you define some resistance between the two "perfect" EMF sources because inserting a resistance is the only way you can avoid infinite current flow between two sources of even slightly different EMFs. Ohm's law still applies. ANY differential voltage, even microvolts, divided by zero ohms (zero internal resistance and zero external resistance) results in infinite current flow. You don't need a Spice simulation to get this answer. Just apply Ohms law.

Now, I'm sure you know that most "EMF sources" can't supply infinite current, because every real power source has SOME internal resistance, however low, and SOME limitation on available current. But you CAN get high enough currents to burn switches, wires, fuses, etc., or have fireworks. In other words, there is no such thing as a perfect EMF source. So you have to consider what the effective internal resistance of the sources may be.

Your teacher's statement that the lower of the two EMFs will appear at the output can only be supported by some unspoken assumption about the relative internal resistances or circuitry of the sources. There is nothing magical about lower EMF source versus higher EMF sources that dictates that the lower source will dominate the output EMF unless you make some assumption about the characteristics of the sources OTHER that simply source EMF and source resistance. The lower EMF source has EXACTLY the same capability to control the output EMF as the higher EMF. The resulting output voltage depends ENTIRELY upon their relative internal resistances unless the sources have some internal nonlinearity that was not described in your presentation of the question. It all comes down to Ohm's law and evaluation of the current capability of the two sources (which comes down to an effective internal resistance)

Throw away your Spice simulator and apply Ohm's law.

By the way, what current did your Spice simulation give you with your nano-ohm resistances?

awright
 
Re: Parallel EMFs

awright said:
lord loh, I think you are seeing the limitation of reliance on simulations to answer your questions when you get into regions that are not adequately described in models or are undfinable. I think it is not to your advantage to be plugging things into a Spice simulation instead of understanding the fundamental concepts. (This is not intended to be insulting to you or to deride your education, but rather to call your attention to the ABSOLUTE NEED to understand the basic concepts and only use the Spice simulations where appropriate.)

I am, not the least insulted. I knew I had got some fundamental concept wrong and asked the question to get it right. This forum has terribley helped me.


awright said:
By the way, what current did your Spice simulation give you with your nano-ohm resistances?

awright

It was 45MA :shock:

:D
 
Ohm's Law

Well, well, we have uncovered some mystery of how the spice simulator handles bizarre conditions.

If we are dealing with presumably pure EMF sources and pure ohmic resistances, Ohm's law tells us that the current will be:

I = E/R

I = {(20V - 5V)/(0.000,000,001 Ohm)}

= {15V/0.000000001}

= 15 billion amps!

Quite a bit different from 45 ma.

Which do you believe, Ohm or Spice?

As I said, I know nothing about Spice simulations. But, either there are some hidden assumptions within your Spice simulator that you are not familiar with or you made an error in data entry or model setup.

Additionally, apparently there is rounding off happening in the simulator because 5 volts and 20 volts connected together via two identical resistors in series should give 12.5 volts, not 12 volts, at the resistor junction.

I think the extremely valuable lesson here is not to put blind faith in a simulation without checking out the result for credibility.

awright
 
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