Re: Effect of the supply internal resistance


"Rich" <notty@xxxxxxxxxx> wrote in message
news:6v5o9lFhqnn1U1@xxxxxxxxxxxxxxxxxxxxx
I think I've figured it.


Youve got to draw a center line down the circuit.

!
| V out
|----------------------|-------------------
| |
| |
| |
R2 R1
| |
| |
|--------B-------- I R ------------------|
X |
!

Looking at the above circuit, because B is ideal battery and zero
resistance, it looks like R1 and R2 are practically in parallel because I
R
(Internal resistance), is 1 Ohm.

But you must look at the left hand side of IR and then the right hand
side.


Add up the voltageson either side and see what IR sees (Z & Y).

| V out
|-------------------
| -
|
|
R1 9V
|
| +
I R ------------------|
Y


| V out
|--------------------
| +
|
|
R2 1V
| -
|
|--------B-------- I R
X - + Z
10.1 V

We see that there are two voltages as far as I R is concerned. And they
are
opposing each other.

On the right, I R sees + 9V

On the left, I R sees 10.1V - 1V = 9.1V.

So, there is *not* 10V across I R. There is 9.1V - 9V = 0.1V

So, there is 0.1V across X and Y. There is 1 Ohm between these points.
Yet
*because of battery B* you cannot say that R1 and R2 are practically in
parallel. Despite the low impedance of the suply source.

And I think this sort of stuff is going to explain Op Amps.

In fact there is something special about point Vout here. It's like a
reference point. Like a ground.

I'm slightly wong.

There is a point along the circuit where positive meets negative.

Going one way is like going into negative, going the other way like being
into positive. Of course anywhere along the circuit one part can be
considered more positive, or negative relative to another.

In the circuit above the center point, which is kind of a reference point is
within R1. It's 40.5 Ohms into R1 on one side. Other side is 40.5 Ohms.
Adding resistances from the center point either way adds up to 50.5 Ohms, so
total is 101 Ohms. True center point, which I suppose is a reference point
is total resistance divided by 2.

So, in my circuit Vout, is not the center point.

And the voltage across I R is + 5.05V on one side, + 4.95V on the other, so
there is 0.1V across I R (internal resistance of battery).

Yet between point X and Y, you would measure 10V, and for a moment it looks
there is 10V across I R, without closer analysis.

.

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