Re: Effect of the supply internal resistance


"Greg Neill" <gneillRE@xxxxxxxxxxxxxxxx> wrote in message
news:498da10c$0$4930$9a6e19ea@xxxxxxxxxxxxxxxxxxxxxxx
Rich wrote:

I know that the actual voltage across I R *is* 0.1V

But looking it another way there is supposed to be 10V across X and Y.
Which
is effectively 10V across I R.

No, there's still 10V across the ideal battery. You can't ignore
that when figuring your voltage drops between X and Y.

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.
.

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