Re: Effect of the supply internal resistance


"Rich" <notty@xxxxxxxxxx> wrote in message
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"Tom Biasi" <tombiasi@xxxxxxxxxxxxx> wrote in message
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"Rich" <notty@xxxxxxxxxx> wrote in message
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I'm looking at voltage dividers. Trying to learn what I probably knew
one time.

Say you have a battery supplying 10v, and it has in internal resistance
of 1 Ohm.

You connect it to a voltage divider network R1 and R2. R1 is 90 Ohms in
series with R2 which is 10 Ohms.

Vcc connects to one end of R1, Ground connects to one end of R2.

No load on the tappinp point.

There is 10V from tap to GND. There is 90V to Vcc.

Okay I get this. But the battery connects between Vcc and GND and that
has in internal resistance of 1 Ohm.

This makes it look like R1 and R2 are also connected together not only
at the tap point, but at their other ends.

Which of course produces a kind of obserdity because that makes irt look
like there is the same voltage across both R1 and R2.

Of course, as far as the battey is concerned, it's 10V across it does
not short through it's own 1 ohm internal resistance.

But anyway, when is that 1 Ohm internal resaistance going to be
significant in any way? I know there will be a voltage drop across it
when current flows and the battery terminal voltage will drop.

But why exactly to you pretty much ignore that internal resistance when
calculating voltages in a two resistor divider network?

But I think in times past for some reason, because of low internal
resistance of the supply, Vcc and GND have been taken in some way as
being connected. I've forgotten now what that was.

Rich,
The battery's internal resistance is in series with its ideal voltage.
The internal resistance is always part of the circuit.
If the load resistance is much larger than the internal resistance then
it won't matter much, if not , its another resistor in your circuit.
Draw the circuit being fed with your supply voltage and a resistor in
series.

Tom

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

B= ideal battery with zero internal resistance.

Lets say the battery voltage is 10.1V open circuit.
1 Ohm Internal Resistence (I R).
R1 = 90 Ohms, R2 = 10 Ohms.

0.1 A of current flows.

10V between X and Y.

9V across R1. 1V across R2.

0.1V across I R. (V = 1A * 0.1 Ohm = 0.1 V.)

Wait, there is 10V across X and Y, yet if B represents an ideal battery
with zero resistance, there should be 10V across I R!


The current is 0.1A a. So, using V= I * R voltage across I R is, 0.1A * 1
Ohm = 1V.


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.

.

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