Re: Effect of the supply internal resistance

On Sat, 7 Feb 2009 21:26:38 -0000, "Rich" <notty@xxxxxxxxxx> wrote:

<snippage>

I'm slightly wong.

No, you're considerably wrong, and/or trying very hard to confuse
yourself (and the rest of us, too), and succeeding.


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.

No, you are forgetting the voltage source within the battery - it
provides 10.1 volts, and there will be 0.1 volts (in the opposite
polarity) across the internal resistance, to give 10.0 volts between X
and Y.

Let me steal the drawing that John Fields made in another post:


+---------+--E1
| |
[1R] [90R] R1
| |
+--Ebat +--E2
|+ |
[BAT] [10R] R2
| |
+---------+
|
GND

I have added a "Ground" label so that we have a reference point to
measure voltages against. You are confusing things by adding your
"center point", rather than selecting a connection point as the zero
volt reference.

In electronics, "Ground" might better be called "Common", or
"Reference point" - it is the point in the circuit that the designer
has decided to call "Zero Volts". In the above circuit, it is the
negative terminal of the battery.

Assume that the unloaded battery voltage Ebat is 10.1 volts.

The total circuit resistance including the battery's internal
resistance is 101 ohms. This means that the current in the circuit
will be 0.1 amp, giving 0.1 volt across the battery's internal
resistance, so the voltage across the battery terminals (between GND
and E1) is 10.0 volts. There will be 9 volts across R1, and 1 volt
across R2.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
.

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