Re: Effect of the supply internal resistance

On Sat, 7 Feb 2009 12:55:08 -0000, "Rich" <notty@xxxxxxxxxx> wrote:

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.

---
Only when there's a load on the battery.
---

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.

---
Its internal resistance is in _series_ with the load since it only drops
voltage when charge is flowing.
---

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?

---
Because the load from the divider is usually small causing the drop
across the internal resistance to be small, as well.
---

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.

---
The internal resistance of a battery is considered to be in series with
the load, so your circuit will look like this: (View in Courier)


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


With a 10V supply and a 101 ohm load the current in the circuit will be:


Ebat 10V
I = ------ = ------ = 0.099A = 99mA
Rt 101R


The voltage drop across the internal resistance will then be:


E(r)= IR = 0.099A * 1R = 0.099V


That will make the voltage across the divider equal to:


E1 = Ebat - E(r) = 10V - 0.099V = 9.9V


and the voltage at the junction of R1 and R2:

E1 * R2 9.9V * 10R
E2 = --------- = ------------ = 0.99V
R1 + R2 90R + 10R


JF
.

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