Effect of the supply internal resistance

[Rich]

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]

"Rich" <notty@emailo.com> wrote in message
news:6v5el3Fi418aU1@mid.individual.net...

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

1V & 9V. :c)

 

[Tom Biasi]

"Rich" <notty@emailo.com> wrote in message
news:6v5el3Fi418aU1@mid.individual.net...

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

 

[John Fields]

On Sat, 7 Feb 2009 12:55:08 -0000, "Rich" <notty@emailo.com> 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

 

[Rich]

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:498d9122$0$25459$607ed4bc@cv.net...

"Rich" <notty@emailo.com> wrote in message
news:6v5el3Fi418aU1@mid.individual.net...
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!

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6v5kldFi4gn0U1@mid.individual.net...

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:498d9122$0$25459$607ed4bc@cv.net...

"Rich" <notty@emailo.com> wrote in message
news:6v5el3Fi418aU1@mid.individual.net...
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.

 

[Greg Neill]

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.

 

[Rich]

"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498da10c$0$4930$9a6e19ea@news.newshosting.com...

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.

 

[Rich]

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.

 

[Ned]

Here is the way I learned to draw circuits:
And simplify them in College:


R(int)=1 ohm
|--/\/\/\-----------
+| |
--- |
| B | \
|10v| R1 / 90 ohm
|___| \
| /
-| |_____________ V(out)
| |
| \ /\
| / |
| R2 \ 10 ohm |
| / |
| | \/
|__________________|____________ 0 volts



V(out) = (output / input) * V(source)

output is I * R2
Input is I *[(int) + R1 + R2]

then V(out) = (10 ohms / 101 ohms) * 10 volts
v(out) = 0.9901 volts

Your equivelent source impeadance will be R2 parallel [R1 + R(int)]

R eq = product / sum
= 10 ohms * 91 ohms / (101 ohms)
= 9.01 ohms

There: at V(out) the voltage will be 0.9901 volts with an equivelent source
resistance of 9.01 ohms

Shaun



"Rich" <notty@emailo.com> wrote in message
news:6v5o9lFhqnn1U1@mid.individual.net...

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.

 

[John Fields]

On Sat, 7 Feb 2009 14:48:34 -0000, "Rich" <notty@emailo.com> wrote:

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

---
Read this:

sa3ro4l4ddcsv1c6mfdd6vv2tgmrpeadh7@4ax.com

JF

 

[Shaun]

"John Fields" <jfields@austininstruments.com> wrote in message
news:tqpro4hr5stk1ufql0hj4qtn1m7fotrj8e@4ax.com...

On Sat, 7 Feb 2009 14:48:34 -0000, "Rich" <notty@emailo.com> wrote:


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

---
Read this:

sa3ro4l4ddcsv1c6mfdd6vv2tgmrpeadh7@4ax.com

JF

That's some wierd email address, and if I click it, Outlook express pops up.

link didn't work

Shaun

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6v5o9lFhqnn1U1@mid.individual.net...

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.

 

[Rich]

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.

40.5 Ohms and 49.5 Ohms.

 

[Shaun]

The supply internal resistance will give you a voltage drop, lowering the
source voltage available at the battery that depends on the current you
voltage divider and load require. V(battery) = V(10 volts) - I *
R(internal). I is the total circuit current with your voltage divider and
whatever is connected to it. V(10 volts) is your open circuit voltage.

Shaun


"Rich" <notty@emailo.com> wrote in message
news:6v5el3Fi418aU1@mid.individual.net...

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.

 

[John Fields]

On Sat, 7 Feb 2009 14:07:18 -0600, "Shaun" <scepp@shaw.ca> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:tqpro4hr5stk1ufql0hj4qtn1m7fotrj8e@4ax.com...
On Sat, 7 Feb 2009 14:48:34 -0000, "Rich" <notty@emailo.com> wrote:


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

---
Read this:

sa3ro4l4ddcsv1c6mfdd6vv2tgmrpeadh7@4ax.com

JF

That's some wierd email address, and if I click it, Outlook express pops up.

link didn't work

---
What a surprise!

It's a USENET message ID, not an email address...

JF

 

[Peter Bennett]

On Sat, 7 Feb 2009 21:26:38 -0000, "Rich" <notty@emailo.com> 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

 

[Shaun]

"Rich" <notty@emailo.com> wrote in message
news:6v6cq1Figba3U1@mid.individual.net...

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.

40.5 Ohms and 49.5 Ohms.

NO! NO! NO!

You have the battery and R(internal), This is your source. Then you have
your voltage divider and you want the voltage across the 10 ohm resistor.
See Neds description of how to work out the circuit, it is the proper method
that is taught in schools and books. You don't use an imaginary reference
point, you're over complicating the circuit, the only reference point to use
is battery neg (common).

Shaun

 

[Rich]

"Peter Bennett" <peterbb@somewhere.invalid> wrote in message
news:acaso45fe5u1fh4tu0m6r5sai253sdsfvn@news.supernews.com...

On Sat, 7 Feb 2009 21:26:38 -0000, "Rich" <notty@emailo.com> 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.

One think I'm shaky about is this concept of reference and ground.
Especially when I try to grasp Op Amps.

Of course, I know that there is 0.1V across 1R (the internal resistance) not
10 Volts.

10.1 Volts at Ebat is referenced to ground. We can draw 1R as connected to
that point. Of course that one side of 1R is 10.1V above ground is not that
significant, if it is above ground is a fact that does not assist with the
understanding of how to calculate the voltage across 1R.

What is significant is the voltage across 1R. In fact what is significant is
what 1R sees as being it's source.The following circuit shows what that
there across the interal resistance of the battery. It's usefullness is in
showing more clearly how only 0.1V PD apears across 1R. It ignores any kind
of reference point to ground, although in practice you may need to know the
voltage across 1R and ground.


+---------+--E2 = 0.1V
| |
[10R] R2 |
| |
+---E1
| |
[90R] R1 |
| [1R]
+--Ebat |
|+ |
[BAT] |
| |
| |
| |
+---------+

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6v7ti7Firo8rU1@mid.individual.net...

"Peter Bennett" <peterbb@somewhere.invalid> wrote in message
news:acaso45fe5u1fh4tu0m6r5sai253sdsfvn@news.supernews.com...
On Sat, 7 Feb 2009 21:26:38 -0000, "Rich" <notty@emailo.com> 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.


One think I'm shaky about is this concept of reference and ground.
Especially when I try to grasp Op Amps.

Of course, I know that there is 0.1V across 1R (the internal resistance)
not 10 Volts.

10.1 Volts at Ebat is referenced to ground. We can draw 1R as connected to
that point. Of course that one side of 1R is 10.1V above ground is not
that significant, if it is above ground is a fact that does not assist
with the understanding of how to calculate the voltage across 1R.

What is significant is the voltage across 1R. In fact what is significant
is what 1R sees as being it's source.The following circuit shows what that
there across the interal resistance of the battery. It's usefullness is in
showing more clearly how only 0.1V PD apears across 1R. It ignores any
kind of reference point to ground, although in practice you may need to
know the voltage across 1R and ground.


+---------+--E2 = 0.1V
| |
[10R] R2 |
| |
+---E1
| |
[90R] R1 |
| [1R]
+--Ebat |
|+ |
[BAT] |
| |
| |
| |
+---------+

BAT is suppossed to be a device with zero resistance. So, you can
erroniously conclude there is 10V across 1R, thinking one part of 1R is at
ground. The voltage stated is a voltage *reference to ground*. One end of 1R
is + 10V above ground or the negative terminal that's for sure. But in fact
the other side of 1R is *not* connected to ground at all through a zero
resistance, so there is not + 10V across R1. This shows how careful one must
be when associating a zero resaistance value to some ideal component. I must
be careful in doing that.


+-----------+--- 10V
| |
| [R1]
[1R| |
| |
| |
[BAT] [R2]
| |
| |
+-----------+
|
GND