Effect of the supply internal resistance

Rich wrote:

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

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.
Click to expand...
No, you can't, because BAT is a device that is *defined* to have
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.

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.
Click to expand...
What's to be careful about? The circuit plainly shows the 1R resistor
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).

+-----------+--- 10V
| |
| [R1]
[1R| |
| |
| |
[BAT] [R2]
| |
| |
+-----------+
|
GND
Click to expand...
 
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498ed579$0$22975$9a6e19ea@news.newshosting.com...
Rich wrote:

+---------+--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.

No, you can't, because BAT is a device that is *defined* to have
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.

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.

What's to be careful about? The circuit plainly shows the 1R resistor
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).



+-----------+--- 10V
| |
| [R1]
[1R| |
| |
| |
[BAT] [R2]
| |
| |
+-----------+
|
GND
Click to expand...
But, when I drew the circuit, BAT and 1R is an equivalent circuit for a
battery. Consisting of a perfect battery (BAT) and an internal resistance
(1R).

Face value you want to say, hey if BAT is a perfect battery with zero
resistance, there's 10V across 1R. Of course that is not true. It's just one
of these issues you get when dealing with perfect components in series with
resistance. 10v is just a statement that that point is 10V above a refrence
point. And there really is not a zero resistance across BAT. It's mixing a
fiction with reality. The equivalent circuit of a batttery is a fiction.
 
Rich wrote:
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498ed579$0$22975$9a6e19ea@news.newshosting.com...
Rich wrote:

+---------+--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.

No, you can't, because BAT is a device that is *defined* to have
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.

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.

What's to be careful about? The circuit plainly shows the 1R resistor
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).



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

But, when I drew the circuit, BAT and 1R is an equivalent circuit for a
battery. Consisting of a perfect battery (BAT) and an internal resistance
(1R).

Face value you want to say, hey if BAT is a perfect battery with zero
resistance, there's 10V across 1R.
Click to expand...
No, because then the 1R would be in parallel with BAT, not in
series as is drawn. You don't get to arbitrarily assign the
stated potential of the voltage source to another passive
component (the internal resistance).

Of course that is not true. It's just one
of these issues you get when dealing with perfect components in series
with
resistance.
Click to expand...
Sorry, but it's not a common problem. You seem to have your own
unique set of confusions that lead to such inferences.

10v is just a statement that that point is 10V above a refrence
point. And there really is not a zero resistance across BAT. It's mixing a
fiction with reality. The equivalent circuit of a batttery is a fiction.
Click to expand...
For all intents and purposes the BAT component has no resistance
associated with it. The resistance of the "real" battery is
lumped into the 1R resistor. Any circuit analysis that is
performed on this circuit *must* consider the potential difference
across the battery.

Note also that when you eventually come to analyse AC circuits
that also have DC sources in them, the DC sources *can* look
like short circuits (zero resistance) to the AC signal components.
The +Vcc voltage supply in an audio amplifier circuit, for
example, will "look" like a ground every bit as much as the actual
DC ground does to the audio signal.
 
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498edc3d$0$22977$9a6e19ea@news.newshosting.com...
Rich wrote:
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498ed579$0$22975$9a6e19ea@news.newshosting.com...
Rich wrote:

+---------+--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.

No, you can't, because BAT is a device that is *defined* to have
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.

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.

What's to be careful about? The circuit plainly shows the 1R resistor
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).



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

But, when I drew the circuit, BAT and 1R is an equivalent circuit for a
battery. Consisting of a perfect battery (BAT) and an internal resistance
(1R).

Face value you want to say, hey if BAT is a perfect battery with zero
resistance, there's 10V across 1R.

No, because then the 1R would be in parallel with BAT, not in
series as is drawn. You don't get to arbitrarily assign the
stated potential of the voltage source to another passive
component (the internal resistance).
Click to expand...
Really, the issue has been what to make of that fact that in the circuit as
drawn, R1 is at a potential of + 10v and it's other side is "seemingly*
connected through BAT to GND.

That can cause you to wonder how to explain why there is not really a 10V
potential across R1. Not that one is unable to calculate the actual voltage
across R1, which I can.


Of course that is not true. It's just one
of these issues you get when dealing with perfect components in series
with
resistance.

Sorry, but it's not a common problem. You seem to have your own
unique set of confusions that lead to such inferences.
Click to expand...
Not really, just trying to see where the error would be in thinking there
was 10V across R1. That I don't think would be an uncommon thing to do.

10v is just a statement that that point is 10V above a refrence
point. And there really is not a zero resistance across BAT. It's mixing
a
fiction with reality. The equivalent circuit of a batttery is a fiction.

For all intents and purposes the BAT component has no resistance
associated with it. The resistance of the "real" battery is
lumped into the 1R resistor. Any circuit analysis that is
performed on this circuit *must* consider the potential difference
across the battery.

Note also that when you eventually come to analyse AC circuits
that also have DC sources in them, the DC sources *can* look
like short circuits (zero resistance) to the AC signal components.
The +Vcc voltage supply in an audio amplifier circuit, for
example, will "look" like a ground every bit as much as the actual
DC ground does to the audio signal.
Click to expand...
Yes, I've noticed that in my ham radio experience.
 
On Sun, 8 Feb 2009 11:21:46 -0000, "Rich" <notty@emailo.com> wrote:

"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] |
| |
| |
| |
+---------+
Click to expand...

---
Note that the battery and its internal resistance have both been placed
in the same box (which is how it works in real life) and look at it like
this:

If you have a 10V battery with an internal resistance of 1 ohm and you
measure its output voltage with a high-impedance voltmeter, then the
current through the voltmeter will be miniscule as will the current
through the battery's internal resistance(since current is everywhere
the same in a series circuit) and the voltmeter will read 10.0V:

.. 10.0V
.. /
.. +-----E1<---+
.. | |
.. |+ |
.. +---+----+ |
.. | | | |
.. | [1R] | |
.. | | | [VOLTMETER]
.. | [10V] | |
.. | | | |
.. +---+----+ |
.. |- |
.. | |
.. +<----------+
.. |
.. GND

Now, if the voltage source and the internal resistance swap sides:

.. ???V
.. /
.. +-----E1<---+
.. | |
.. |+ |
.. +---+----+ |
.. | | | |
.. | [10V] | |
.. | | | [VOLTMETER]
.. | [1R] | |
.. | | | |
.. +---+----+ |
.. |- |
.. | |
.. +<----------+
.. |
.. GND

will the voltmeter still read 10.0V?

Since the voltmeter is still connected exactly as it was earlier and it
doesn't care what's going on in box, what do you think?


Now,if we take the next step and hook up the voltage divider:

..
.. +----------+--E1<-------+
.. | | |
.. |+ | |
.. +---+----+ [90R] |
.. | | | | |
.. | [1R] | | |
.. | | | +--E2 [VOLTMETER]
.. | [10V] | | |
.. | | | | |
.. +---+----+ [10R] |
.. |- | |
.. | | |
.. +----------+--0V<-------+
.. |
.. GND

Notice that the 10V is now pushing 99mA through a 101 ohm load, with
that current making the voltage at E2 = 0.990V, the voltage at E1 = 9.9V
and that last 0.1V across the battery's internal resistance, for a total
voltage of 10.0V out of the voltage source.


Finally, if we swap the internal resistance and the voltage source:

..
.. +----------+--E1<-------+
.. | | |
.. |+ | |
.. +---+----+ [90R] |
.. | | | | |
.. | [10V] | | |
.. | | | +--E2 [VOLTMETER]
.. | [1R] | | |
.. | | | | |
.. +---+----+ [10R] |
.. |- | |
.. | | |
.. +----------+--0V<-------+
.. |
.. GND

What will happen to E1 and E2?

JF
 
Rich wrote:
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498edc3d$0$22977$9a6e19ea@news.newshosting.com...
Rich wrote:
"Greg Neill" <gneillRE@MOVEsympatico.ca> wrote in message
news:498ed579$0$22975$9a6e19ea@news.newshosting.com...
Rich wrote:

+---------+--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.

No, you can't, because BAT is a device that is *defined* to have
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.

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.

What's to be careful about? The circuit plainly shows the 1R resistor
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).



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

But, when I drew the circuit, BAT and 1R is an equivalent circuit for a
battery. Consisting of a perfect battery (BAT) and an internal
resistance
(1R).

Face value you want to say, hey if BAT is a perfect battery with zero
resistance, there's 10V across 1R.

No, because then the 1R would be in parallel with BAT, not in
series as is drawn. You don't get to arbitrarily assign the
stated potential of the voltage source to another passive
component (the internal resistance).

Really, the issue has been what to make of that fact that in the circuit
as
drawn, R1 is at a potential of + 10v and it's other side is "seemingly*
connected through BAT to GND.

That can cause you to wonder how to explain why there is not really a 10V
potential across R1. Not that one is unable to calculate the actual
voltage
across R1, which I can.
Click to expand...
There is no ambuiguity. One *end* of the 1R resistor is fixed at
a potential of Vbat with respect to ground (taking the battery
negative as ground). This says nothing about the other end of
the 1R resistor until you look at the rest of the circuit and
hence the current flowing through the 1R resistor.

If the circuit happens to be open (no path to ground via the other
end of the 1R resistor), then *both* ends of the 1R will be fixed
at Vbat above ground; there will still not be a potential of Vbat
*across* the 1R.

Of course that is not true. It's just one
of these issues you get when dealing with perfect components in series
with
resistance.

Sorry, but it's not a common problem. You seem to have your own
unique set of confusions that lead to such inferences.

Not really, just trying to see where the error would be in thinking there
was 10V across R1. That I don't think would be an uncommon thing to do.
Click to expand...
The error is in assigning the properties of a voltage source
(battery) to a resistor. This is not a 'legal' operation.
An ideal resistor is a passive component that contains no
sources.

A resistor presents across its terminals a voltage that
depends upon the amount of current flowing through it:

v(I) = I*R.

Without a current there is no voltage across the resistor.

An ideal voltage source presents the same voltage across
its terminals irrespective of the current flowing through it:

v(I) = V

10v is just a statement that that point is 10V above a refrence
point. And there really is not a zero resistance across BAT. It's mixing
a
fiction with reality. The equivalent circuit of a batttery is a fiction.

For all intents and purposes the BAT component has no resistance
associated with it. The resistance of the "real" battery is
lumped into the 1R resistor. Any circuit analysis that is
performed on this circuit *must* consider the potential difference
across the battery.

Note also that when you eventually come to analyse AC circuits
that also have DC sources in them, the DC sources *can* look
like short circuits (zero resistance) to the AC signal components.
The +Vcc voltage supply in an audio amplifier circuit, for
example, will "look" like a ground every bit as much as the actual
DC ground does to the audio signal.

Yes, I've noticed that in my ham radio experience.
Click to expand...
 
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

Shaun
Click to expand...
Shaun, if you are using a newsgroup aware program (as I do) it should
regonize the address as a newsgroup (usenet) ID. All I need to do is
double click it and a search takes place looking for the article, for
example. First within my existing local database, then next online
with my newsgroup server via NNTP (one of many protocols, this one
called "network news transfer protocol," if memory serves.) If found
online, the article is downloaded and stored locally, automatically,
and then displayed for me. (It's possible that the server I use no
longer stores the article, though.)

For example, the message ID for your own message (the one I'm
responding to) was found by looking at the message headers for your
message as:

Message-ID: <MSljl.14757$Tt1.9590@newsfe07.iad>

Every newsgroup message that gets posted, I think, has one of these
fields present in the headers. I don't know how it is generated or
who does the generation such that they are unique, but there must be
an RFC (request for comment), or more than one, that documents it.

If I double-click the message ID above, for example, your message
immediately shows in my newsgroup browser -- flipping me away from
this message I'm composing. Then I just click back to the tab for
this message and continue writing.

Jon
 
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