"chung" wrote in message
vers.com...
afh3 wrote:
"chung" wrote in message
vers.com...

You can have an extremely low impedance, and yet not be able to source
a
lot of current. An op amp is a classical example.

Yes. But without the ability to source a lot of current, I'm sure you
can
see why we don't use 741's for output stages of amplifiers. I neither
stated
nor implied that one required the other -- simply that a good power
amplifier design would have both.

Regarding the SET amp not being popular, try rec.audio.tubes with that
line...



And I'm sure somewhere there is a forum or newsgroup where the
flat-earth theory is very popular...


I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance. There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.
While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value, the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

So, I call biwiring "Hogwash". I don't do it. Do you?

-afh3

"Bob Saccamano" wrote in message
...
I have a pair of speakers that support bi-wiring and bi-amping. I
only have
one aplifier. Will I appreciate any improvements in sound by
bi-wiring
alone?

No. Draw out the equivalent circuit, add the known wiring series and
shunt impedances, do the signal analysis and it will be obvious.

Cheers,

Roger

"Bob Saccamano" wrote in message
...
I have a pair of speakers that support bi-wiring and bi-amping. I
only have
one aplifier. Will I appreciate any improvements in sound by
bi-wiring
alone?

No. Draw out the equivalent circuit, add the known wiring series and
shunt impedances, do the signal analysis and it will be obvious.

Cheers,

Roger

"Bob Saccamano" wrote in message
...
I have a pair of speakers that support bi-wiring and bi-amping. I
only have
one aplifier. Will I appreciate any improvements in sound by
bi-wiring
alone?

No. Draw out the equivalent circuit, add the known wiring series and
shunt impedances, do the signal analysis and it will be obvious.

Cheers,

Roger

"Bob Saccamano" wrote in message
...
I have a pair of speakers that support bi-wiring and bi-amping. I
only have
one aplifier. Will I appreciate any improvements in sound by
bi-wiring
alone?

No. Draw out the equivalent circuit, add the known wiring series and
shunt impedances, do the signal analysis and it will be obvious.

Cheers,

Roger

In article ,
(Bruce Burke) wrote:

On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote:

"flint" wrote in message
...



Here is my attempt at the "good logical argument."

It's called the application of Kirchoff's laws.

Since the bi-wired speakers ultimately connect to the same set of output
terminals on the voltage source (amplifier/receiver/whatever) all of the
points on both pairs of conductors will be at the same electrical
potential at any one point in time.

Excluding the negligible resistance of the conductors

Actually, you can't. That small change is exactly what is being
counted on to create the isolation improvement between the halves of
the crossover. The idea is to work that increased resistance or more
properly, impedance, against the very, very low output impedance of
the majority of today's solid-state amplifiers as a Tee network.

The possibility of the difference being audible is pretty slim.
Regards,

Bruce
Hitting reply is futile, use the following:
).

The losses in the speaker cable are less than 1/1000th of those in the
crossover and the speaker itself. That very tiny wire loss also has
zero distortion. An air core inductor, like speaker wire, can not cause
intermodulation distortion. If you're trying to avoid IM distortion
from the crossover inductors, buy a less crappy crossover.

In article ,
(Bruce Burke) wrote:

On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote:

"flint" wrote in message
...



Here is my attempt at the "good logical argument."

It's called the application of Kirchoff's laws.

Since the bi-wired speakers ultimately connect to the same set of output
terminals on the voltage source (amplifier/receiver/whatever) all of the
points on both pairs of conductors will be at the same electrical
potential at any one point in time.

Excluding the negligible resistance of the conductors

Actually, you can't. That small change is exactly what is being
counted on to create the isolation improvement between the halves of
the crossover. The idea is to work that increased resistance or more
properly, impedance, against the very, very low output impedance of
the majority of today's solid-state amplifiers as a Tee network.

The possibility of the difference being audible is pretty slim.
Regards,

Bruce
Hitting reply is futile, use the following:
).

The losses in the speaker cable are less than 1/1000th of those in the
crossover and the speaker itself. That very tiny wire loss also has
zero distortion. An air core inductor, like speaker wire, can not cause
intermodulation distortion. If you're trying to avoid IM distortion
from the crossover inductors, buy a less crappy crossover.

In article ,
(Bruce Burke) wrote:

On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote:

"flint" wrote in message
...



Here is my attempt at the "good logical argument."

It's called the application of Kirchoff's laws.

Since the bi-wired speakers ultimately connect to the same set of output
terminals on the voltage source (amplifier/receiver/whatever) all of the
points on both pairs of conductors will be at the same electrical
potential at any one point in time.

Excluding the negligible resistance of the conductors

Actually, you can't. That small change is exactly what is being
counted on to create the isolation improvement between the halves of
the crossover. The idea is to work that increased resistance or more
properly, impedance, against the very, very low output impedance of
the majority of today's solid-state amplifiers as a Tee network.

The possibility of the difference being audible is pretty slim.
Regards,

Bruce
Hitting reply is futile, use the following:
).

The losses in the speaker cable are less than 1/1000th of those in the
crossover and the speaker itself. That very tiny wire loss also has
zero distortion. An air core inductor, like speaker wire, can not cause
intermodulation distortion. If you're trying to avoid IM distortion
from the crossover inductors, buy a less crappy crossover.

In article ,
(Bruce Burke) wrote:

On Tue, 02 Mar 2004 22:13:39 GMT, "afh3" wrote:

"flint" wrote in message
...



Here is my attempt at the "good logical argument."

It's called the application of Kirchoff's laws.

Since the bi-wired speakers ultimately connect to the same set of output
terminals on the voltage source (amplifier/receiver/whatever) all of the
points on both pairs of conductors will be at the same electrical
potential at any one point in time.

Excluding the negligible resistance of the conductors

Actually, you can't. That small change is exactly what is being
counted on to create the isolation improvement between the halves of
the crossover. The idea is to work that increased resistance or more
properly, impedance, against the very, very low output impedance of
the majority of today's solid-state amplifiers as a Tee network.

The possibility of the difference being audible is pretty slim.
Regards,

Bruce
Hitting reply is futile, use the following:
).

The losses in the speaker cable are less than 1/1000th of those in the
crossover and the speaker itself. That very tiny wire loss also has
zero distortion. An air core inductor, like speaker wire, can not cause
intermodulation distortion. If you're trying to avoid IM distortion
from the crossover inductors, buy a less crappy crossover.

On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

No there aren't - not one single one.

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

So, I call biwiring "Hogwash". I don't do it. Do you?

It's hogwash, but not for the reasons you incorrectly claim.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

No there aren't - not one single one.

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

So, I call biwiring "Hogwash". I don't do it. Do you?

It's hogwash, but not for the reasons you incorrectly claim.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

No there aren't - not one single one.

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

So, I call biwiring "Hogwash". I don't do it. Do you?

It's hogwash, but not for the reasons you incorrectly claim.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

No there aren't - not one single one.

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

So, I call biwiring "Hogwash". I don't do it. Do you?

It's hogwash, but not for the reasons you incorrectly claim.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The voltage
at
the end of one of the conductor pairs will be exactly the same as the
voltage at end of the other conductor pair, minus the difference in the
voltage drop across each pair of conductors REGARDLESS of what passive or
reactive component is connected to each end. Read Kirchoff's laws if this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand this,
so I've got to believe that you either WAY misread what I wrote above,
or something is fundamentally wrong with how I was communicating my point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The voltage
at
the end of one of the conductor pairs will be exactly the same as the
voltage at end of the other conductor pair, minus the difference in the
voltage drop across each pair of conductors REGARDLESS of what passive or
reactive component is connected to each end. Read Kirchoff's laws if this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand this,
so I've got to believe that you either WAY misread what I wrote above,
or something is fundamentally wrong with how I was communicating my point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The voltage
at
the end of one of the conductor pairs will be exactly the same as the
voltage at end of the other conductor pair, minus the difference in the
voltage drop across each pair of conductors REGARDLESS of what passive or
reactive component is connected to each end. Read Kirchoff's laws if this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand this,
so I've got to believe that you either WAY misread what I wrote above,
or something is fundamentally wrong with how I was communicating my point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The voltage
at
the end of one of the conductor pairs will be exactly the same as the
voltage at end of the other conductor pair, minus the difference in the
voltage drop across each pair of conductors REGARDLESS of what passive or
reactive component is connected to each end. Read Kirchoff's laws if this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand this,
so I've got to believe that you either WAY misread what I wrote above,
or something is fundamentally wrong with how I was communicating my point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--


Oh ferchristsakes, of fricking course they will be different -- who ever
said they would not be???

Read the sentence. It says that the

"difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair"

You either agree or you don't. Which is it? I can't tell from your
non-sequitor reply.


Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--


Oh ferchristsakes, of fricking course they will be different -- who ever
said they would not be???

Read the sentence. It says that the

"difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair"

You either agree or you don't. Which is it? I can't tell from your
non-sequitor reply.


Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--


Oh ferchristsakes, of fricking course they will be different -- who ever
said they would not be???

Read the sentence. It says that the

"difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair"

You either agree or you don't. Which is it? I can't tell from your
non-sequitor reply.


Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Wed, 03 Mar 2004 19:25:06 GMT, "afh3" wrote:


"Stewart Pinkerton" wrote in message
.. .
On Wed, 03 Mar 2004 17:12:33 GMT, "afh3" wrote:

The statement I made was an assertion that the voltage drop across the
conductor is proportional to the resistance of the conductor. The
voltage
at
the end of one of the conductor pairs will be exactly the same as
the
voltage at end of the other conductor pair, minus the difference in
the
voltage drop across each pair of conductors REGARDLESS of what passive
or
reactive component is connected to each end. Read Kirchoff's laws if
this
is
unclear.

No, it won't, because the currents in the two legs will be quite
different. Read Ohm's Law if this is unclear.

Stewart, I've read your posting here before and I know you understand
this,
so I've got to believe that you either WAY misread what I wrote
above,
or something is fundamentally wrong with how I was communicating my
point.

It is simply a fact that (to paraphrase directly what I said previously):

"The voltage at the end of one of [a] conductor pair will be exactly
the
same as the voltage at end of the [same] conductor pair, minus the
difference in the voltage drop across [the] pair of conductors REGARDLESS
of
what passive or reactive component is connected to each end."

The reason for this is that biwiring is just one pair of conductors with
a
voltage supply in the middle and a load on each end. The only possible
difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the
other
section of conductor pair.

And those voltage drops will be different if the currents are
different - which they are.
--


Oh ferchristsakes, of fricking course they will be different -- who ever
said they would not be???

Read the sentence. It says that the

"difference in voltage at either end of the conductor pair is due to the
voltage drop of that section of conductor pair minus the drop of the other
section of conductor pair"

You either agree or you don't. Which is it? I can't tell from your
non-sequitor reply.


Stewart Pinkerton | Music is Art - Audio is Engineering

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET
amps
out there that offer pretty great performance.

No there aren't - not one single one.

Opinion of course. I don't want one, but there are many others with Bel
Canto designs (among others) who would disagree with your assessment.


There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I was of course, referring the to the use of negative feedback in the sense
that is typically applied when speaking of amplifier design - EXTERNAL
negative feedback - just like about 99% of people would assume was meant in
this context.

I don't care for tubes. Others do. So be it.


I still prefer solid state, but even among those you have to spend quite
a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

I could find no references to the output impedance or damping factor for
this model, so I will assume your statement is accurate -- very much unlike
the one you make below.


While Krells, Meridians, CJs and other quality solid-state amps can
handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

Krell KAV500i = 0.044 ohms
Krell FPB-300c = 0.069 ohms

Are these modern enough? Should I go on?

-afh3

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET
amps
out there that offer pretty great performance.

No there aren't - not one single one.

Opinion of course. I don't want one, but there are many others with Bel
Canto designs (among others) who would disagree with your assessment.


There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I was of course, referring the to the use of negative feedback in the sense
that is typically applied when speaking of amplifier design - EXTERNAL
negative feedback - just like about 99% of people would assume was meant in
this context.

I don't care for tubes. Others do. So be it.


I still prefer solid state, but even among those you have to spend quite
a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

I could find no references to the output impedance or damping factor for
this model, so I will assume your statement is accurate -- very much unlike
the one you make below.


While Krells, Meridians, CJs and other quality solid-state amps can
handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

Krell KAV500i = 0.044 ohms
Krell FPB-300c = 0.069 ohms

Are these modern enough? Should I go on?

-afh3

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET
amps
out there that offer pretty great performance.

No there aren't - not one single one.

Opinion of course. I don't want one, but there are many others with Bel
Canto designs (among others) who would disagree with your assessment.


There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I was of course, referring the to the use of negative feedback in the sense
that is typically applied when speaking of amplifier design - EXTERNAL
negative feedback - just like about 99% of people would assume was meant in
this context.

I don't care for tubes. Others do. So be it.


I still prefer solid state, but even among those you have to spend quite
a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

I could find no references to the output impedance or damping factor for
this model, so I will assume your statement is accurate -- very much unlike
the one you make below.


While Krells, Meridians, CJs and other quality solid-state amps can
handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

Krell KAV500i = 0.044 ohms
Krell FPB-300c = 0.069 ohms

Are these modern enough? Should I go on?

-afh3

"Stewart Pinkerton" wrote in message
...
On Thu, 04 Mar 2004 00:58:54 GMT, "afh3" wrote:

I'm no tube bigot, but there are certainly some very fine Class A SET
amps
out there that offer pretty great performance.

No there aren't - not one single one.

Opinion of course. I don't want one, but there are many others with Bel
Canto designs (among others) who would disagree with your assessment.


There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Triodes have internal negative feedback, and soft clipping is a sign
of inadequate linearity below the clipping point. You may *like* that
sound, but it ain't high fidelity!

I was of course, referring the to the use of negative feedback in the sense
that is typically applied when speaking of amplifier design - EXTERNAL
negative feedback - just like about 99% of people would assume was meant in
this context.

I don't care for tubes. Others do. So be it.


I still prefer solid state, but even among those you have to spend quite
a
bit of money to get output impedance values below or approaching .1 ohms.

No, you don't, a Yamaha AX-592 has plenty low output impedance.

I could find no references to the output impedance or damping factor for
this model, so I will assume your statement is accurate -- very much unlike
the one you make below.


While Krells, Meridians, CJs and other quality solid-state amps can
handily
beat even that value,

Actually, modern Krells do *not* have particularly low output
impedance.

Krell KAV500i = 0.044 ohms
Krell FPB-300c = 0.069 ohms

Are these modern enough? Should I go on?

-afh3

afh3 wrote:
"chung" wrote in message
vers.com...
afh3 wrote:
"chung" wrote in message
vers.com...

You can have an extremely low impedance, and yet not be able to source
a
lot of current. An op amp is a classical example.

Yes. But without the ability to source a lot of current, I'm sure you
can
see why we don't use 741's for output stages of amplifiers. I neither
stated
nor implied that one required the other -- simply that a good power
amplifier design would have both.

Regarding the SET amp not being popular, try rec.audio.tubes with that
line...



And I'm sure somewhere there is a forum or newsgroup where the
flat-earth theory is very popular...


I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

Performance is in the eye of the beholder, I guess. But you do agree
that class A SET's are not popular designs, by any measure? They are at
best a tiny niche. What is the sales figure for SET's compared to solid
state amps?

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Sure, there are always those who love euphonic distortion. Softer
clipping also means distortion at a lower level. But SET's being popular
in the overall scheme of things?


I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No. Unless your definition of big money is different than mine.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value, the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

Depends on your definition of isolation. If you use that ASCII picture
of yours, these amps isolate in the sense that Z1 does not see Z2.


So, I call biwiring "Hogwash". I don't do it. Do you?

My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

afh3 wrote:
"chung" wrote in message
vers.com...
afh3 wrote:
"chung" wrote in message
vers.com...

You can have an extremely low impedance, and yet not be able to source
a
lot of current. An op amp is a classical example.

Yes. But without the ability to source a lot of current, I'm sure you
can
see why we don't use 741's for output stages of amplifiers. I neither
stated
nor implied that one required the other -- simply that a good power
amplifier design would have both.

Regarding the SET amp not being popular, try rec.audio.tubes with that
line...



And I'm sure somewhere there is a forum or newsgroup where the
flat-earth theory is very popular...


I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

Performance is in the eye of the beholder, I guess. But you do agree
that class A SET's are not popular designs, by any measure? They are at
best a tiny niche. What is the sales figure for SET's compared to solid
state amps?

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Sure, there are always those who love euphonic distortion. Softer
clipping also means distortion at a lower level. But SET's being popular
in the overall scheme of things?


I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No. Unless your definition of big money is different than mine.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value, the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

Depends on your definition of isolation. If you use that ASCII picture
of yours, these amps isolate in the sense that Z1 does not see Z2.


So, I call biwiring "Hogwash". I don't do it. Do you?

My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

afh3 wrote:
"chung" wrote in message
vers.com...
afh3 wrote:
"chung" wrote in message
vers.com...

You can have an extremely low impedance, and yet not be able to source
a
lot of current. An op amp is a classical example.

Yes. But without the ability to source a lot of current, I'm sure you
can
see why we don't use 741's for output stages of amplifiers. I neither
stated
nor implied that one required the other -- simply that a good power
amplifier design would have both.

Regarding the SET amp not being popular, try rec.audio.tubes with that
line...



And I'm sure somewhere there is a forum or newsgroup where the
flat-earth theory is very popular...


I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

Performance is in the eye of the beholder, I guess. But you do agree
that class A SET's are not popular designs, by any measure? They are at
best a tiny niche. What is the sales figure for SET's compared to solid
state amps?

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Sure, there are always those who love euphonic distortion. Softer
clipping also means distortion at a lower level. But SET's being popular
in the overall scheme of things?


I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No. Unless your definition of big money is different than mine.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value, the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

Depends on your definition of isolation. If you use that ASCII picture
of yours, these amps isolate in the sense that Z1 does not see Z2.


So, I call biwiring "Hogwash". I don't do it. Do you?

My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

afh3 wrote:
"chung" wrote in message
vers.com...
afh3 wrote:
"chung" wrote in message
vers.com...

You can have an extremely low impedance, and yet not be able to source
a
lot of current. An op amp is a classical example.

Yes. But without the ability to source a lot of current, I'm sure you
can
see why we don't use 741's for output stages of amplifiers. I neither
stated
nor implied that one required the other -- simply that a good power
amplifier design would have both.

Regarding the SET amp not being popular, try rec.audio.tubes with that
line...



And I'm sure somewhere there is a forum or newsgroup where the
flat-earth theory is very popular...


I'm no tube bigot, but there are certainly some very fine Class A SET amps
out there that offer pretty great performance.

Performance is in the eye of the beholder, I guess. But you do agree
that class A SET's are not popular designs, by any measure? They are at
best a tiny niche. What is the sales figure for SET's compared to solid
state amps?

There are those who will have
nothing else (negative feedback being completely evil and all) because of
the softer clipping and numerous other reasons.

Sure, there are always those who love euphonic distortion. Softer
clipping also means distortion at a lower level. But SET's being popular
in the overall scheme of things?


I still prefer solid state, but even among those you have to spend quite a
bit of money to get output impedance values below or approaching .1 ohms.

No. Unless your definition of big money is different than mine.

While Krells, Meridians, CJs and other quality solid-state amps can handily
beat even that value, the ability for this to provide audible "isolation" in
a biwired configuration is questionable at best.

Depends on your definition of isolation. If you use that ASCII picture
of yours, these amps isolate in the sense that Z1 does not see Z2.


So, I call biwiring "Hogwash". I don't do it. Do you?

My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:



Yes, you are right. What I was trying to say here is that the
amount
of
resistance of the conductor doesn't change based on the load
value,
and
therefore the voltage drop on the conductor will always be
proportional
to
it's own resistance (and equal to the resistance times the
current).

Strictly speaking, that is not correct. For example, if the load is
a
short-circuit, then the voltage across the conductor is the same as
the
open-circuit votage of the source, *regardless* of the conductor
impedance.

Strictly speaking, as I stated above, it *is* equal to the current
times
the
conductor impedance, *regardless*.

Do the math to see.


If you do the math, you will see that in the case of a short-circuit
load, the drop across the conductor is *not* proportiional to its
resistance as you were saying. Here is what you said, to refresh your
memory: "therefore the voltage drop on the conductor will always be
proportional to it's own resistance".

If the short applied is a true (z=0) short, then the only remaining
resistance in the circuit is the conductor (ideal voltage sources having
zero impedance and all), right?

The amount of current flowing will be equal to the source voltage
divided by
the conductor resistance, right?

So the amount of voltage present across the conductor (necessarily equal
to
the source voltage) will be the current through the conductor times it's
resistance, right?

If that ain't proportional, I guess I don't the meaning of the word.


Here a simple way to look at it. If you double the resistance of the
wire, the voltage across it does not double, if the load is a short (or
an open). Simple enough?

Yes, simple enough - and the current will be half, and the voltage will
still equal the current times the resistance.

Thanks for re-iterating the point I made several posts ago.


I think (hope) we're saying the same thing.



When you said voltage drop is purely proportional to the resistance, you
are implying a linear relationship, like V(drop)=kRc. I just showed to
you that it is not a linear relationship. In fact, it is

V(drop)= Vs*Rc/(Rc+Rload).

That is not a linear relationship between V(drop) and Rc. In fact, wehn
Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing?

afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:



Yes, you are right. What I was trying to say here is that the
amount
of
resistance of the conductor doesn't change based on the load
value,
and
therefore the voltage drop on the conductor will always be
proportional
to
it's own resistance (and equal to the resistance times the
current).

Strictly speaking, that is not correct. For example, if the load is
a
short-circuit, then the voltage across the conductor is the same as
the
open-circuit votage of the source, *regardless* of the conductor
impedance.

Strictly speaking, as I stated above, it *is* equal to the current
times
the
conductor impedance, *regardless*.

Do the math to see.


If you do the math, you will see that in the case of a short-circuit
load, the drop across the conductor is *not* proportiional to its
resistance as you were saying. Here is what you said, to refresh your
memory: "therefore the voltage drop on the conductor will always be
proportional to it's own resistance".

If the short applied is a true (z=0) short, then the only remaining
resistance in the circuit is the conductor (ideal voltage sources having
zero impedance and all), right?

The amount of current flowing will be equal to the source voltage
divided by
the conductor resistance, right?

So the amount of voltage present across the conductor (necessarily equal
to
the source voltage) will be the current through the conductor times it's
resistance, right?

If that ain't proportional, I guess I don't the meaning of the word.


Here a simple way to look at it. If you double the resistance of the
wire, the voltage across it does not double, if the load is a short (or
an open). Simple enough?

Yes, simple enough - and the current will be half, and the voltage will
still equal the current times the resistance.

Thanks for re-iterating the point I made several posts ago.


I think (hope) we're saying the same thing.



When you said voltage drop is purely proportional to the resistance, you
are implying a linear relationship, like V(drop)=kRc. I just showed to
you that it is not a linear relationship. In fact, it is

V(drop)= Vs*Rc/(Rc+Rload).

That is not a linear relationship between V(drop) and Rc. In fact, wehn
Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing?

afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:



Yes, you are right. What I was trying to say here is that the
amount
of
resistance of the conductor doesn't change based on the load
value,
and
therefore the voltage drop on the conductor will always be
proportional
to
it's own resistance (and equal to the resistance times the
current).

Strictly speaking, that is not correct. For example, if the load is
a
short-circuit, then the voltage across the conductor is the same as
the
open-circuit votage of the source, *regardless* of the conductor
impedance.

Strictly speaking, as I stated above, it *is* equal to the current
times
the
conductor impedance, *regardless*.

Do the math to see.


If you do the math, you will see that in the case of a short-circuit
load, the drop across the conductor is *not* proportiional to its
resistance as you were saying. Here is what you said, to refresh your
memory: "therefore the voltage drop on the conductor will always be
proportional to it's own resistance".

If the short applied is a true (z=0) short, then the only remaining
resistance in the circuit is the conductor (ideal voltage sources having
zero impedance and all), right?

The amount of current flowing will be equal to the source voltage
divided by
the conductor resistance, right?

So the amount of voltage present across the conductor (necessarily equal
to
the source voltage) will be the current through the conductor times it's
resistance, right?

If that ain't proportional, I guess I don't the meaning of the word.


Here a simple way to look at it. If you double the resistance of the
wire, the voltage across it does not double, if the load is a short (or
an open). Simple enough?

Yes, simple enough - and the current will be half, and the voltage will
still equal the current times the resistance.

Thanks for re-iterating the point I made several posts ago.


I think (hope) we're saying the same thing.



When you said voltage drop is purely proportional to the resistance, you
are implying a linear relationship, like V(drop)=kRc. I just showed to
you that it is not a linear relationship. In fact, it is

V(drop)= Vs*Rc/(Rc+Rload).

That is not a linear relationship between V(drop) and Rc. In fact, wehn
Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing?

afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:

"chung" wrote in message
vers.com...
afh3 wrote:



Yes, you are right. What I was trying to say here is that the
amount
of
resistance of the conductor doesn't change based on the load
value,
and
therefore the voltage drop on the conductor will always be
proportional
to
it's own resistance (and equal to the resistance times the
current).

Strictly speaking, that is not correct. For example, if the load is
a
short-circuit, then the voltage across the conductor is the same as
the
open-circuit votage of the source, *regardless* of the conductor
impedance.

Strictly speaking, as I stated above, it *is* equal to the current
times
the
conductor impedance, *regardless*.

Do the math to see.


If you do the math, you will see that in the case of a short-circuit
load, the drop across the conductor is *not* proportiional to its
resistance as you were saying. Here is what you said, to refresh your
memory: "therefore the voltage drop on the conductor will always be
proportional to it's own resistance".

If the short applied is a true (z=0) short, then the only remaining
resistance in the circuit is the conductor (ideal voltage sources having
zero impedance and all), right?

The amount of current flowing will be equal to the source voltage
divided by
the conductor resistance, right?

So the amount of voltage present across the conductor (necessarily equal
to
the source voltage) will be the current through the conductor times it's
resistance, right?

If that ain't proportional, I guess I don't the meaning of the word.


Here a simple way to look at it. If you double the resistance of the
wire, the voltage across it does not double, if the load is a short (or
an open). Simple enough?

Yes, simple enough - and the current will be half, and the voltage will
still equal the current times the resistance.

Thanks for re-iterating the point I made several posts ago.


I think (hope) we're saying the same thing.



When you said voltage drop is purely proportional to the resistance, you
are implying a linear relationship, like V(drop)=kRc. I just showed to
you that it is not a linear relationship. In fact, it is

V(drop)= Vs*Rc/(Rc+Rload).

That is not a linear relationship between V(drop) and Rc. In fact, wehn
Rload=0, V(drop) is *independent* of Rc. Now are you saying the same thing?

"chung" wrote in message
rvers.com...


My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

Yes, yes, yes, mea culpa. I did say that I should not have included the word
"only" when I made that statement. I wasn't using the term to exclude
current from the equation but rather to exclude the load from any bearing on
the resistance of the conductor -- and therefore it's voltage drop
characteristics.

"chung" wrote in message
rvers.com...


My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

Yes, yes, yes, mea culpa. I did say that I should not have included the word
"only" when I made that statement. I wasn't using the term to exclude
current from the equation but rather to exclude the load from any bearing on
the resistance of the conductor -- and therefore it's voltage drop
characteristics.

"chung" wrote in message
rvers.com...


My argument was with your statement that "voltage drop in a conductor is
only dependent upon its resistance". I see that you have now understood
that error.

Yes, yes, yes, mea culpa. I did say that I should not have included the word
"only" when I made that statement. I wasn't using the term to exclude
current from the equation but rather to exclude the load from any bearing on
the resistance of the conductor -- and therefore it's voltage drop
characteristics.