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#281
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Bi-wiring - Hogwash?
"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 |
#282
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Bi-wiring - Hogwash?
"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 |
#283
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Bi-wiring - Hogwash?
"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 |
#284
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Bi-wiring - Hogwash?
"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 |
#285
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Bi-wiring - Hogwash?
"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 |
#286
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Bi-wiring - Hogwash?
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#287
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Bi-wiring - Hogwash?
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#288
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Bi-wiring - Hogwash?
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#289
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Bi-wiring - Hogwash?
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#290
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Bi-wiring - Hogwash?
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 |
#291
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Bi-wiring - Hogwash?
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 |
#292
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Bi-wiring - Hogwash?
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 |
#293
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Bi-wiring - Hogwash?
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 |
#294
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Bi-wiring - Hogwash?
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 |
#295
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Bi-wiring - Hogwash?
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 |
#296
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Bi-wiring - Hogwash?
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 |
#297
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Bi-wiring - Hogwash?
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 |
#298
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Bi-wiring - Hogwash?
"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 |
#299
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Bi-wiring - Hogwash?
"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 |
#300
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Bi-wiring - Hogwash?
"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 |
#301
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Bi-wiring - Hogwash?
"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 |
#302
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Bi-wiring - Hogwash?
"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 |
#303
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Bi-wiring - Hogwash?
"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 |
#304
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Bi-wiring - Hogwash?
"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 |
#305
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Bi-wiring - Hogwash?
"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 |
#306
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Bi-wiring - Hogwash?
"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 |
#307
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Bi-wiring - Hogwash?
"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 |
#308
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Bi-wiring - Hogwash?
"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 |
#309
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Bi-wiring - Hogwash?
"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 |
#310
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Bi-wiring - Hogwash?
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. |
#311
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Bi-wiring - Hogwash?
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. |
#312
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Bi-wiring - Hogwash?
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. |
#313
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Bi-wiring - Hogwash?
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. |
#314
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Bi-wiring - Hogwash?
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? |
#315
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Bi-wiring - Hogwash?
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? |
#316
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Bi-wiring - Hogwash?
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? |
#317
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Bi-wiring - Hogwash?
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? |
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Bi-wiring - Hogwash?
"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. |
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Bi-wiring - Hogwash?
"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. |
#320
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Bi-wiring - Hogwash?
"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. |
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