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#11
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1997 Savana Van has my mechanic STUMPED!
On Jun 4, 11:41 pm, Erik > wrote:
> In article >, > > "M.M." > wrote: > > Noozer wrote: > > > > A high resistance circuit will draw more amps than normal. > > > How so? > > > By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a > > more or less constant voltage as you'd have in a car, as R increases, I > > decreases. > > M.M.'s right... > > When these pumps 'fail', are you guys checking to be sure power is > actually being delivered to the pump, and that a good reliable ground is > provided? > > As another poster mentioned, are the connector/s in good shape. > > I have a feeling this one is going to be something simple along those > lines... > > Erik Thanks Erik! Yeah, I feel the same, that it will probably be a simple thing, its just finding it!! Bill |
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#13
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1997 Savana Van has my mechanic STUMPED!
On Mon, 04 Jun 2007 19:09:22 -0700, "M.M." > wrote:
>Noozer wrote: >> >> A high resistance circuit will draw more amps than normal. >> >> > > >How so? > >By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a >more or less constant voltage as you'd have in a car, as R increases, I >decreases. So what happens when the V drops, R stays the same, I does what?? SteveL |
#14
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1997 Savana Van has my mechanic STUMPED!
On Tue, 05 Jun 2007 01:25:57 GMT, Noozer > wrote:
>> If the ground were bad, wouldn't you expect the current to be low, on the >> average, >> rather than high? >A high resistance circuit will draw more amps than normal. exactly wrong |
#15
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1997 Savana Van has my mechanic STUMPED!
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#16
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1997 Savana Van has my mechanic STUMPED!
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#17
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1997 Savana Van has my mechanic STUMPED!
"M.M." > wrote in message ... > wrote: >>> By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a >>> more or less constant voltage as you'd have in a car, as R increases, I >>> decreases. >> >> So what happens when the V drops, R stays the same, I does what?? >> > > Well, let's see...suppose it was 12v and 12 ohms, just to make it easy. > That would make I be 12/12 = 1 amp. > > Now, suppose the voltage drops to 6v, I would be 6/12 = .5 amp. > > Looks to me that I would decrease then, too. > > At least, that's the way I remember it from Circuits 101, way back when > they taught us about vacuum tubes, but I don't think the laws of physics > have changed much since then... Yaya... I get it. I was wrong. So why do you need to use a larger gauge wire in a longer extension cord? Obviously there is more resistance in 500' of 14gauge wire than there is in 50'. According to what you're saying, you should be able to carry more current over a longer distance due to the resistance. |
#18
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1997 Savana Van has my mechanic STUMPED!
Noozer > wrote:
> > >So why do you need to use a larger gauge wire in a longer extension cord? >Obviously there is more resistance in 500' of 14gauge wire than there is in >50'. According to what you're saying, you should be able to carry more >current over a longer distance due to the resistance. This is a voltage drop issue. Your load is in _series_ with the resistance of the wire. The total resistance of the system is (Rwire + Rload) so the total current through the system can be found with Vbattery = I(Rwire+Rload) Now the idea here is that that additional resistance is _reducing_ the total amount of current that can flow through the circuit. Another way of looking at this is by saying that the voltage the load sees is the voltage the battery produces MINUS the voltage drop due to resistance. --scott -- "C'est un Nagra. C'est suisse, et tres, tres precis." |
#19
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1997 Savana Van has my mechanic STUMPED!
Noozer wrote:
> "M.M." > wrote in message > ... >> wrote: >>>> By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a >>>> more or less constant voltage as you'd have in a car, as R increases, I >>>> decreases. >>> So what happens when the V drops, R stays the same, I does what?? >>> >> Well, let's see...suppose it was 12v and 12 ohms, just to make it easy. >> That would make I be 12/12 = 1 amp. >> >> Now, suppose the voltage drops to 6v, I would be 6/12 = .5 amp. >> >> Looks to me that I would decrease then, too. >> >> At least, that's the way I remember it from Circuits 101, way back when >> they taught us about vacuum tubes, but I don't think the laws of physics >> have changed much since then... > > Yaya... I get it. I was wrong. > > So why do you need to use a larger gauge wire in a longer extension cord? > Obviously there is more resistance in 500' of 14gauge wire than there is in > 50'. According to what you're saying, you should be able to carry more > current over a longer distance due to the resistance. He addressed what happens when resistance stays the same and something else changes. You've somehow parlayed that into what happens when resistance changes and the other parameters are unspecified. If you'll specify what happens to either voltage or current, we'll tell you which way the other parameter goes. |
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