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Fwd: [vox-tech] ohms law
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Fwd: [vox-tech] ohms law




Begin forwarded message:

From: jim stockford <jim@well.com>
Date: February 2, 2007 7:01:12 PM PST
To: evesautomotive@charter.net
Subject: Re: [vox-tech] ohms law


it's a matter of a voltage divider.
   normally, hot and return legs have no impedance.
the device presents its designed impedance to the
supplied voltage (again, assuming all is normal).
   if either hot or cold leg has significant impedance
(take your case, the cold leg), then the supplied
voltage traverses through the device and its
impedance then through the return (cold) leg and
its impedance--power (P = E x I) is divided.
   assume device presents 100 ohms to normal
10 Volts, you get 0.1 amp and all is well.
   given a return with 50 ohms, the current is about
0.06 amp and voltage across the device is about
6.6 Volts and voltage across the leg's impedance
is about 3.3 Volts.
   0.06 A x 100 ohms is 6 Volts,
   0.06 A x 50 ohms is 3 Volts,
   3 V + 6 V = 9 V (close enough for electronics)
   In other words, the device onlygets voltage from
the supply side to the top of the return leg.
hope it makes sense.


On Feb 2, 2007, at 12:34 PM, Jimbo wrote:

Greetings:
This is a technical question that concerns dc voltage so hopefully someone that has knowledge in this area can help me with this.

I know that computers use low dc voltage of 12, 5 and 3.3 volts so hopefully this will fit the mail list criteria.

I am a mechanic by trade. I am good at diagnosing electrical and drivability. I have seen a few times that high resistance in the negative leg of a circuit can take out components like computers, modules and even not-so-complicated devices like bulbs and switches. What I don't understand is why. Ohm's law states that E=IXR. If this is the case then if resistance is high it will decrease amperage. I would tend to think that just the opposite would happen...component would just lose power and not fry.

Please enlighten me,

Jimbo

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