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# Re: Dupla Cables

```
>From: KB Koh <KB_Koh at ccm_ipn.intel.com>
>
>>From: chapman at SEDSystems_ca
>>Date: Wed, 6 Mar 1996 10:36:48 -0600
>>Subject: Dupla Cables (The whole message)
>>
>>
>>In the first place, electricity does not just take the path of least
>>resistance -- it takes every available path.  You can see that this is true
>>from your own experiences -- you can plug several different items into a
>>power bar, and electricity flows into all of them, even though each has a
>>different resistance.
>
>Electric current actually take the path of least resistance. If one of your
>several items has a short circuit to ground, all the current will flow
there and
>none to the rest of your items. The sum of current to all items is equal to
the
>input current. Something got to do with Kirchoff(sp?) Law.
>
>>The purpose of the ground wire is to make a low resistance path to ground
>>available so that if a fault occurs, a large current will flow and result in
>>a blown fuse or circuit breaker.  If the cable has an internal fuse, as the
>>salesperson implied, that is its purpose.  No fuse can tell if current is
>>flowing through your body or not.
>
>Partly true because the path to ground has the least resistance. All currents
>flow to the ground and none through your body. The ELCB would then trip. The
>fuse would blow only if the ELCB malfunction or the ground wire is open, as
>mentioned by the sales person.
>

As an electrical engineer, I must agree with the first poster, that
"electricity takes all available paths."  Given a voltage source with
parallel paths to ground, electricity will flow through all paths, with
current varying inversely to the resistance in each path.

The reason the ground wire works is that no voltage source is an ideal
voltage source.  If we have a ground path with near zero resistance, the
current drawn by that path will be so great that either:

(a) the voltage drop across the internal resistance of the voltage source
(every non-ideal voltage source has a non-zero internal resistance)
approaches the total voltage, so the output voltage is reduced to near zero.
It won't be zero, because there it will reach an equilibrium determined by
the ratio of the resistances (load/ground vs. internal).  Since the output
voltage is so low in this stage, the current through your body will be
unnoticeable (although non-zero), but it will most probably be in this state
a very short period of time, because...

(b) the large total current causes a fuse to blow.

Here's an ASCII representation of the whole circuit:

--------/\/\/\-----o ------------------------------------
|     Internal                     |                     |
|     Resistance   |               |                     |
|                                  \                     \
Voltage  (V)              Output      Ground /             Your    /
source   |              voltage       wire  \             body    \
|                        (very low /            (high    /
|                  |    resistance)|          resistance)|
|________ _________o_____________________________________|
~         |
Fuse      ___  Ground
_

***************************************************
Hoa G. Nguyen
NRaD Code 531               Email: nguyenh at nosc_mil
San Diego, CA 92152-7383    Tel.:  (619) 553-1871
URL: http://www.nosc.mil/robots/people/nguyenh.html
***************************************************

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