Section Electrical Safety : Shock Current Path

Discussion in 'General Electronics Chat' started by foolios, Jan 1, 2010.

  1. foolios

    Thread Starter Active Member

    Feb 4, 2009
    160
    1
    There was some explanation about common points of a wire in another section. That taking a voltage reading at two points on the same wire will result in no voltage difference.

    So in this case where the person is grounded along the same wire as a ground, they are common to each other which prevents him from being shocked; no voltage difference.

    But he is in parallel to the wire so I imagine the only real reason he's not getting shocked is because the ground isn't going to be as good a conductor in this scenario as the wire of the circuit.


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    But now let's replace this person and ground with a thicker wire. Even though this thicker gauge wire is connected in common to the original wire at two points, current will travel across this thicker(less resistant) wire rather than the original.

    So common testing is for determining whether you're on the same wire. You'll get no voltage difference. But after a load, there will be a voltage drop. That begins a second wire and a difference in potential.

    This part is fuzzy, why does a difference in potential exist now in this second wire after the load even though it's in series with the original wire. I don't understand what this drop actually is that allows us to measure this difference at this point of the circuit(just after the load) as opposed to on the original wire. Does the load act as a blocker between wire 1 and wire 2? Because if we short circuit the source with 1 wire we get a difference in potential. But now wire 1 added to the circuit with load and wire 2, we lose that difference in potential on wire 1 but why?

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    Last edited: Jan 23, 2010
  2. Paulo540

    Member

    Nov 23, 2009
    188
    0
    When you measure the potential across the battery, you are not short circuiting it. The meter has an input resistance.

    The load resistor becomes a drop, based on its resistance, so when you measure across it, you are measuring how much it is 'using', in this case it would be 10v since its the only resistor. If you had different values in series, you would measure a different voltage across each one, but they would all still add up to the 10v source.

    http://www.allaboutcircuits.com/vol_1/chpt_1/2.html

    this might help with the basics.
     
  3. foolios

    Thread Starter Active Member

    Feb 4, 2009
    160
    1
    So then when you are connecting the meter to two points on the same wire 1, what is happening is that even the slight resistance in the meter will prevent any voltage reading because the power will not flow through it as a result of it having an alternate almost purely non-resistant path to travel through(wire 1) thus avoiding the meter.

    Then I am guessing the reason why we will get a reading on the meter connecting to a point before(wire 1) and also after the load(wire 2) is that now the circuit has enough resistance that the meter's internally small resistance will experience power that seeks to go through it(the meter) as a result of a much higher resistance being experienced along the entire path that include wire 1, load and wire 2.

    Does this seem to be the case?
     
  4. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Try using math. It is the language of electronics.

    If you have 0.2Ω resistance in a wire that is carrying 10A, it will drop 2V. This wire is pretty flawed BTW. If you put a 2V across a meter with 10MΩ input resistors (the norm for DVMs) you have 0.2µA (0.0000002A).

    Most wires would have millivolts drop however, so these numbers go way down.
     
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