Path of least resistance question (related to book)

Discussion in 'General Electronics Chat' started by DGex, Nov 13, 2012.

  1. DGex

    Thread Starter New Member

    Nov 13, 2012
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    Specifically this chapter:

    http://www.allaboutcircuits.com/vol_1/chpt_3/3.html

    So this book so far has helped me untangle some misconceptions but I am still stuck on a few things.

    1. Something being common means that its resistance (I believe) in connection with a wire is 'close' enough where they are considered connected. So let's talk about this picture:

    [​IMG]

    and the quote from the book

    Isn't there a path though from the circuit, through the guy and back to the earth though? What's the difference where the end of the path is if the ground is 0V at the human's leg to ground and the circuit has a higher voltage, won't the electrons have a desire to flow through?

    2. [​IMG]

    Aren't multiple paths to ground here going to create a possible scenario of electrons still flowing to the guy?

    Finally, a small side question when voltage drop is mentioned in some chapters.

    In a series circuit, it is stated that the total voltage drop across all loads must equal the voltage of the source. Let's say I have a DC 9V battery and two wires in series connected to a load. There is a voltage drop of 4V across the load. I fail to see how the above statement is true considering that only one voltage drop of 4V exists and it is certainly not equal to source voltage alone. The positive common wire set (wires after the load) will now read as 5V and the negative common wire set (from the source) will read as 9V.
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
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    1. The negative voltage side of the source is not connected to ground, so it doesn't matter if the person is....

    2. The presence of the person creates a large resistance from ground to... well, ground... large resistor vs 0(theoretical resistance) in the circuit's return path, the electrons will travel through the wire, not the person....

    3. the only way you should see this is if you either used high resistance wires in your circuit, or your resistor is comparable to the internal resistance of the battery...see here
     
  3. DGex

    Thread Starter New Member

    Nov 13, 2012
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    Thanks for the replies.

    For your answer to number one though, isn't there going to be a difference in voltage so some of the electrons would move through the person? To me, it would seem that there might be, on the positive side, a certain amount of voltage after the electrons pass through the load. So, with there being 'x' amount of voltage on the positive side and 0 volts leading to ground from the person, there would exist a potential difference. This potential difference, to me, would result in some of the electrons going through the person and into the ground. Where am I going wrong with this idea?
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
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    There well may be a voltage, and, if the voltage were high enough, it will cause the medium to breakdown and conduct.

    However, assuming that we are not working with incredibly high voltages, a voltage doesn't mean current flow. Conversely, a current flow does not mean voltage. The only way a current path can be made between the person and the negative side of the voltage source is if the source is connected to earth ground in some fashion.

    Now, if the person held the other side of the cable and attempted to pet the pretty bird standing on the wire, then the person would probably have a nice lunch after(assuming the person survived :)), though that would be dependent on the load resistance...
     
  5. DGex

    Thread Starter New Member

    Nov 13, 2012
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    In the first picture though, the person's hand is holding the wire after the load (the positive side). Sure the current originated from the source but wouldn't it go wherever there is some difference to drive it? After the load, there would still be a current and a voltage. It has two places to go, one being the end of the circuit and the other being to ground through the person. Why wouldn't the electrons output from the load go through to ground at 0V (through the person since there is a potential between that wire and the ground) versus just going to the end of the circuit after the load?
     
  6. davebee

    Well-Known Member

    Oct 22, 2008
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    In picture 1, there is no voltage specified between the circuit and ground, so you can't say what would happen.

    It's true that in real life, there might be a voltage, and the person could get a shock. But in a teaching diagram, you can't assume conditions that are not specified.
     
  7. tshuck

    Well-Known Member

    Oct 18, 2012
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    Perhaps I made it a little more confusing than need be... :p

    You should re-read this page to cure your notion of voltage -> current or current -> voltage...

    Why would there be a voltage after the load with respect to the source? The load has dropped the source voltage, the only way that it doesn't is if there is another load in series with it, which, there isn't...

    The person could very well be holding on to the source itself and not have current move through his body.

    You tell me, then, why would the current go to ground? Would it not need to find a way back to the source? There is no specified connection between the source and ground, therefore, there is infinite resistance between earth ground(attached to the person) and the source. R = ∞ -> I = 0
     
  8. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    The figure is a theoretical discussion that assumes PERFECT isolation between the voltage and the earth ground. Your intuition that it may not be completely safe to touch this anyway is a very wise thing - I wouldn't touch it, though an ungrounded person can touch a grounded circuit, and frequently does so in high tension line repairs.

    I would need a link to where you are looking to fully comment on the 9V circuit. Indeed, if you have a 9V source and 4 V at the load then 5V is getting dropped somewhere!

    Do keep in mind that real world conductors are not perfect and may (do actually) have drops along the wire. If the wire is small enough and the load large enough then you may very well loose 4 volts in the wires alone. That's one reason why in items such as solar cells the connecting wires are way oversized to give a minimal voltage drop and thus get the maximum useful power out of the cell.
     
  9. Six_Shooter

    Member

    Nov 10, 2012
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    I think the confusion is that in many circuits the return path or negative side of the circuit is also referred to as "ground" which in many cases is not entirely correct, it's just a convention that many people go by.

    In the above pictures, lets call the node at the top "A" and the node at the bottom "B".

    In the top picture, the node B, is only connected to the voltage source and the load, one path for current, no more. The fact that the person is touching node A is irrelevant, the person and in this case the "ground" has the same potential as node A. If you were to measure the voltage between the node A and the ground at the persons feet, you would read 0V, because they share the same potential. Node B in the first picture is not connected to anything but the voltage source and the load, there is no path to connect it to "ground." This is why the person has no path for current to flow.

    In the second picture, it is the same scenario, with the difference being that the person is touching what is the node B, or 0V reference of the circuit.

    In the 1st picture, the only way for there to be current through the person is to connect the node B to ground, or to get picky, for the person to also be touching node B.
    In the second picture, a similar situation would have to apply, the person would have to be touching both node A and node B, for there to be a current path through the person.

    In the end it's very important to realize that just because something may be connected to the negative side of a voltage source, it does NOT automatically mean that it is also connected to ground, it may be, and would as suggested not take that for granted, when working with electronics.

    You will get to a point where you will see that just because a node may be connected to the negative side of a voltage source, does not mean that it is always equal to 0V. What I mean by that is that you can have voltage sources connected in such a way that a node that shows 0V in reference to another node, that same node can show a positive voltage in reference to third node, possibly a node named "C", but we'll leave that idea for now, just get used to the idea that the negative side of a voltage source, or current source for that matter, does not automatically equal ground. There needs to be a complete path from positive to negative for current to flow, any branches off either node, that have no connection on the other will not have current flowing through them.
     
  10. BreadCrum6

    New Member

    Aug 17, 2011
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    Just to simplify what has already been said.

    As I see it, the assumption in the first photo is that the source is isolated from Earth ground. If that is true, ideally, the only way the Guy can get shocked is if he was in series with the wires or in parallel with the resistor. You can equate the Guy as a large resistor that touches the circuit but is neither in series or parallel with the circuit.


    In the second photo, the Guy (very large resistor) is within the current path of the source but he is in parallel with an ideal wire therefore the current takes the path of least resistance. This is similar to the bird in the first photo.
     
  11. Six_Shooter

    Member

    Nov 10, 2012
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    You have the first picture correct, but not the second.

    The second picture has nothing to do with the possible resistance value of the person, but simply being at the same potential as the node, with no path for current.
     
  12. tshuck

    Well-Known Member

    Oct 18, 2012
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    Agreed.

    (this is padding to ensure 10 characters :))
     
  13. BreadCrum6

    New Member

    Aug 17, 2011
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    You're right. There's no complete path for current.
     
  14. tshuck

    Well-Known Member

    Oct 18, 2012
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    You are assuming nothing! This schematic does not say they are connected. You don't assume two terminals of a voltage source are connected when a schematic shows no such connection.
     
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