Damn these electrically common points! xD

Discussion in 'General Electronics Chat' started by Michael De Carvalho, Jul 17, 2017.

  1. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    What's up guys, so I'm pretty new here and I am having trouble understanding a few things, hoping you guys can help me out, and yes I know there are so many posts of people asking similar questions, they just not making sense to me.

    eeeee.png
    So from what I understand points are electrically common if they are connected by the same wire and there is practically no voltage across them(zero if superconductor) , that does't make much sense to me as to why that is the case, from what I gather it is because of the low resistance, but that brings me to a question, I've read that V is proportional to R , so in the image attached why is there 10V across the hanging lines and 2390V by the earth? Is it because earth has a lot of resistance? I mean how in a cable there is "no" voltage then it reaches a resistor then there is all of a sudden high voltage?

    And if you become electrically common, how do you obtain the same voltage as the wire?

    Sorry if I am asking stupid questions, just really want to understand.Thanks <3
     
  2. LesJones

    Active Member

    Jan 8, 2017
    643
    132
    It is just two resistors in series with 2400 volts connected between the ends. The The wire is one of these resistors (Which has 10 volts drop across it.) and the other resistor is the earth between the negative end of the supply and the point where the end of the broken wire makes contact with the ground. (We sould really take into account the voltage drop in the length of wire between the third pole and the end of the wire that touches ground.) If we ignore the resistance of the wire from the top of the third pole to ground we can say that the resistance of the ground path is 239 times the resistance of the wire as the current through the ground is the same as the current through the wire.

    Les.
     
  3. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    2,018
    350
    I think that is a very poor picture of what's going on.

    Let's TRY to put some sense into the picture. Design criteria for transmission lines is to have 3 to 5% drop from one end to another. HV feeders, not sure what it is. But the power company does generate an AC voltage relative to earth.

    Since the overhead transmission line has resistance, I think they arbitrarily chose 10V as the voltage drop.
    5% of 2400 V is 120 V.

    A pole with a transformer on it will have a good ground and that's probably the path that's going to be taken.

    So resistivity varies a LOT. See https://en.wikipedia.org/wiki/Soil_resistivity

    I picked a few numbers out of my hat and said a 5 x 10 cm shoe (cross sectional area) and a reasonable foot distance, I got 250 ohms between two metal pads about the size of shoes sitting on the Earth. That does not include the shoes, and the human body resistance up the leg and back down.

    Staying away from downed power lines is still a good idea. If we suppose it's raining and there is a downed line.
    We really don;t know the path. Trees have moisture. The wind blows, so it;s not a good idea to get close to a downed wire.

    if a concrete pad is not poured correctly around a pool, you can get shocked in a lightning storm. bare feet, voltage are higher. The ground potential is different.

    There CAN be a voltage drop across the Earth and it does occur with thunderstorms.

    Ohms Law is not a bad first approximation. Temperature can change the behavior. AC circuits with inductances and capacitances can make it not work. For now, believe it. R=V/I

    There is a material property called resistivity p and R=pL/A where Rho is a material property called resistivity in ohm-cm, L is length in cm and A is the cross sectional area in sqcm.

    Electric shock is very hard to quantify: https://en.wikipedia.org/wiki/Electric_shock

    See: http://www.apscservices.info/pdf/14/14-069-c_20_1.pdf

    and

    https://en.wikipedia.org/wiki/Earth_potential_rise
     
  4. profbuxton

    Member

    Feb 21, 2014
    313
    131
    The 10 volts across the line on the poles is as a result of the current flowing through the LOW(relatively) resistance of the wire. This same current flows through the fallen line back(direction optional, its how you chose it) through the resistance of the soil which may(will) have a much larger resistance than the wire. Therefore there will(possibly) be a 2390v drop in that soil section. If you are standing on the soil where the current is flowing in bare feet and legs spread you MAY experience a voltage across your body.
    Thats why when designing high voltage switchyards the earth mat under ground must be carefully designed to reduce possibly dangerous"step" potentials if a fault or lighting strike occurs.
    Note that in the case you have shown since there is still full 2390 volts drop across the soil path the power source(battery or power station) will see this as normal load(but trip breakers on earth fault). In a real situation the soil resistance is usually low enough to cause the protection gear to switch off the power.
     
    Michael De Carvalho likes this.
  5. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    thanks guys, but I'm still confused about my initial questions though, you guys had some long replies which I have no idea what you mean, I think you assume I have more knowledge than I do xD
     
  6. dl324

    Distinguished Member

    Mar 30, 2015
    4,753
    1,031
    Welcome to AAC!
    The 10V drop to the 3rd pole is from resistance in the wire. Only superconductors have "zero" resistance. Transmission lines are often aluminum because they're lighter than copper; but have higher resistance.

    The picture is also showing that there is a voltage gradient from the point where the fallen line contacts the ground back to what the picture is calling GND.

    If you ever find yourself in that situation, you want to keep your feet together and hop away from the fallen line.

    The picture is a little misleading because it shows a 10V drop to the 3rd pole, but no drop for the segment from there to the 4th pole (which is now touching the ground). It also shows electron current flow which is backwards from the way most of us think about (conventional) current flow.
     
  7. crutschow

    Expert

    Mar 14, 2008
    16,220
    4,336
    The power line makes a complete circuit when it touches ground.
    There are two resistors in series in this complete circuit, the line resistance and the ground resistance.
    The ground has a much higher resistance than the power line.
    Voltage drop for resistors in series is proportional to their relative resistance.
    For the relative values picked for this example only 10V is dropped across the power line resistance and the rest of the voltage is dropped across the ground resistance.
    This means the ground potential at the line-down point jumps to the line voltage minus 10V (2390V) relative to the generator ground voltage.
    This ground voltage slowly drops from 2390V to 0V as you travel across the distributed resistance from the line-down point to the generator ground point.

    Make sense?
     
    Michael De Carvalho likes this.
  8. MrChips

    Moderator

    Oct 2, 2009
    14,296
    4,198
    Ever been struck by lightning and lived to tell the tale?

    At low voltages (<30V maybe) the current path in a circuit is somewhat predictable.
    As you get into higher voltages (>100V, 1000V and beyond) a lot of unpredictable things happen.

    Motto: Don't mess around with high voltages.
     
    cmartinez likes this.
  9. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    2,018
    350
    Really good advice, At work I used to mess with high voltage stuff:
    a) A 15 kV, 1.5 Amp regulated power supply (e-beam evaporator)
    b) A 100 kV, 0.1 A unregulated supply (X-ray diffraction PS)
    c) A 1000 W RF transmitter (3000 VDC on the plate for RF sputtering)

    No mishaps.
     
  10. LesJones

    Active Member

    Jan 8, 2017
    643
    132
    I have re read your question. I think the point taht you are not understanding is that a voltage is always measured between two points. You can't say the voltage at a point is a specific value. It has to be with reference to some other point. It would be like sayng London is 200 miles. You would have to say London is 200 miles from Liverpool for example. We will call the power source a battery so make the explanation simpler. The negative terminal is connected to ground at the power source location as well as the negative power line. The positive of the battery is +2400 volts with respect to ground at the source location. As there is current flowing through the wire and the wire has some resistance there will be a voltage between the battery positive end and the end that is in contact with the ground.( We will consider this to be 10 volts as there is an error in the drawing.) This 10 volts is subtracted from the 2400 volts so the voltage WITH RESPECT to the battery negative is + 2390 volts.
    Does that help your undersatnding ?

    Les.
     
    Michael De Carvalho likes this.
  11. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    Yea, but that brings me to a question I forgot to ask, I don't understand how being on 2 feet makes a difference(meaning 2 separated feet) ? Is it that being on one foot you become electrically common and then being on 2 completes the circuit? If so that just confuses me more, how would that make a difference? You grounded either way?

    And thanks for the reply :D
     
  12. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    Ooooh okay, no that does make sense thank you, I didn't understand because I misunderstood voltage drop, but that makes sense now..Although I am confused still about kirchhoff's voltage law, how does it always end up equaling 0v by the end of the circuit? what if the earth had a different resistance, why would it still equal 0v?
     
  13. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    Yes it does, thank you for your efforts , they are much appreciated <3
     
  14. crutschow

    Expert

    Mar 14, 2008
    16,220
    4,336
    It always equals 0V around the loop irrespective of the relative impedances.
    If it didn't equal 0V, where would you put the left over voltage? :confused:
     
  15. dl324

    Distinguished Member

    Mar 30, 2015
    4,753
    1,031
    If you place your feet together, there's no (actually there is a little) voltage difference, so you can't get shocked just standing there.

    If you separate your feet, there will be a voltage difference between your feet and current will flow from one foot to the other.

    If you were standing at the 2000V "point" with your feet together and reached out and touched something that was at a lower potential, current would flow from your feet to your hand.
     
  16. atferrari

    AAC Fanatic!

    Jan 6, 2004
    2,830
    928
    Had the drawing included arrows showing eventual current flowing from one foot to the other, through the body (maybe frying or just cooking delicate accesories) it would make things a little easier to grasp.
     
    cmartinez and KeepItSimpleStupid like this.
  17. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    Ahh, alrighty, thank you man. <3
     
  18. Michael De Carvalho

    Thread Starter New Member

    Jul 13, 2017
    7
    0
    Ohh... Please tell me if I am understanding it correctly , feet together the voltage flows through both feet so voltage isn't dropped as if 9same voltage in both feet, but feet separated as it flows from one leg to another voltage is dropped and that difference in voltage is what causes a person to shock, right?

    Sorry for being such a nuisance, but that makes me wonder, so if you do not touch something you become electrically common with the ground but if you do touch something (like a car) then you complete the circuit and you get shocked?
     
  19. dl324

    Distinguished Member

    Mar 30, 2015
    4,753
    1,031
    Yes. The voltage difference between the feet causes current to flow.
    Not exactly. Your feet and hands would be at different potentials; that's what causes the shock.

    These days, few cars have grounding straps; so the car would be isolated by the rubber tires. When you touched the car, it would assume the potential as your body.
     
  20. crutschow

    Expert

    Mar 14, 2008
    16,220
    4,336
    The ground looks like one long continuous resistance, thus the voltage drops along the length of ground from the line touching the ground back to the generator ground.
    This voltage gradient is what can cause a shock.
    Thus suppose the ground gradient was 100V per foot of distance where you were standing.
    If you feet were 1 foot apart there would be a 100V difference and if they were 2 feet apart there would be a 200V difference.
    Thus the farther your feet are apart, the more voltage between your feet and the stronger the shock.
     
    Kjeldgaard likes this.
Loading...