Volt drop in a cable

Discussion in 'Homework Help' started by RdAdr, Jan 29, 2016.

  1. RdAdr

    Thread Starter Member

    May 19, 2013
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    Consider a constant supply voltage VS and a consumer with a power rating of P watts.
    Then they calculate the voltage drop on the cable connecting the supply with the consumer using Ohm's law like this:

    1. Calculate current drawn by consumer: I = P/VS
    2. Calculate voltage drop on cable: VD = R*I, where R = resistivity*length/area

    So they assume that the consumer will draw that current. But the current through the circuit is I = VS/(RC+R) and not I=VS/RC = P/VS, RC being the resistance of the consumer.

    Why they first assume that the current will go through the consumer like all the supply voltage will fall on it? The voltage that the consumer will see is less than the supply voltage. Thus the current drawn is smaller too.
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,136
    1,786
    The resistance of the cable in both directions is added to the consumers load and the current draw is limited by the magnitude of three resistances in series. As in all series circuits, the voltage drop across the load will be less than, sometime much less than, the VS of the constant voltage supply.
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,094
    3,033
    That is a simplifying assumption, that the drop across the supply wiring will be very low compared to the drop at the load. Like all simplifying assumptions, they are useful but not always true.
     
  4. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    I understand that.

    I will just go with the simplifying assumption.

    Thanks for answers.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,421
    490
    Hi,

    The power company also loses money when the line voltage drops.
     
    RdAdr likes this.
  6. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    In the book it also says:

    "The choice of cable size depends on the current drawn by the consumer. The larger the cable used then the smaller the volt drop in the circuit, but the cable will be heavier. This means a trade-off must be sought between the allowable volt drop and maximum cable size. "

    The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?
     
  7. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    I think it refers to the cross-sectional area, not its length. The choice of words is very poor.
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,421
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    Hi,

    The key is in the first sentence. You dont adjust the *length* of the cable based on the current drawn, you adjust the diameter. Therefore that idea carries over to the next sentence where 'larger' would mean a larger diameter. They could have made it more clear though for someone who doesnt already know how this works.
     
    anhnha likes this.
  9. profbuxton

    Member

    Feb 21, 2014
    233
    68
    "The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?"


    WRONG: Larger cable is LESS resistance=Lower volt drop. Look up cable supplier tables of resistance per metre.
    Usually maximum allowable volt drop is 10%.(in OZ).
     
  10. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    I know but english is not my first language. "Larger" in my language means both "longer" and "larger in diameter", but mainly "longer". Hence my confusion.
     
  11. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    In that post, for me "larger" meant "longer". The bigger the length, the bigger the resistance.
     
  12. wayneh

    Expert

    Sep 9, 2010
    12,094
    3,033
    It gets worse. A "larger" (greater diameter) cable will have a smaller gauge number. A 10 gauge wire is much fatter than a 20 gauge wire.
     
  13. profbuxton

    Member

    Feb 21, 2014
    233
    68
    Understood. Here we mean bigger cross-sectional area when we say "larger". When we say "longer" is means just that., longer length.
     
    djsfantasi likes this.
  14. MrAl

    Well-Known Member

    Jun 17, 2014
    2,421
    490
    Well see if you trusted what you knew already you would know they were not talking about the length. You already knew that an increase in length would cause a higher resistance, thus they could only be talking about the diameter.
    But for someone who did not already know that they could be confused, so the writer could have done better.

    I see this kind of thing quite often on the web. The writers are not very succinct.
    The best way to write a technical piece is to try to think of all the ways it might be interpreted and then make it clear enough so that it can only be taken one way. This sometimes requires redundancy which means more typed characters for the authors which they try to avoid most likely. What i like to do is start with a little thing like, "In other words..." and then restate the information in a different way. This helps to avoid a misinterpretation.
    For example, "In other words, when the diameter increases the resistance goes down.".

    ...going back to sleep now :)
     
    Last edited: Feb 1, 2016
  15. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    The best way for me is saying those things in terms of formulas. And after the formulas are stated, words can be used to describe those formulas.
    Using only words can be annoying, especially if I do not know the formulas before. Especially if I do not know the whole subject before. You try to understand those words and what the author wants to say and somewhere else you see someone explaining the same thing in terms of formulas and everything clicks. And then you reread the words that you could not understand initially and say: 'yeah, it is easy".
     
  16. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yes, formulas are always nice, and ones that show relative relationships are nice too.
    For example, the resistance goes down as the square of the diameter but goes up in direct proportion to the length:
    R=K1*(d1/d2)^2
    R=K2*Length
     
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