problem with ohms law

Discussion in 'General Electronics Chat' started by Nwuyag, Nov 4, 2014.

  1. Nwuyag

    Thread Starter New Member

    Sep 12, 2013
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    Hello everyone, I have a little problem with resistance and ohms law.

    I know that resistance of a cconductor is pl/A where
    p is resistivity
    l is length
    A is surface area.

    In power transmission, or even fuses principles, I hear that voltage is stepped up and current is stepped down. So as to use a conductor with smaller diameter.

    Wouldnt this increase its area thereby increasing its resistance.
    By that, will this not also increase its I square R loss which is not wanted?
     
  2. sirch2

    Well-Known Member

    Jan 21, 2013
    1,008
    351
    Increasing the voltage reduces the current for a given power. The cross sectional area of the wire is proportional to current for given losses (i.e. resistivity in your equation) so lower current means smaller cross sectional area.
     
  3. Nwuyag

    Thread Starter New Member

    Sep 12, 2013
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    Wouldn't the reduced area(from reduced conductor diameter) not increase its resistance?
     
  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,762
    924
    forget the wirr diameter for a minute.

    losses are calculated by I(sq) x R

    by raising the voltage and lowering the current, losses are reduced at a greater rate than the equivalent reduction of resistance could lower them. so once the losses are reduced by stepping up voltage, the company can further reduce costs by increasing loss via increasing resistance(smaller cables). By lowering current enough losses are still reduced even when cables get smaller.
    Current squared. Resistance.
     
  5. Nwuyag

    Thread Starter New Member

    Sep 12, 2013
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    I believe you meant "inrease in resistance" in your sixth line.

    I think I have gotten you well now.
    Losses are reduced by stepping up voltage and stepping down current.
    But cost is reduced by reducing size of cable diameter at the expense of increased resistance. Tho power loss is at a different (reduced) rate.

    Did I get it?
     
  6. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    you have it. I did mean decrease though.

    decreasing the current reduces losses faster than decreasing resistance would.
     
  7. Nwuyag

    Thread Starter New Member

    Sep 12, 2013
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    Thanks. I ve gotten it now
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    488
    Hi,

    BTW it is the cross sectional area not the surface area that is used to calculate resistance.

    If R1 is the initial wire resistance and I1 the initial current, and we reduce the cross sectional area of the wire by a factor N and decrease the current by a factor of N, then the new power dissipated in the wire is:
    P2=I1^2*R1/N

    and since the original power was:
    P1=I1^2*R1

    that means P2/P1=1/N , so we've reduced the power lost in the wire by a factor of 1/N.
     
    Last edited: Nov 4, 2014
    Nwuyag likes this.
  9. Nwuyag

    Thread Starter New Member

    Sep 12, 2013
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    I meant cross sectional area. Thanks for the corretion.

    Your explanation is wonderful.
    Thanks again
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
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    488
    Hi again,

    You're welcome, and i actually thought that was what you meant i just wanted to make sure we were all thinking along the same lines :)
     
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