calculating R1+R2

Discussion in 'Electrics UK' started by sime27, 11 Feb 2007.

  1. sime27

    sime27

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    whats the method used to calculate R1+R2 from r1 rn r2 readings on a ring final?
     
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  3. davelx

    davelx

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    Easy to work out using basic circuit analysis principles - just draw out the circuit with lumped component values and apply the Ohm's Law to get the equivalent resistance.

    Or look at page 70 in the on-site guide.
     
  4. rlc

    rlc

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    end - 2 end resistances for the phase, neutral and cpc being r1, rn & r2 resp.

    Taking some dummy values for say 2.5/1.5mm cable:

    r1=0.54ohms
    ------|===|------

    +
    r2=0.54*1.67 = 0.90ohms
    ------|===|-----

    note 1.67 being scaler based on 2.5/1.5 with no units.

    R1 + R2 for the ring:

    0.54 + 0.90
    (r1+r2)/4 = ------|===|------------|===|----- = 1.44/4 = 0.36 ohms

    alternatively you can calculate using Product /Sum or 1/Rt method but your measured highest value is the one you record:

    Prod/Sum = (0.54 * 0.90)/(0.54+0.90) = 0.486/1.44 = 0.3375 ohms

    Resistance for a 2.5mm^2 @70degC ~18 milli-ohms / metre and note the lower the temperature the lower the resistance will be due to the temp coefficient.
     
  5. davelx

    davelx

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    But, does the OP understand why it's (r1+r2)/4? IMO he should. And I don't mean "because it's a ring", I mean does he understand how to calculate the equivalent value of resistances in series and parallel combinations. That level of understanding should be a requirement for all sparks, IMO.
     
  6. plugwash

    plugwash

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    and what about the understaning needed to show that the furthest point on the ring is the worst case?
     
  7. ricicle

    ricicle

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    Or a long spur ;)
     
  8. rlc

    rlc

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    sime27: does any of this information help you or do you require further detailed explanation - as pointed out it's a fundamental requirement to understand the basic principles, lumped circuit model, nodal analysis, measured / calculated results and what affect they have on the circuit.
     
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