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Confused about the answer given by vcaa on electric power

  1. Nov 8, 2012 #1
    hello!
    I was going over some questions on the vcaa site and there was this question in the year 2006 physics exam two. It was question ten.

    Here's the question (page 7 question 10)
    http://www.vcaa.vic.edu.au/Documents/exams/physics/2006physics2-w.pdf

    The answer said that in order to find the current at point R, I=100KW/22000V. I don't get why they used 22000V. Shouldn't they deduct the voltage loss from that value?

    here's their answer! (page 3 question 10)
    http://www.vcaa.vic.edu.au/Documents/exams/physics/physics2_assessrep_06.pdf

    please help!
     
  2. jcsd
  3. Nov 8, 2012 #2

    SammyS

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    What voltage loss ?
     
  4. Nov 8, 2012 #3

    gneill

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    According to the question they are looking for an approximate value for the current. Take the approximate value and estimate the potential drop that results due to the power line resistance. If you then deduct that drop from the supply voltage and recalculate the current using this reduced voltage, what is the "new" value for the current? Has it changed significantly?
     
  5. Nov 8, 2012 #4
    sammyS- the voltage loss due to the resistance in the wire (2 ohms)
    gneil- ah, i see.. the answer for the current after voltage loss deduction is 4.55A also. But in an exam how would you know for certain that you don't need to deduct the voltage loss first? would you have to write ' the voltage loss is negligible' The question is a bit vague.. right?
     
  6. Nov 8, 2012 #5

    gneill

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    It's simple enough to check to see if the calculated current will cause a significant potential drop. So write a statement like, "Check to see if power line potential drop is significant", then show the calculation: plug in the change and see if it steps on the significant figures of the result.
     
  7. Nov 8, 2012 #6
    okay, i see. thanks! it'll be a pain if the loss was significant.. :P
     
  8. Nov 8, 2012 #7

    gneill

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    Not too painful. If 100kW is being drawn then E*I = 100kW at point R (where E is the potential there). So E = 100kW/I . But the potential there is also the the source potential less the resistive line loss, or E = 22000V - I*2Ω. Equate the E's and solve for I. Yes, it'll be a quadratic and you'll have to sort out which root yields the reasonable result.
     
  9. Nov 8, 2012 #8
    mm.. yes.. i see. By using the quadratic formula, i achieved two answers - the desired 4.55A and 10995.5A. To eliminate the second answer, can you say that it is not possible for 10995.5A being larger than 400A (which is the current of the secondary coil of the step down transformer) since the transformer is a step down transformer?
     
  10. Nov 8, 2012 #9

    gneill

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    Why not check to see what the line voltage drop would be at 10995.5A?
     
  11. Nov 9, 2012 #10
    lol... thanks so much for your help! ^^
     
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