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Find transformer primary and secondary voltage/current

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data
    (in picture)

    2. Relevant equations
    V=IZ
    All the ratio formulas (i.e. V2/V1 = N2/N1)

    3. The attempt at a solution
    I thought V1 would be 1000 Vrms. Well the answers are shown and it's not. So I thought maybe I would multiply Vrms (1000) by root 2 to find Vpeak (Vrms = Vpeak/root 2). Didn't work. I know if I find V1, the rest will fall into place because from there, it's simple ratios. But it's getting the answer of 790.6cos(18.43) that I'm having trouble with . . .
     

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  3. Apr 5, 2015 #2

    NascentOxygen

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    Staff: Mentor

    V1 is not 1000V because some of Vs is lost in the 1000 ohm series resistance, by Ohms Law.
     
  4. Apr 5, 2015 #3
    Right. But there's only one resistor, so the voltage always comes out to 1000. I tried doing voltage division this way:

    V1 = (Vs(RL+XL))/(1010 + j20)

    And didn't get the correct result either. I used the (10 + j20) combination as the load in my division.
     
  5. Apr 5, 2015 #4

    NascentOxygen

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    You need to have an expression for the current in order to determine how much of Vs is lost across that series resistance. So what is an expression for the primary current in terms of other parameters?
     
  6. Apr 5, 2015 #5

    gneill

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    As far as we can tell from the problem statement the transformer is ideal and has a 10:1 turns ratio. What does that tell us about the relationships between V1 and V2, I1 and I2?
     
  7. Apr 5, 2015 #6
    Um . . . I don't know . . .

    I know the ratio (V1/V2 = N1/N2, and inverse for current), but I need the V1 first . . .
     
  8. Apr 5, 2015 #7

    gneill

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    That's why you need to write loop equations (KVL). The now known relationships between V1 and V2, I1 and I2, allows you to eliminate two of the variables, leaving two equations in two unknowns.
     
  9. Apr 5, 2015 #8

    NascentOxygen

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    You can often start on the secondary side, and work back towards the primary.
     
  10. Apr 5, 2015 #9
    Go on . . . Because I thought of that but have no voltage or current to use.

    1000 + 1000I1 +V1 = 0
    (10 + j20)I2 - V2 = 0

    V1/V2 = I2/I1

    These the equations you're referring to?
     
  11. Apr 5, 2015 #10

    gneill

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    Staff: Mentor

    Almost. The first equation is not correct. Check the signs of the terms.

    And you are given the transformer turns ratio, so use it.
     
  12. Apr 5, 2015 #11
    -1000 + 1000(I1) + V1 = 0
    (10 + j20)I2 - V2 = 0

    I2/I1 = 10 (Or N1/N2 if you'd prefer).

    So I can turn I2 into 10I1 and V2 into 0.1V1?
     
  13. Apr 5, 2015 #12

    gneill

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    Staff: Mentor

    Yup.
     
  14. Apr 5, 2015 #13
    Boom. Took forever to find a Ti-84 App to do the trick but got one. Did not think you had to do that with such a simple transformer setup.
     
  15. Apr 6, 2015 #14

    NascentOxygen

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    You need an app to solve an equation in one unknown?? :wink:
     
  16. Apr 6, 2015 #15
    Two unknowns. But when we only have 50 minutes for an exam, I need to find one that doesn't break my calculator! AHHH.
     
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