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AC transformer question

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    The sinusoidal voltage source in the circuit shown in Fig. 2 is developing an RMS voltage of 2000V. The 4ohm load in the circuit is absorbing 4 times as much average power as the 25ohm load. The two loads are matched to the sinusoidal source that has an internal impedance of 500<0 ohms. (please see attached circuit)
    Specify the numerical values of a1 and a2

    2. Relevant equations
    (V1/V2)/(N1/N2)=(I1/I2)=a

    3. The attempt at a solution
    I have no idea on how to do this :( can someone please help me? We haven't done much on questions like this. thanks

    Circuit diagram
     

    Attached Files:

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      fig2.jpg
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  3. Sep 15, 2011 #2

    gneill

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    Can you specify the currents in the secondaries in terms of the a's and I1? How about the power being dissipated in the secondary resistors?
     
  4. Sep 15, 2011 #3
    ummm.... The I2=a1*I1
    and I3=a2*I1?

    and is the power being dissapated: P=(I2^2)*R2
    and P=(I3^2)*R3?
     
  5. Sep 15, 2011 #4

    gneill

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    Yes, so what are the powers in terms of I1? Also, you're given a relationship between those powers, so determine the relationship between the a's.
     
  6. Sep 15, 2011 #5
    You now need to find out how to use the givens in the problem statement.

    From this statement you can compute a number for I_1 and find the voltage across the internal impedance and the voltage across the left coils of the transformers. You need to consider what it means for the loads to be matched to the source. From the answer for I_1, you can find the power being put into the left side of the transformers.


    Knowing the power input to the transformer and the above statement, you should be able to figure how much power is being spent in each loop on the right. Knowing the powers can you now find I_2 and I_3?

    Knowing all the currents allows you finally to find a1 and a2.
     
  7. Sep 15, 2011 #6
    ok, I tried that and I got a ratio of a2=(5*a1)/2

    now do I sub this back into one of the power equations? and is this ratio even right?
     
  8. Sep 15, 2011 #7

    gneill

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    Can you show your work for finding the ratio of the a's? It's very nearly right...

    Once you've determined that relationship between the a's you need to find another so that you will have two equations in two unknowns. This second equation involves the "matched" property of the loads and source impedance. We'll get to that :smile:


    By the way, you can produce superscripts and subscripts by using the x2 and x2 icons in the reply box header.
     
  9. Sep 16, 2011 #8
    ok, I've attached my working out... as you can tell I'm not very good at simplifying equations, so please bear with me.
     

    Attached Files:

  10. Sep 16, 2011 #9

    gneill

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    I think that there's a small problem in your end game :smile:

    attachment.php?attachmentid=38946&stc=1&d=1316226449.gif
     

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  11. Sep 17, 2011 #10
    Hi i have the same problem

    i am unsure how to find the value of I1 and how to calculate the power dissipated in the left loop. Do you use V1*I1=V2*I2? and therefore V2=sqrt(V1*I1*R2)?

    Also i am a bit confused of the current in the left hand loop.

    Is I1 = (V1)/(R1) meaning I1= (2000<0)/(500<0) = 4<0 A? Or does the internal resistance of the source come into play meaning that for the left hand loop, I1 = (2000<0)/(500<0)+(500<0) = 2<0 A *500 ohm resistor + 500 ohms of internal resistance*

    Could you please tell me what the Current is in the left hand loop (I1) and what the transformer voltages are for V2 & V3. I think would you then use V2=sqrt(V1*I1*R2) & V3=sqrt(V1*I1*R3) to calculate the voltage, but im not 100% sure.




    Note:
    I know that it should be in terms of impedance (Z), not resistance (R). For simplicity im assuming they are the same.

    for the previous post
    a2 = 5 a1, is what i assume your getting at.
     
  12. Sep 17, 2011 #11

    gneill

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    Rather than go directly for the current in the primary loop, it may be easier to consider the given claim that the loads are matched to the source's impedance. Transformers have the property that impedances in the secondary circuits are 'reflected' back into the primary circuit. So a resistance R in a secondary loop of an N:1 transformer will cause the primary loop to behave as though there was a series resistance N2R in it.

    What should be the total resistance reflected into the primary circuit if the loads are matched to the source impedance (given as 500 Ohms)?
     
  13. Sep 17, 2011 #12
    Okay, ive taken the reflective advice and matched it with what i already have figured i think ive got it.

    Anyway, i get
    a1=2 a2=10
    I2= 8 A I3=40 A
    I1= 4 A

    For the 25 ohm resisior in circuit 2
    P=1600W V=200 V

    For the 4 ohm resistor
    P=6400W V=160 V

    If its not to much of you, could you check to see if i am correct? i would like to be as certain as possible.
     
  14. Sep 17, 2011 #13
    ohhhh, yeah. ok, I didn't expand the 2(a2I1) and I get a2I1=5. next I did some working (attached) and I found that a2=5a1. What do I do after? I tried to sub it into some equations, but I can't seem to get a decent answer... is the working out I've done anywhere close to right?
     

    Attached Files:

  15. Sep 17, 2011 #14
    Just realised, does the internal impedance of the 2000<0 V have an effect on the left hand circuit? i cant decide if the current I1 is equal to 2<0 A or 4<0 A. I mention it before when i said

    *Is I1 = (V1)/(R1) meaning I1= (2000<0)/(500<0) = 4<0 A? Or does the internal resistance of the source come into play meaning that for the left hand loop, I1 = (2000<0)/(500<0)+(500<0) = 2<0 A *500 ohm resistor + 500 ohms of internal resistance*
     
  16. Sep 18, 2011 #15

    gneill

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    The source's internal impedance is in series with the source voltage, and you can't avoid or ignore it. The load impedances (resistances in this case) will be reflected into the primary circuit, too, and those reflected impedances will have to be added. So the primary's current will be
    [tex] I_1 = \frac{2000V}{Internal + Reflected} = \frac{2000V}{500 \Omega + 500 \Omega} [/tex]
     
  17. Sep 18, 2011 #16
    Got it now, thanks a lot.
     
  18. Sep 18, 2011 #17
    can someone please answer my question too? thanks, I'm nearly finished this question...
     
  19. Sep 18, 2011 #18
    Hope this helps
     

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