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Load resistance, dissipated power (AC)

Discussion in 'Electronics Homework Help' started by evol_w10lv, Apr 1, 2013.

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  1. evol_w10lv

    evol_w10lv

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    Feb 19, 2013
    Hi!
    I hope that someone will help me to calculate few tasks.

    1. [​IMG]

    We know alternating voltage = 120V, Resistance of wire = 0.3 Omh as we can see in scheme.
    I have to solve voltage in load resistance, dissipated power in wires and dissipated power in load resistance.



    2. [​IMG]

    The same situation, but there are added two ideal transformers and relation 1:5 and 5:1 as in scheme. Need to solve voltage in load resistance, dissipated power in wires and dissipated power in load resistance.

    Which forumulas I have to use? How to start this tasks?
     
  2. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    Starting with 1.
    The wire reistance is not 0.3R, it is 0.6R.

    You can calculate the total resistance, 120V, 100A
    You can therefore find the load resistance.
    You can then calculate the dissipation in each component.
     
  3. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Because these are ideal transformers, given the value of current in one loop you know the value of current in all loops. Look at the middle loop -- What is the value of the voltage source on the other side of the transformer? What is the value of the load resistance reflected back through the transformer? (hint: reflected resistance is related to the square of the turns ratio) The resistance seen in the middle loop is the wire resistance (Rw) plus the reflected resistance (Rr). So the formula is E/I=2Rw+Rr and from that calculate the value of the load resistance (Rl) from Rr. The simplest way to get the voltage in the load resistance is V=IRl=100Rl (hint: it is somewhat less than 120 V).
     
  4. evol_w10lv

    evol_w10lv

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    Feb 19, 2013
    At last I had time to get back this task.
    I tried first task:

    Rtotal = V/I = 120/100 = 1.2 Ohms
    Rl = Rtotal - 2*Rw = 1.2 - 2*0.3 = 0.6 Ohms
    Vl = I * Rl = 100 * 0.6 = 60 V
    Ptotal = V^2/Rtotal = 120^2/1.2 = 120000 W
    Pl = Vl^2/Rl = 60^2/Rl = 60^2/0.6 = 6000 W
    Vw1 = Vw2 = I * Rw = 100 * 0.3 = 30 V
    Pw = Pw1 + Pw2 = (Vw1^2/Rw1) + (Vw2^2/Rw2) = (30^2/0.3)/(30^2/0.3) = 6000 W
    Efficiency = Pl/Ptotal = 6000/120000 = 0.5

    Is it correct? And internal resistance is it something simular wire resistance? When it's need to use formula Pl = (V^2 + Rl) / (Rl + r)^2 ?
    Later I will try to solve second task.
     
    Last edited: Apr 5, 2013
  5. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Internal resistance might be considered similar to wire resistance when the internal portion of a device comprises wire; however, when internal resistance is due to leakage factors it would be modeled as a parallel resistance rather than being in series with other elements of the model.

    It seems (to me) that the purpose of this exercise is to demonstrate that high voltage power transmission is more efficient than low voltage transmission over the same wires due to I^2R losses, so it may be more appropriate to think of Rw as transmission line rather than internal resistance.

    I believe your calculations are correct (not 120 kW) but there is a more direct method to calculate power:

    Ptotal=ExI=120x100= 12,000 W
    Pw=I^2xR=100x100x0.6= 6,000 W
    Pl=Ptotal-Pw=12,000-6,000= 6,000 W
     
  6. evol_w10lv

    evol_w10lv

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    Feb 19, 2013
    Now speaking about the second task.
    It means that there Ptotal = 12000 W and Pw = 6000 W as in the first task, because in the all three loops are same current I = 100 A?

    Then between first and second loop Rr = 0.04*Rl, using formula [(1/n)^2 * Rl] and between second and third loop Rr = 25*Rl. Where or how can I get connection between formulas to get Rl for all scheme?
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    Start from the end.

    If the output current of the last transformer is 100A, what is the input current?
     
  8. evol_w10lv

    evol_w10lv

    73
    0
    Feb 19, 2013
    If we are using ratio, that current of the first transformer should be 20 A
    And If the current is increased as we can see, I guess, then the voltage in the load is decreased.
    How I have to use the square of the turns ratio to get Rreflected?
     
  9. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    If your textbook doesn't cover how to do that, then we should not do it here. So consider another approach. As before, the total power is the power dissipated in the wire plus the power dissipated in the load. But you already know the current through the wires and the current through the load, so calculate the I^2R power dissipation.

    Ptotal=ExI=Pw+Pload=I^2x0.6+(5xI)^2xRload

    That will get you the value of the load resistance. But do you need the value of the load resistance, or just the power dissipated in the load?
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Or perhaps just the power lost in the resistors which simulate long wires?
     
  11. evol_w10lv

    evol_w10lv

    73
    0
    Feb 19, 2013
    Actually, I have to calculate dissipated power in the load (Pl), dissipated power in the wires (Pw), total dissipated power (Pt) and voltage in load (Vl).

    I made my attempt using your suggestions:
    Il = 100A
    Iw = Il/5 = 100/5 = 20A
    Pw = Iw^2 * Rw = 20^2 * 0.6 = 240W
    I(at start) = Iw/0.2 = 20/0.2 = 100A
    Pt = I * U = 100*120 = 12000 W
    Pl = Pt - Pw = 12000 - 240 = 11760 W
    Rl = Pl/I^2 = 11760/100^2 = 1.176
    Vl = Rl * Il = 1.176 * 100 = 117.6 V
    Efficiency = Pl/Ptotal = 11760/12000 = 0.98

    Is it correct?
     
  12. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    You got the same answer as myself. Does that make it correct?

    However, I calculated the load resistance a different way from the loop voltage equation for the inner loop using the reflected load resistance and source voltage of 600 volts at 20 amps.

    600 = 20x(0.6+5^2xRload) ..... Rload = (30-0.6)/25 = 1.176
     
  13. evol_w10lv

    evol_w10lv

    73
    0
    Feb 19, 2013
    :)
    In electronics I believe you more than I believe myself.
    Thank you for helping me.
     
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