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AC Circuits (RL circuit)

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data

    In the circuit shown in Figure Q2(b), two loads A and B are connected to the source. The cable connecting the power supply to the load is modeled as series R and L of values R = 0.1Ω, L = 0.1 mH as shown in the figure.

    The loads have the following specifications:
    Load A: 230V, 50Hz, 1kW, pf=0.5 lagging.
    Load B: 230V,50Hz, 1kW, pf=0.87 lagging

    Find the series R-L equivalent of the two loads together.

    Figure 2(b):
    j9q00p.jpg

    2. Relevant equations

    ZL = jωL
    Pactive = Vrms*Irms*cos∅ (∅ means phi)
    Preactive = Vrms*Irms*sin∅ (∅ means phi)

    3. The attempt at a solution

    Given the specifications for loads A and B, am I right to say that as long as I supply 230V, 50Hz to those 2 loads, I will get a power output of 1kW? Just wondering, does "Find the series R-L equivalent of the two loads" mean I have to find the impedances of Load A and B and convert them from parallel to series connection (just like 2 parallel batteries)?

    Since the loads have lagging power factor, they are inductors.

    As I am not given more information on the loads, I can only deduce L from Pactive = Vrms*Irms*cos∅, and vL(t) = L(diL(t)/dt)). Am I right to say this? However, I am stuck as I don't know Vrms and Irms.

    May I have some hints please?

    Thank you. :)
     
  2. jcsd
  3. Nov 10, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Yes, 1kW from each.
    Convert from 2 inductors and 2 resistances to a single equivalent L-R.
     
  4. Nov 10, 2014 #3
    Thank you. :) What do you mean by 2 inductors and 2 resistances? Just to confirm, Load A and B are not purely inductors right? (I am getting a bit suspicious because it dissipates active(real) power, as stated, "1kW".)

    If that is the case, I will have ZA = R + jX, where R is resistance and X is the inductance (or imaginary part), right?
     
  5. Nov 10, 2014 #4

    NascentOxygen

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    Each load can be considered to comprise one R and one L, and the combination determines that load's power factor.

    The R accounts for real power.
     
  6. Nov 10, 2014 #5
    I see. Thanks. I shall go and try it and let you know if I can get the answer, prolly a few hours later as I have to run now. :)
     
  7. Nov 11, 2014 #6
    I am still not getting what does "find the series R-L equivalent of the 2 loads together mean". Does it mean I find the total impedance and then do 1/Zeff = 1/Z1 + 1/X2 ?

    i.e.: For Load A, I do:

    Pactive = Vrms*Irms*cos∅ (phi).

    Hence, Pactive = Vrms*(Vrms / Z)*cos∅, so substituting all values in, we have:

    1000 = (2302 / ZA)*cos∅.

    However, I was wondering, the active power will not be the same since the inductor and resistor took up some voltage before it reaches the node of Load A. Am I right to say this? And may I know if my working above is correct?

    Thanks!
     
  8. Nov 11, 2014 #7

    NascentOxygen

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    Yes, you can sum the reciprocals of the impedances, the loads are parallel. Your method looks like it should work.

    Inclusion of the cable impedance gives the problem realism. You are correct in saying the loads will not receive their rated 230V, but that is not relevant to the specific task you are tackling here.
     
  9. Nov 11, 2014 #8
    I see. I didn't know the cable will have impedance. I suppose it is linked to resistance in the cable and passing an AC voltage changes the term resistance to 'impedance', right?

    May I know why is it not a concern that the loads are not receiving 230V? I thought they will not give the max rated power output of 1kW if they are not supplied with 230V of p.d. across them.

    Thank you. :)
     
  10. Nov 11, 2014 #9

    NascentOxygen

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    The calculations you are doing concern load specifications, and these are specified at 230V.

    A cable has resistance, also inductance, as well as capacitance to earth, so it has a characteristic impedance.
     
    Last edited: Nov 11, 2014
  11. Nov 11, 2014 #10
    I see. But if I don't supply 230V as in this case, won't I be wrong if I use 1kW as the power output? Sorry. I am a bit confused at the moment. :(
     
  12. Nov 11, 2014 #11

    gneill

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    The loads are defined by Specification:

    That means that these loads, if connected to an ideal 230 V, 50 Hz source without any intervening cable would draw the specified power at the specified phase angle. This gives you enough information to characterize these loads in terms of their impedances. The power cable is irrelevant for this problem.

    Perhaps there are other sections to the problem which you've not mentioned yet where the cable impedance becomes important?
     
  13. Nov 11, 2014 #12

    NascentOxygen

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    Staff: Mentor

    In practice, the supply voltage is never going to be exactly what the load was nominally specified to operate from. This is a fact of life, and power engineers accept this. Sure, it does mean that the load probably won't be drawing exactly 1,000.0 watts of power whenever its voltage departs from 230.0V but loads are designed to be tolerant of voltage errors (and may compensate for it by various means if ever it's important). If precise calculations are needed then that can be done, but here you are basing your calculations on the nominal design voltage, 230V. So long as the cable impedance is within acceptable limits, the loads will work just fine on whatever the voltage ends up being.
     
  14. Nov 11, 2014 #13
    I see. Thanks for your explanation. :)

    Hmm. There are 2 small question parts related to this question. They are:

    Part ii) Find the power factor as seen from the source.
    Part iii) Find the value of the capacitor to be connected across the source so that the source voltage and the total current drawn from it are in phase with each other.

    I think the questions above should not interfere with how I see/factor in the RL of the cable. :)
     
  15. Nov 11, 2014 #14
    I see. Alright. I shall try it out later. :) Thanks! I am just a bit puzzled about the voltage issue but other than that, I should be fine. :)
     
  16. Nov 11, 2014 #15

    gneill

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    Okay, these two questions are where the cable impedance comes into play. The source sees the cable impedance as part of its total load.
     
  17. Nov 11, 2014 #16
    I see. Thanks! I will take a look again tomorrow and if I have any problems I will post here again. :)
     
  18. Nov 12, 2014 #17
    I got ZA = 26.45 Ohms after working out the above. However, how do I express this in terms of Z = R + jX = R + jωL?

    I realised my teacher somehow arrived at 13.23 Ohms for R. How and why did my teacher do that? *puzzled* :S
     
  19. Nov 12, 2014 #18

    gneill

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    26.45 is the magnitude of the "A" impedance. What's its angle?
     
  20. Nov 12, 2014 #19
    From power factor = cos∅, cos∅=0.5, hence, by cos-1 0.5 = 60°. Angle = 60°
     
  21. Nov 12, 2014 #20

    gneill

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    Yup. So you have the polar form of that impedance. Convert to rectangular.
     
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