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Real Power in AC circuit:cry:

  1. Jul 31, 2007 #1
    hello all
    just studying to that crazy electrical engineering 2 exam coming up friday
    encountered something weird about power dissipation.

    Given is an AC circuit, source of 10V f=100Hz Z=20 Ohm
    Matched to give maximum power is the load impedance which will be 20 Ohm as well.

    now my problem:
    I have assumed to calculate the current (10V/40 Ohm = 0.25A) and then easily apply I^2*V which gave 1.25W now, in the answer sheet is given solution just half of it : 625mW
    now ... why did they divide it by two? how should I know when to do that ? is that a different method I should apply for DC / AC circuits ? does it make any difference that it is called impedance and not ressistance

    Thank you all for you time people
    Greets, Amit.
    Last edited: Jul 31, 2007
  2. jcsd
  3. Jul 31, 2007 #2
    And one more thing later on...

    Load of 20 Ohm is actually combined from R||(C+L)
    where R=50Ohm C=159.15microF xL=omega*L=10Ohm --> L=pi/20 ?? (omega=200pi)

    so, after the dissipated real power over the load has found 625mW , which i still dont know why. weve been asked to find the power over the resistor(50Ohm) which is parallel to the CL combinatioin and is found to be Zero

    left me with my mouth open :surprised with no direction to start and guess this answer

    any ideas ????
    sorry for bothering, thanx for reading&answering
  4. Jul 31, 2007 #3


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    I guess you're finding the power delivered to the load? It seems that 1.25W is is the product of IRMS and VRMS. But what if we want to find the power across the 20ohm impedance? Shouldn't it be IRMS2*20ohm? That gives 625mW.
  5. Jul 31, 2007 #4


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    Well from the data you've listed, it seems that 100Hz is the resonant frequency for L and C i.e. both the inductor and capacitor have the same reactance (10ohms). When you look at the impedance graph for a capacitor and inductor, what do you notice? Don't they oppose each other? This means...?
  6. Jul 31, 2007 #5
    Thanx 4 the answers, ranger !

    but still .... few points arnt so clear to me.

    1. Irms: is that simply I/sqrt(2) or is it more complicated then that ? when do I use it, in all AC circuits ? in DC
    do i use regular current or RMS as well ?

    2. I would appreciate if you can put some more light or even link me
    to more information about the second point where L&C oppose each
    other and there fore no power is dissipated over the parallel resistor.

    Thanks for the patience (-:
    Last edited: Jul 31, 2007
  7. Jul 31, 2007 #6


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    Correct. To get the rms current/voltage, just divide by sqrt(2).
    This translates to:
    v(t) = 10*sin wt
    Given the frequency, one can find omega. The form which its stated above is small signal variation (AC). Here, 10 represents the maximum voltage (peak). It should be obvious why dividing by sqrt(2) gives RMS.
    Let me ask you something, of what use is it to use RMS (effective) power as opposed to simply using peak?

    Umm...I can't find a decent link with graph, but how would you express the impedance of both a capacitor and inductor in polar notation (no need to use specific values for Xl and Xc)?
    Last edited: Jul 31, 2007
  8. Jul 31, 2007 #7
    well AC means alternating current and DC means direct current. AC changes with time sin wave. thus has frequency while DC is nonfluctuating current and is straight line graph parallel to axis.

    Thus when we calculate problems in AC circuits we use rms values called root mean square value as current changes with time (current changes its direction) thus we take average current value to calculate the power same way we use rms values of voltage and thus power=Irms*Vrms/2.

    While in DC the current does not change its value and thus we straight way use I value given to us and not rms values.
  9. Jul 31, 2007 #8

    Hey !!
    well ... i thinki understand the part where they cancel each other out, but i wonder how and why does it influencing the power on R making it zero
    to make it easier i attached the original exam, from it is the first problem


    Attached Files:

  10. Jul 31, 2007 #9


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    Well if the C and L cancel out each other, then that makes the impedance on the branch which consists L and C...?
  11. Jul 31, 2007 #10
    0 .... which means all current going through that branch and not through the resistor ?
  12. Jul 31, 2007 #11
    Ahhh... i think i can flood this forums now and make its members busy for so long, as long as im still studying for exam. sun is shining outside, everybody going to swim in the lake .... and im here.
    well at least not alone, huh ?
  13. Jul 31, 2007 #12
    i meant these forums and not this forums .... too tired to spell, heheh....
  14. Jul 31, 2007 #13


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    You can think of it that way. Or you can say since the resistor (R2) is in parallel with 0ohm (a idealized piece of wire), the equivalent impedance is 0ohms. Therefore all the power is dropped across the potentiometer (R1) and the internal source resistance. Since we wish to get maximum transfer of power from source to load, the load should have a resistance equal to that of the source (which you already pointed out). Since this value is 20ohms, it should now be obvious why it seemed like they divided the total power by two.
  15. Jul 31, 2007 #14
    Quick problem

    From problem no1 in the file attached here.

    1e) Which capacitance to make the apparent power real.

    My direction was to try and make the Total impedance real, which would make
    the Current real (voltage is real).
    but im kinda stuck, not able to proof in this way, im pretty sure its an 'kinda-easy-if-you-get-the-hang-of-it' question ... but, i guess i didnt

    any ideas ?

    Attached Files:

  16. Jul 31, 2007 #15


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    Interesting problem. Are you aware of the following:
    [tex]S = P + jQ[/tex]
    where S is apparent power, P is real power and Q is reactive power.

    So you are trying to make S = P right? You can read this for more help on the issue:
    http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html [Broken]
    Last edited by a moderator: May 3, 2017
  17. Jul 31, 2007 #16
    Thanks for the link, it looks interesting, just started to read
    one thing bothered me there
    They relate to amplitude of V or I as RMS, just as it is and without dividing
    it by sqrt(2)

    for example, at the explanation coming directly after the RL series circuit
    they say its a source of 120V RMS ... or does that mean it actually has the following formula v(t)=120*sqrt(2)*cos(wt+a)
    or am i just competly mixed ... hopefully reading on will clear some stuff.
  18. Jul 31, 2007 #17


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    The highlighted section implies that you are indicating a time varying signal - not RMS! That 120V volt in that expression is the peak.

    When we work with AC circuits, and we need to work with ohms law, power law, voltage divider, etc, we must use complex notation. A majority of these require use to divide and multiply quantities. You should know that polar notation is best suited for this. And of course the magnitude in polar notation is always RMS.

    EDIT: sorry. I missed when you multiplied 120V*sqrt(2). In that case, its in peak. And it fits in properly to express small signal variations. So you're correct.
    Last edited: Jul 31, 2007
  19. Jul 31, 2007 #18
    Very intereseting article, so much that i would never get from
    my sleepy self-bored-to-death professor.
    Anyways, too much already over 10 hours today
    so i'll retire for now, see you tomrrow (for sure! hehehe....)

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