Phase difference of two current generators in a circuit with switch.

Discussion in 'Homework Help' started by cdummie, Jan 10, 2016.

  1. cdummie

    Thread Starter Member

    Feb 6, 2015
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    1
    So i have a circuit with three receivers whose impedances are ##Z_1 = (125 + j375)\ohm , Z_2 = (700 + j100)\ohm , Z_3 = (500 - j500)\ohm ## and with two current generators. Effective value of second generator is ## I_{g2}=40mA## while Ig1 is completely unknown. Apparent power of second receiver, when the switch is off is ## S_2 = \frac{\sqrt{2}}{2} VA ##. When the switch is on, then, active power(P) of all receivers is twice smaller than it was before it was on. Find the phase difference of these two current generators.
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    When the switch is off i know the value of S2=U2I2 but i don't know how that can help me to determine the value of Ig1 since i have product of effective values of voltage and current in the second branch.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hi,

    The way to proceed with circuits is generally the same way for any only solving for a different parameter, unless of course you know a short cut theory.

    This means do something like Nodal Analysis, then solve for the unknown current. In this case since it has a switch, you'll probably have to do Nodal on the circuit before it's closed, and then again after it's closed.
    See if that helps.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    When the switch is off, you can ignore Ig2, right?

    If you know the apparent power in Z2, can you find the current through it and the voltage across it?

    If you know the voltage across Z2 can you find the current in Z3?

    If you know the current in Z2 and the current in Z3, can you find the current in Z1?

    If you know the current in Z1 can you find the current in Ig1?

    Don't forget to take phases into account!
     
  4. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    I completely understand what are you saying, that's what i tried, but i don't understand how can i find the current and the voltage across Z2, because S2= sqrt(2)/2 = U2I2 so i have product of effective values of current and voltage on Z2, but i don't know anything else, i mean ok, U2I2= sqrt(2)/2 but i know nothing about I2 and U2 so i can find infinite number of values for I2 for which i could find corresponding U2 and it would fit into U2I2=sqrt(2)/2. I mean I2 could be 1/2 in that case U2=sqrt(2) or maybe U2=1 then I2=1/sqrt(2) and so on. Maybe i am missing something here, but i really don't see how can i get through this.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You know the apparent power in Z2.

    You know the impedance of Z2.

    What is the current in Z2?

    What is the voltage across Z2?
     
  6. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    Well, apparent power is known, and it's the product of effective values of U2 and I2, since U2=Z2I2 (but effective value of Z2, not Z2 as a complex number) it means that S2=Z2I2^2 so I2=sqrt(S2/Z2) , if this is correct, then i have effective value of current in second branch, but i don't have it's argument, which means that i don't know it's actual complex value. Since i don't know how to determine it's argument i can't go any further.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    But what does the argument of the current represent?

    It merely tells you the phase angle of the current relative to some arbitrary reference point (i.e., t = 0). So pick your own reference point such that the phase angle of the current in this component is zero. Let it serve as your time reference.
     
  8. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    I think i know what you mean, the way i understood it, i should choose the argument of the current I2 to be zero, but it would be correct even if i assumed it was Pi or something, since difference between phases matters, and only difference between the phases affects the way the circuit works. Since signals in AC are sinusoidal they will go from 0 to 2Pi all the time, but difference between the arguments of the two currents will remain the same. Is this correct?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    For the most part you've got it.

    The one point I would make a correction to is when you say that sinusoidal signals go from 0 to 2pi all the time. While this is true (that the angle of a sinusoid is constant varying), that is unrelated to the issue at hand. The "phase" (distinct from "angle") of a sinusoid refers to the difference between the angle of that sinusoid and the angle of some (possibly non-existent) reference sinusoid. You get to pick the reference and doing so merely adds a constant phase offset to ALL sinusoids in the system. So by choosing the argument of I2 to be zero, all you are doing is choosing a reference sinusoid that happens to be in phase with I2.
     
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  10. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    Thank you for explanation, i tried to solve it and this is what i did:

    first, S2=U2I2 = Z2I2^2 so I2 = sqrt(U2/Z2) , Z2 = sqrt(700^2 + 100^2) = 500sqrt2, I2=1/sqrt(1000) =sqrt(10)/100

    now that i have argument of I2 i can assume that it's phase is zero, so complex I2= sqrt(10)/100 (i don't know how to add underline when i want to note that it's complex number i am talking about so i will indicate it)

    Now that i have I2 complex i can find U2 complex since i have impedance Z2, U2=Z2I2 = (700 + j100)sqrt(10)/100 = [7sqrt(10) + jsqrt(10)] V since the voltage is the same on Z2 as it is on the Z3 it means that U2=Z3I3 so I3= U2/Z3 = [3sqrt(10) + j 4sqrt(10)]/500 A , now sum of these two currents in the current in the branch with first generator(considering that I2 and I3 are going "away" from node located between the three impedances) since the switch is off so Ig1= I2 + I3 = [8sqrt(10)/500 + j4sqrt(10)/500]A, but when i am trying to find the argument of this current i get Ig1=sqrt(2)/25[2sqrt(5)/5 + jsqrt(5)/5] A , so i don't have actually that good value since this is approximately 26 degrees, and i had to use a calculator to find the argument, so i thought i made i mistake somewhere
     
  11. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    I explained what i did in post #10 , i am not sure if you saw it, but i needed help because i had a problem with finding the argument of Ig1, it turned out to be approximately 26 degrees, but i found that out that when i used calculator, which is a problem, because if i got to the point where i can't find the argument easily on exam, i wouldn't be able to use calculator, so i thought i maybe made a mistake. I would really be thankful i you could check out if i am missing something.
     
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