complex power supplied

Discussion in 'Homework Help' started by tunasengok, Feb 18, 2012.

  1. nishu_r


    Jun 2, 2012
    Yes Sir,it is clear now.
  2. WBahn


    Mar 31, 2012
    Okay, here's a few for you.

    Same circuit.

    1) Find L such that the complex power in the voltage source is 42.56W at an angle of -28.6 degrees.

    2) Replace the unknown L with just an unknown reactance (i.e., could be either an inductor or a capacitor), find the component type and value that results in the complex power being purely real.

    3) Find all types and values of the unknown reactance that result in the power in the voltage source being purely reactive.
  3. nishu_r


    Jun 2, 2012
    Sir, sorry for the delay in my response. I had some college work I had to attend to. Anyways, I describe my solution to the first question

    Method i used:
    1. used the complex power to calculate current in that branch
    2. using the current in the branch found voltage drops across the 14 Ohm resistor and j7 Ohm inductor (freq domain), and found the total voltage across the node since V=(-40+drops in the 14+j7 resitor combination) ?
    3. then using this voltage and the KCL, found value of L

    S=42.56 VA at (-28.6) deg<br />
S=(37.37-j20.37) VA<br />
    Since the voltage source V supplies the complex power,
    I_2^*= (S/V) =\frac{(37.37-j20.37)VA}{40V}= (0.93-j0.51) A<br />
I_2 = (0.93+j0.51) A = 1.064 A at (-28.6) deg<br />

    Since V = {(I_2(14+j7) - 40V)V}<br />
V = (-30.48+j13.51) V = 33.34 V at 156.1 deg
    this being the voltage across all the branches.
     I_{(12+j8L) \Omega || 20 \Omega}= I_1 + I_2<br />
I=\frac{V}{Z}<br />
Z=\frac{(j8L+12)20}{(j8L+22)}\frac{\Omega^2} {\Omega }<br />
    equating both the sides<br />
\frac{(j8L+12)20}{(j8L+22)}\Omega=\frac{(-30.48+j13.51)V}{(2+0.93+j0.51)A}<br />
(j160L+240)(2.93+j0.51)\Omega A=(-30.48+j13.51)(j8L+22)V<br />
V=\Omega A<br />
(j468.8L-81.6L+703.2+j122.4)V=(-j243.84L-670.56-108.08L+j297.22)V<br />
    L=(-0.093+j2.161) this is also confusing, since i am with no dimensions left at the end. Sorry sir

    I doubt my work, since i feel i have gone horribly wrong somewhere, please point it out so that i can correct it. Thank you. I will proceed to the next two questions once i find my errors here.
    Last edited: Jul 9, 2012
  4. WBahn


    Mar 31, 2012
    Tell you what, go back and put in the units properly and then I will look over the work.
  5. nishu_r


    Jun 2, 2012
    Sir please consider my answer. I have added the units
    Last edited: Jul 10, 2012
  6. WBahn


    Mar 31, 2012
    Your edits to include the units properly are much closer, but still not there (as you note when you say that L is coming up dimensionless).

    The first place where you have units errors is you sixth line. The first term is a current times a complex number. So that has units of A. You then add that to -40V. You then multiply that impossible combination by V. This is an easy one to clean up.

    The second one, and the one that haunts the rest of the work, is a couple lines later where you introduce (12+j8L). Since L has units of henries, that either means that 8 has to have units of inverse henries or that 12 has to have units of henries. Otherwise, you can't add them together.

    In point of fact, the '8' has units of radians/sec.

    Recall that:

    v = L (di/dt)

    this means that

    (1/dt) = v/(L di) = (v/di) / L

    Thus, dimensionally, we have

    (radian/sec) = Ω/H

    Remember, radian is dimensionless and can come and go as it makes sense from a clarity standpoint without affecting things.

    Aside: You'll also find that radian/sec has units of 1/(FΩ).

    Since you factored out the Ω, the '8' within the parenthesis does, in fact, carry units of H^-1

    Which is another point that I don't think you've quite gotten down (you're doing better). Units aren't notations that you put here and there, they are as much a part of the value as the number. Consider the following:

    x = ay + z

    Now let's say you want to factor out the 'a', would you go:

    x = (y + z)a ?

    Of course not! If you factor out an 'a' from the second term, you have to have an 'a' to factor out! So you would, at least mentally and subconsiously, go:

    x = ay + (a/a)z
    x = ay + a(z/a)

    x = (y + z/a)a

    The same is true with units. You can't factor out Ω unless there is an Ω to factor out. If there isn't, then you have to introduce it just like we introduced the 'a' into the second term.

    If you want to make a units notation out to the side of an equation, that's fine. But make sure that it is separate from the equation and is NOT a part of it.

    Similarly, do NOT supply units for symbolic quantities. The L carries its own units, as does every other symbol. By supplying units in addition, you are creating inconsistent units in the equation. This is essentially what you are doing when you have that Ω multiplying the j8L (because my guess is you would have done the same thing if it had been 12Ω+jwL and you would have written it as (12+jwL)Ω.