Mesh/supermesh problem

Discussion in 'Homework Help' started by Mnimetic, Mar 13, 2011.

  1. Mnimetic

    Thread Starter New Member

    Mar 13, 2011
    Have completed a half a dozen attempts at this problem and the power delivery/absorption always comes out uneven and I can't figure out where at along the way I'm messing up. The teacher taught us the mesh method in class (which wasn't included in the text book) with one example, but from what I'm reading the problem I came up in pspice is a supermesh problem and I'm still not understanding where I'm going wrong or what the process is exactly for a supermesh. I'm using clockwise circles to calculate all of the equations.

    Top left is Ia, top middle is Ib, top right is Ic, elongated bottom touching all the other meshes is Id, and the bottom middle is Ie.

    V1 = 5V
    V2 = 10V
    V3 = 15V
    V4 = 20V
    V5 = 25V

    R1 = 5Ω
    R2 = 10Ω
    R3 = 15Ω
    R4 = 20Ω
    R5 = 25Ω
    R6 = 30Ω
    R7 = 35Ω
    R8 = 40Ω

    The equations I've come up with match the values for all the resistors and power sources but I believe I must have a mistake with a negative sign somewhere when computing all the current values for the voltages and resistors to calculate the power delivered/absorbed by them.

    EQ1: -5 = 25Ia - 20Id
    EQ2:35 = 40Ib - 30Ic
    EQ3:-20 = -30Ib + 80Ic - 35Id
    EQ4:-10 = -20Ia -35Ic +80Id
    EQ5:-15 = 40Ie

    Using matrices I get the values (values aren't precise but close enough, I computed the fractions out by hand as well but these should suffice):
    Ia ~ -0.3921
    Ib ~ 0.8476
    Ic ~ -0.0366
    Id ~ -0.2406
    Ie ~ -0.3766

    I assume when calculating the power delivered/absorbed I'll be taking the absolute value of whatever difference (I believe that's what we did in class) there is for resistors/voltage sources bordering two meshes.

    Current through items (Amps):
    V1 = abs(-0.3921) = 0.3921
    V2 = abs(0.8476--0.3921) = 1.2397
    V3 = abs(-0.2406--0.3766) = 0.136
    V4 = abs(-0.0366) = 0.0366
    V5 = abs(-0.2406-0.8476) = 1.0882

    R1 = abs(-0.3921) = 0.3921
    R2 = abs(0.8476) = 0.8476
    R3 = abs(-0.0366) = 0.0366
    R4 = abs(-0.2406--0.3921) = 0.1515
    R5 = abs(-0.2406) = 0.2406
    R6 = abs(-0.0366-0.8476) = 0.8842
    R7 = abs(-0.2406--0.0366) = 0.204
    R8 = abs(-0.3766) = 0.3766

    Power delivered by voltage sources = Ʃ(Δv*I) ~ 44.33
    Ex for V1: 5V*0.3921A = 1.9605 Watts

    Power absorbed by resistors = Ʃ(Δv*I) = Ʃ(R*I^2) ~ 40.46?
    Ex for R1: (5Ω*0.3921A)*(0.3921A) = 0.7687 Watts

    It's weird because the problem we did in class also had a voltage source bordering two meshes and I did it once and the powers were equivalent. Even using precise fractions my powers are off by roughly 4 watts.
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    Your problem is that you have taken the absolute value of the currents through the voltage sources. You can't do that because when you do that it comes out that all the voltage sources are delivering power. But, one of them is absorbing power.

    The sign of the current in V1 is different from the sign of all the other currents through the other voltage sources.

    Fix that problem and you'll get the right answer.
  3. Mnimetic

    Thread Starter New Member

    Mar 13, 2011
    Ok, the absolute values I can take care of, the values I'm getting for the currents I can't seem to get anything that makes a difference. I must be setting up the equations wrong? Setting them up exactly as we did in class, getting those values every time.

    This is confusing, the way we did it in class didn't seem to take into account the resistor polarities (or perhaps was simple enough to where it didn't matter). Is there a specific way you determine the polarity on resistors in a circuit?
    Last edited: Mar 13, 2011
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    I got the same currents you did. That much of your work is correct.

    You don't have to take into account resistor polarity. You have to take into account current polarity through the voltage sources.

    You have .3921 amps in v1; just change the sign of the product of V1 times .3921 when you add it up with all the other products of currents and voltages of sources, and you will get 40.46 instead of 44.33
  5. Mnimetic

    Thread Starter New Member

    Mar 13, 2011
    Ah, wasn't sure what you meant so I took another look at the polarity and the assumed direction of the currents (clockwise). I think I get it now; if the values attained for the current are negative, flop the circular arrow symbolizing the current to the counterclockwise direction then calculate the power through the voltage sources by assigning positive or negative values depending on which direction they go through the voltage source. (At least that made a hell of a lot more sense to me anyway)

    P(V1) = 5V*(-0.3921)
    P(V2) = 10V*(0.8476+0.3921)
    P(V3) = 15V*(0.3766-0.2406)
    P(V4) = 20V*(0.0366)
    P(V5) = 25V*(0.8476+0.2406)

    Ptot ~ 40.391 W