AC mesh analysis and power calculations

Discussion in 'Homework Help' started by regexp, Jan 2, 2011.

  1. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hi,

    In this circuit.
    [​IMG]

    I need to determine the power dissipated in the resistance and also the reactive power delivered to the source V1.

    Starting by writing the mesh equations for I1:

    8I_{1} -I_{1}\cdot J2 + I_{2}\cdot J2 = 40

    I_{2}J2 -I_{1}J2 +I_{2}J4 = -j20

    Subtracting the second from the first gives
    8I_{1} -I_{2}J4 = 40+j20

    The correct answer is supposed to be 8I_{1}-J4I_{1} = 40+J20

    I don't really see where i went wrong =/
     
  2. edgetrigger

    Member

    Dec 19, 2010
    133
    19
    your answer is right. Even i am getting the same result. Its almost ten years that i dealt with mesh analysis, but i don't see why i shouldn't concur with you.
     
  3. bglazierjr

    New Member

    Apr 2, 2011
    6
    1
    I know this is a little late but it seems to me that you've mixed up your signs on the second equation

    If written using the format approach you will get -I1(-j2)+I2(-j2+j4) = -20j
    which simplifies to be I1(j2)+I2(j2) = -20j

    Now if you use a calculator such as the ti-89 or ti-86 you can enter the coefficients in directly to solve for the current I1 and I2. A lot of people complain about not being able to do this with an ti-83 or 84 but you actually can with an extremely simple program. If you want I can type the source out.

    I1= 3 + 4j -----> 5A< 53.13 degrees
    I2= -13-4J -----> 13.6A< -162.9 degrees

    I assume you mean the power dissipated in the resistor.
    VR1 = 8ohms * 5A< 53.13 degrees = 40 V < 53.13 degrees
    40V < 53.13 degrees * 5A< 53.13 degrees = 200 Watts < 106.26 degrees

    PV1 = 40V * 5A< 53.13 degrees = 200watts < 53.13 degrees
     
    Last edited: Apr 2, 2011
  4. bglazierjr

    New Member

    Apr 2, 2011
    6
    1
    Opps I mean the power is 5/sqrt(2) * 40V / sqrt(2) * cos (0) = 100Watts
    and PV1 is 5/sqrt(2) * 40V / sqrt(2) * cos(53.13) = 60 watts

    where x= phase shift between voltage and the current for cos(x)
     
Loading...