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AC circuit analysis

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the currents through ##R_3, i_1(t)## and ##R_2, i_2(t)##.
    I attached a diagram of the circuit (diagram.pdf).

    In the original diagram the currents I'm trying to find is marked going top to bottom, there's also no directions on the voltages in the original diagram but from the currents I guess they're in the direction drawn

    ##u_1(t) = 2\sin(200t+30^\circ)##
    ##u_2(t) = 3 \cos(200t + 20^\circ)##
    ##R_1 = 100 \Omega##
    ##R_2=300 \Omega##
    ##R_3 = 250 \Omega##
    ##C=10 \mu F##
    ##L=300mH##




    2. Relevant equations
    N/A

    3. The attempt at a solution
    Lets first simplify at bit by calculating the impedance.
    We then have
    ##Z_1 = R_3+\frac{1}{j\omega C} = 250-j500##
    ##Z_2 = R_2+j\omega L = 300+j60##

    We then have the diagram (diagram2.pdf)
    We also write the voltages in complex form
    ##u_1(t) = 2e^{200t+30^\circ}##
    ##u_2(t) = 3e^{200t+110^\circ}.## (add ##90^\circ## to write the ##\cos## in ##\sin## form.)

    Node analysis:
    I ground the bottom node then I use KCL on the top Node, let's call it ##A## and it's potential be ##U_A##.
    We then get the equation
    ##\frac{U_A-u_1(t)}{R_1} + \frac{U_A}{Z_2}+\frac{U_A+u_2(t)}{Z_1}=0.##
    Rearanging
    ##U_A\left( \frac{1}{R_1} +\frac{1}{Z_2} + \frac{1}{Z_1} \right) = \frac{u_1}{R_1}-\frac{u_2(t)}{Z_1}.##
    Then ##U_A## can be calculatee and the currents are then
    ##i_1(t) = \frac{U_A+u_2(t)}{Z_1}## and
    ##i_2(t) = \frac{U_A}{Z_2}##.

    To check that I didn't make any mistakes I calculated this in Matlab
    Code (Text):

    R=100;
    Z1 = 250-500i;
    Z2 = 300+60i;
    u1 = 2*exp(i*pi*30/180);
    u2 = 3*exp(i*pi*110/180);
    R3 =250;
    Zc = -500j;

    temp = u1/R-u2/Z2;
    UA = temp/(1/R+1/Z2+1/Z1)
    I1 = (UA+u2)/Z1
    I2 = UA/Z2
    A1 = abs(I1) %amplitude of i_1(t)
    phi1 = 180/pi*atan(imag(I1)/real(I1)) %phase constant of i_1
    A2 = abs(I2)
    phi2 = 180/pi*atan(imag(I2)/real(I2))
    Using this I get the output
    ##UA = 1.3377 - 0.0697i##
    ##I1=-0.0041 + 0.0027i##
    ##I2 = 0.0042 - 0.0011i##
    ##A1 = 0.0049##
    ##A2 = 0.0044##
    ##phi1 = -33##
    ##phi2 = -14.3##
    So the currents should be
    ##i_1(t) =4.9 \sin (200 t -33^\circ) mA## and
    ##i_2(t) = 4.4 \sin (200t - 14.3^\circ mA##.

    However the answer says that it should be
    ##i_1(t) = 5.13\sin(200t + 15.7^\circ )## mA and
    ##I_2(t) = 3.73 \sin (200t + 26.2^\circ )##mA.

    I've been at this question for two days and I can't see where I'm going wrong.
     

    Attached Files:

    Last edited: May 25, 2016
  2. jcsd
  3. May 25, 2016 #2

    Merlin3189

    User Avatar
    Gold Member

    Only thing I've noticed so far:

    ACtemp.png
     
  4. May 25, 2016 #3
    Nice catch! That together with changing the sign of one of the voltage sources fixes it. Apparently one of them was meant to be the other way around.
     
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