Inductors and Reistors in parallel

Discussion in 'Homework Help' started by eta1308, Apr 22, 2008.

  1. eta1308

    Thread Starter New Member

    Apr 22, 2008
    i'm attempting to find the current for two inductors and two resistors that are in parallel my question is, is current still additive in parallel when it's not just resistors because i've added up my totals and there not the same as the total current i had calculated. here are my totals...

    120Vac 10MHz
    L1=60\muH R1=10k\Omega
    L2-75\muH R2=15k\Omega

    Lt=33.33\mu Rt=6k\Omega
    Z=1.97k\Omega It=60.91mA
    IL1=31.83mA Il2=25.48mA
    IR1=12mA IR2=8mA

    adding all of the current together with the totals i came up with from IL1&2 and IR1&2 i came up with 77.31 is this correct or is there something i'm doing wrong?...
  2. Caveman

    Active Member

    Apr 15, 2008
    You can add currents in parallel, but you have to add it using phasors to get the total magnitude right.

    Impedance of the two inductors in parallel is 2*PI*f*L = 2094.4 Ohms < 90deg.
    So the current through the inductors is
    I = (120<0deg)/(2094.4<90deg) = 57.30mA <-90deg.

    The current through the resistors is 20mA < 0deg.

    So add them up, SQRT(57.30^2 + 20^2) = 60.69mA.
    The angle is -atan(57.3/20) = -70.76deg.

    Got it?
  3. Dave

    Retired Moderator

    Nov 17, 2003