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\(\mu\)H R1=10k\(\Omega\)
L2-75\(\mu\)H 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?...

 

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?

 

Further to Caveman's answer, can I point you towards to the relevant section in the AAC e-book: http://www.allaboutcircuits.com/vol_2/chpt_3/index.html

You will find further examples for both RL-series and RL-parallel circuits.

Dave