Recent content by dsmlights

ok great I am back in track, thank you so much for the help. I am sure you will be hearing from me much more .

hey Zryn thank you for all of your help, specially having patience with walking me through this. I was looking for online tutorials for days and could not figure it out. My AC circuit books does not really explain this that good. I have learned more in 2 days in this forum than the last week...

ok so I am a bit confused so is the answer 2525 - j3750 or 2525 - j3750 --------------- 55-j50

(600 - j2000)*********35 ---------------+ ----------------= (55 - j50)************1 (600 - j2000)*****35(55 - j50) ---------------+ ----------------= (55 - j50)**********(55 - j50)

ok finally I am getting somewhere so now with the Z= Z1+Z2 I got the answer of (2225-j3750)/55-j50 is this correct ? Now do i just cover it into polar form and then just divide in polar form? i

ok so using this formula Ztot= Z1 Z2 / (Z1 + Z2 ) my answer was (600-j2000)/55-j50 am I way off ?

ok so I got this far from reading some stuff in the web. 15-j50 35-j0 40-j0 now I have to find the Z in parallel 1/(1/z+1/z) now how do I do this with these numbers

Yes I am familiar with both forms but not as in depth. so please tell me if I am right till this point 35-j0 15-j0 40-j0 0-j50 now where do I go from here ? add all of them together ?

well the impedance of a series circuit is found by Zt=SQr(R^2+Xc^2) You add impedances in series to get the total Impedance. and in parallel you use, 1/((1/Z)+(1/Zn) I am still stuck though ?

Homework Statement here is the circuit I am trying to figure out Homework Equations I have to find the total impedance for the circuit. The Attempt at a Solution I know how to find the total impedance in a series circuit and in a parallel circuit when I have just the Resistor...