tazman171,
From the sound of his question, he wants the parallel impedance or admittance. If so, then polar/rectangular coordinates cannot be avoided.
Of course there will be, unless the capacitive and inductive reactances are equal and cancel out.
Impedance and admittance are complex quantities, but they are not vectors or phasors.
You are calculating the series impedance, the OP wants the parallel impedance.
That is the incomplete version, as resistance and inductance cause a phase change between voltage and current in both series and parallel circuits.
Ratch
MY bad, I mis interpreted his statement. What I gave though does get you the answer with out showing the phase relationship, simple like I said.
I meant no phase shift between IL and IC as there was XL and did say that voltage 90 deg out of phase with current, with current leading by 90 deg for XL. The phase angles of resistive and inductive impedance are always 0 deg and +90 deg, respectively, regardless of the given phase angles for voltage or current. So the IL will be -37.016 deg out while the VL is 52.984deg, 90 deg leading for the inductor.
Just trying to find total impedance though right?
Purely a simple method that's damn close without messing with any phase relationship, Square root of (1/(1/100 squared + 1/50 squared))= 44.721 Ohms.
Saw no mention of needing more that that, but I also missed it being parallel.
You can run out the polar and rectangular values (and will need to, complex mode on your calc will be your friend) but seeing as he said that they had only done series I was assuming that they weren't to that level of instruction yet as we had covered the basics of both parallel and series XL and XC separate and combined before we ever got into polar and rectanguler methods of doing the calculations. Didn't want to throw all the stuff at him that you guys did, so I threw the "simplified" way of finding the impedance.
So what ever, I did it for series and just showed the value for parallell using a simple method with out worrying about the phase angles and got just about the same answer as the other method..
In the real world though, the phase angle of your load are extremly important on everything from sizing your wire to circuit protection to what enclosures you'll need to buy to put it all in.