Parallel RL Circuits with Impedance Magnitude

Hi,

Sorry, I have to ask for the third time. My problem this time have to do with parallel RL circuits and impedance magnitude. Here is my attempt on the question:

"R=8200 Ohms, L=29 mH & e is a 240 V, 23Hz power supply. What is the magnitude of the impedance of the circuit?"

The solution:

6.28x29x23=4190.8846 ohms

( 8200 x 4190.8846 ) / [ ( 8200^2 + 4190.8846^2 ) ]^0.5=63657

This is one of the questions that was given to me at the practice problems section of the book and it gives me the answer 3732 Ohms. However, I got a different answer. How can I really get 3732 Ohms in a parallel RL circuit? Thanks.

Cast_Member,

Sorry, I have to ask for the third time. ....

Click to expand...

Tsk, tsk. Still don't remember from the last time where I showed you that impedance is a phasor quantity?

R = 8200 ohms
Xl = 4191/_90°

Z = 8200*4191/_90°/(8200+4191/_90°) = 3731/_62.93° ohms

Y =√(1/8200²+1/4191²)=2.68E-4 siemans

Z = 1/Y = 3731 ohms

θ = arctan((1/4191)/(1/8200)) = 62.93°
Y = (1/8200)/cos(θ) = 2.68E-4 siemans
Z = 1/Y = 1/2.68E-4 = 3731 ohms

Ratch