SOLUTION: Three impedances are connected in parallel, Z1 = 2+j2 Z2 = 1+j5 and Z3 = j6 Find the equivalent admittance Y Where: Y= 1/Z1 + 1/Z2 + 1/Z3 Express in both rect

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Three impedances are connected in parallel, Z1 = 2+j2 Z2 = 1+j5 and Z3 = j6 Find the equivalent admittance Y Where: Y= 1/Z1 + 1/Z2 + 1/Z3 Express in both rect      Log On


   

Question 719085: Three impedances are connected in parallel,

Z1 = 2+j2
Z2 = 1+j5 and
Z3 = j6

Find the equivalent admittance Y

Where:

Y= 1/Z1 + 1/Z2 + 1/Z3

Express in both rectangular and polar forms.

I am currently struggling on Z3= j6,but clarification on the rest would be a great help.
Thanks
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Y=1%2F%282%2Bj2%29%2B1%2F%281%2Bj5%29%2B1%2F%28j6%29
Each term can be transformed by multiplying numerator and denominator times the conjugate of the denominator.

936=26%2A36=13%2A2%2A4%2A9=13%2A8%2A9=8%2A%289%2A13%29=8%2A117 is a good common denominator

That can be expressed as
Y=270%2F936-%28570%2F936%29j=15%2F52-%2895%2F156%29j= approximately0.288-0.609j
Unless your problem is just a math exercise, you would use 0.288-0.609j
which provides enough decimal places for practical purposes.
For the polar form we know that the absolute value would be
r=approximatelysqrt%280.288%5E2%2B0.609%5E2%29= approximately0.674
If this was just a math exercise, and an exact value was required,

The angle theta would be such that
tan%28theta%29=-570%2F270=-19%2F9= approximately-2.111
That corresponds to approximately
theta=-1.128 in radians or theta=-64.7%5Eo