SOLUTION: Hi, I am having trouble with this question, and understanding the methodology. B) Three impedances are connected in parallel. Z1 = 2-5j, Z2 = 6-3j, Z3 = 5j. Find the equivalent

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi, I am having trouble with this question, and understanding the methodology. B) Three impedances are connected in parallel. Z1 = 2-5j, Z2 = 6-3j, Z3 = 5j. Find the equivalent       Log On


   

Question 1205575: Hi, I am having trouble with this question, and understanding the methodology.
B) Three impedances are connected in parallel. Z1 = 2-5j, Z2 = 6-3j, Z3 = 5j.
Find the equivalent admittance Y where : Y+=+%281%2Fz1%29%2B%281%2FZ2%29%2B%281%2FZ3%29.
Express the admittance in both rectangular and polar forms.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, I am having trouble with this question, and understanding the methodology.
B) Three impedances are connected in parallel. Z1 = 2-5j, Z2 = 6-3j, Z3 = 5j.
Find the equivalent cross%28admittance%29%5E%28impedence%29 Y where : : Y+=+%281%2Fz1%29%2B%281%2FZ2%29%2B%281%2FZ3%29.
Express the cross%28admittance%29%5E%28impedence%29 in both rectangular and polar forms.
EE and Electronics students use j for what mathematics students use the
imaginary unit i.  That's because i (for "intensity") was already used for
amperes of current before the application of complex numbers to alternating
current was discovered. So, rather than change the normal variable for amperes
of current, it was easier to change the imaginary unit from i to j.

Y+=+%281%2Fz1%29%2B%281%2FZ2%29%2B%281%2FZ3%29

Y+=+%281%2F%282-5j%29%29%2B%281%2F%286-3j%29%29%2B%281%2F%285j%29%29









Y+=+%28%282%2B5j%29%2F%2829%29%29%2B%28%286%2B3j%29%2F%2845%29%29-%28j%2F5%29

Y+=+2%2F29%2Bexpr%285%2F29%29%2Aj%2B6%2F45%2Bexpr%283%2F45%29%2Aj-expr%281%2F5%29%2Aj

Y+=+0.0690%2B0.1724%2Aj%2B0.1333%2B0.06667%2Aj-0.2%2Aj

Y+=+%280.0690%2B0.1333%29%2B%280.1724%2Aj%2B0.06667%2Aj-0.2%2Aj%29

Y+=+%280.0690%2B0.1333%29%2B%280.1724%2B0.06667-0.2%29%2Aj%29

Y+=+0.2023%2B0.03908j

Polar form:

r=sqrt%280.2023%5E2%2B.03908%5E2%29=0.2060

theta=arctan%280.03908%2F0.2023%29

theta=10.93%5Eo

Polar form: Z+=+0.2060%28cos%2810.93o%29%5E%22%22%2Bsin%2810.93%5Eo%29%2Aj%29 

which is normally written by EE students:

Z = 0.2060/10.93o

Edwin