Impedance of practical Capacitor

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bigu01
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Homework Statement


A practical capacitor can be modeled by an ideal capacitor in parallel with a resistor.Find the impedance of practical capacitor at the radian frequency ω=377rad/s.Known C1=0.1 x 10^-6F R1=1 *10^6.

Homework Equations



1/Z=1/R1 + jCω

The Attempt at a Solution

I am first determining the Z, by adding the impedance of resistor and the capacitor in parallel.
And I am getting Z1(377)=10^6/(1+j37.7)= 2.6516*10^4 ∠-1.519 However book solutions gives ∠-1.5443.I would like to get some help about where my mistake is,or how should I approach the question. I know that finding the angle we should do θ=arctan(y/x) where y is the imaginary part and x is the real part.Hope I was clear and did not complicate it.
 
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R//C Z=R//1/jωC = R1/(1+jωC1R1) inputing numbers I get =10^6/(1+j37.7) multiplying by 1+j37.7's conjugate I got 703.09-26506.54j , tried to find the length got the 26515.86 , finding the angle theta=arctan(-26506.54/703.09) doesn't give me same as book result. I get -1.519 book got -1.5443
 
Looking at my procedure everything looks fine.I am wondering if book has done any mistake,or there is something else I should do, which I have missed,and it is making me do the mistake.
 
bigu01 said:
R//C Z=R//1/jωC = R1/(1+jωC1R1) inputing numbers I get =10^6/(1+j37.7) multiplying by 1+j37.7's conjugate I got 703.09-26506.54j , tried to find the length got the 26515.86 , finding the angle theta=arctan(-26506.54/703.09) doesn't give me same as book result. I get -1.519 book got -1.5443

Your numbers all look fine except for the result of the arctan(). When I enter that argument and take the arctan I get the book's result.
 
gneill said:
Your numbers all look fine except for the result of the arctan(). When I enter that argument and take the arctan I get the book's result.

You're saying that arctan(26506.54/703.09) gives 1.554 , either my two calculators are wrong or I am doing something wrong.
 
bigu01 said:
You're saying that arctan(26506.54/703.09) gives 1.554 , either my two calculators are wrong or I am doing something wrong.

That's what I'm saying :smile:
 
Okay then, what about this one 101.92+j90.38,what do you get as your ∠? The book answer is ∠0.723 and about this one I have no clue how did they get it
 
bigu01 said:
Okay then, what about this one 101.92+j90.38,what do you get as your ∠? The book answer is ∠0.723 and about this one I have no clue how did they get it

The numbers are probably rounded somewhat. I get 0.725 radians as the angle.
 
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Multiple threads are confusing... it looks like the error was here:
bigu01 said:
I was using the degree mode, I should have used the radian angle unit one

I have no idea how you get -1.519 in degree mode... well does not matter.