Impedance of practical Capacitor

[bigu01]

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.

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.

 

[mfb]

I can confirm the solution of the book. How did you calculate -1.519? Maybe just a rounding error?

 

[bigu01]

I can confirm the solution of the book. How did you calculate -1.519? Maybe just a rounding error?

I was using the degree mode, I should have used the radian angle unit one

 

[Mark44]

Don't post the same problem in multiple threads. I am closing this thread.

 

[rezeew]

Your approach is correct, but there may be a small calculation error in your answer. The correct answer for the impedance of the practical capacitor at ω=377rad/s should be Z1(377)=2.6516*10^4 ∠-1.5443. This is because the angle should be in radians, not degrees. So, when calculating the arctan, make sure to use the radian mode on your calculator.

Additionally, make sure you are using the correct values for C1 and R1. In the problem statement, it states that C1=0.1 x 10^-6F, which means C1=0.0000001F. It also states that R1=1*10^6, which means R1=1000000Ω. Double check these values in your calculation to ensure accuracy.

Overall, your approach is correct and it seems like a small calculation error is causing the difference in the angle compared to the book solution. Keep practicing and double checking your calculations to improve your accuracy.