How Does Frequency Affect Impedance in a Parallel RLC Circuit?

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In a parallel RLC circuit with a 1000 ohm resistor, 500 picofarad capacitor, and 2 millihenry inductor, impedance calculations depend on frequency, specifically at 10 kilocycles and 10 megacycles per second. Admittances should be considered since they add in parallel, leading to a complex number for total impedance. The complex impedance is the reciprocal of the complex admittance, and the modulus should be taken after determining the complete complex impedance. It is clarified that taking the modulus before the reciprocal could lead to incorrect results. Therefore, the correct approach is to compute the complex impedance first and then find its modulus.
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A 1000 ohm resistor, 500 picofarad capacitor, and 2 millihenry inductor are all conducted in parallel. What is the impedance if the frequency is 10 kilocycles per second? 10 megacycles per sec? At what frequency is the absolute value of impedance the greatest?

I know I start by considering admittances since those add in parallel. I will end up with a complex number. Can I take its modulus and then the reciprocal? Or do I take the reciprocal to get impedance and then take the modulus? Or do I take the reciprocal to get impedance and leave it as the complex number?
 
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You have a complex admittance.
The complex impedance is the reciprocal of the complex admittance
"Impedence" and "complex impedance" can be used interchangeably: you'll have to use the context of your classes to decide if the modulus is expected or not (that's how I'd interpret it.)

note; if ##y=a+jb##
then, using ##|z|=|1/y|## (i.e. reciprocal then modulus):$$z=1/y = 1/(a+jb) = (a-jb)/(a^2-b^2)\Rightarrow |z| = \sqrt{a^2+b^2}/(a^2-b^2)$$... something like that - what happens if you do it in the other order (i.e. modulus then reciprocal). Is ##|1/y|=1/|y|##)?

See:
http://en.wikipedia.org/wiki/Electrical_impedance
 
Ahah! I was certainly wrong to think I could do the modulus and then take the reciprocal, that would give something like sqrt(a^2+b^2), and it would ignore the denominator that you got in the modulus of z itself.

So if I were to take the modulus, it should ALWAYS be after I've gotten all the way to the end of the problem in complex numbers?
 
It usually works out simpler that way anyhow.
 
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