Calculate J Notation Impedance of Network w/ Capacitor & Inductor

AI Thread Summary
The discussion focuses on calculating the J notation impedance of a network comprising a capacitor in parallel with an inductor, which is then in series with a resistor and another capacitor. The main confusion arises regarding the signs of the J notation for the capacitor, where participants discuss the implications of using positive versus negative values. The impedance of the parallel circuit is expressed using the product over sum method, leading to questions about the consistency of signs in the calculations. Clarification is provided that the position of J in the equations affects the sign, with the relationship between -j and 1/j being highlighted. Overall, the thread seeks to resolve the confusion around the correct application of J notation in the impedance calculations.
lubo
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Homework Statement



We have a capacitor in parallel with an inductor. These are both in series with a resistor and capacitor.

Calculate the J notation Impedance of the network. I only want the initial basic solution.

The problem I have is that sometimes the J notation of C is -ve and sometimes +ve ?

Homework Equations





The Attempt at a Solution



Product/sum of the parallel cct:

jwL x 1/jwC/jwL*1/jwC This is the Inductor and capacitor impedance equation.

The above will be added to:

R -j(1/wC)

My question is therefore, why in the above example at product over sum would it be ok to say jwL x 1/jwC/jwL*1/jwC i.e. * a +ve 1/jwC

When below it I can add it to R and -j(1/wC)

I hope this makes sence, thanks for any help in advance.
 
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hi lubo! :smile:
lubo said:
My question is therefore, why in the above example at product over sum would it be ok to say jwL x 1/jwC/jwL*1/jwC i.e. * a +ve 1/jwC

When below it I can add it to R and -j(1/wC)

ah, but the first j is on the bottom, while the second is on the top …

and -j = 1/j :wink:
 
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