Born2bwire
Diamond Member
- Oct 28, 2005
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Originally posted by: Gibson486
welll....that is what sinosoidal capacitance looks like. You do not repalce C with 1/(jwC)....1/(jwc) is the defination of impedance for a cap in sinosoidal analysis.Originally posted by: Born2bwire
Can't imagine how he would know to replace C with 1/(j\omega C) and L with j\omega L without knowing Laplace transform.Originally posted by: Gibson486
Originally posted by: Born2bwire
Originally posted by: Fiveohhh
Originally posted by: Born2bwire
Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1
From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.
s = j\omega, not j\nu.
Not following ya.
1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap
Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.
I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes
You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.
he may have not been introduced to laplace yet.....if he has been, it would make this so much easier
Laplace tells you to replace anything related to the imaginary part with s.
The Laplace/s trasform is how you generate the complex impedances from the original differential forms. It seemed more natural to me that they would have introduced the general theory on how the phasor forms were derived first. But I guess they could do it either way.
Originally posted by: Fiveohhh
Originally posted by: Born2bwire
Originally posted by: Fiveohhh
I think I see what you're saying. It is supposed to be 1000 rads/sec. I just drew that up and didn't change the units.
Well in that case your answer is 0.16 \mu F and 0.04 \mu F.
Does it make sense that there are two solution? I'm still new to most of this AC, but I would think that as the circuit progresses from inductive to capacitive that it would only be in phase once.
In what sense is it progressing? You're changing the value of the capacitance and the general nature of the circuit, whether it is capacitive or inductive, is dependent upon the entire circuit as a whole. Depending upon the relationships, increasing the capacitance in one part of a circuit can end up making the entire circuit more inductive as is the case here.