Recent content by abaz

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    MHB How can the second term be mathematically turned into -j or 1/j?

    AFAIK They're just vectors. Since the angles between them in the electrical game is usually the phase difference they call them phasors. As for them being complex, I don't know. It's probably a question for another thread. This is one of the reasons I'm looking at $X_C$, and $X_L$ for that...
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    MHB How can the second term be mathematically turned into -j or 1/j?

    Funny thing that. I'm an electrician, did 3yr trade course and I'm almost certain nothing complex was ever mentioned! Yet here we were, day in, day out, drawing phasor's on the x/y plane. Probably why I find j terms so elusive
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    MHB How can the second term be mathematically turned into -j or 1/j?

    I actually plotted $\frac{\sin \left ( \theta\right )}{\cos \left (\theta\right )}$ and said Hey that's tan!. Substituted sin and cos with (opp/hyp)/(adj/hyp) and ended up with opp/adj Thought I was so :cool: LOL ... forgotten trig id's
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    MHB How can the second term be mathematically turned into -j or 1/j?

    Thank you very much Mark and I Like Serena (is just ILS or Serena ok?) I have read a little on Euler's number $e$ since this post and it's significance is impressive to say the least, so I will spend a while learning more... In my original post I canceled out the $t$'s when I shouldn't have...
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    MHB How can the second term be mathematically turned into -j or 1/j?

    Thank you I like Serena (your avatar had me in stitches!) It would seem that you and zzephod are both steering me toward Euler I may be over thinking this (or being to simplistic) but if Euler's formula yields $X_{c}=\frac{-j}{\omega C}$ and that assuming the partial derivation that I showed is...
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    MHB How can the second term be mathematically turned into -j or 1/j?

    Hi all I am trying to go through and understand the derivation of \[X_{c} = \frac{1}{j\omega C}\] Where $X_{c}$ is capacitive reactance in ohms, $\omega$ is the angular velocity or $2\pi frequency$ and $C$ is capacitance in FaradsTo start with we already have $I=C\frac{dv}{dt}$...
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