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 August 11th, 2011, 04:11 PM #1 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Confusion. So, we start with: $Z=\frac{\frac{1}{i\omega C}\cdot(R+i\omega L)}{R+i\omega L+\frac{1}{i \omega C}}$ Then we get: $\frac{R+i\omega L}{i\omega CR-w^2LC+1}$ by multiplying the equation by $\frac{i\omega C}{i\omega C}$ Wolfram Alpha does not agree with this: http://www.wolframalpha.com/input/?i=Z%3D%5Cfrac{%5Cfrac{1}{i%5Comega+C}*%28R%2Bi%5Comega+L%29}{R% 2Bi%5Comega+L%2B%5Cfrac{1}{i+%5Comega+C}} Why? Same result if I approach it a different way: \begin{align} Z&=\frac{\frac{R+i\omega L}{i\omega C}}{\frac{Ri\omega C-\omega^2LC+1}{i\omega C}}\\ &=\frac{R+i\omega L}{i\omega C} \cdot \frac{i\omega C}{Ri\omega C-\omega^2 LC+1}\\ &=\frac{R+i\omega L}{Ri\omega C-\omega^2 LC+1} \end{align}
 August 11th, 2011, 04:18 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Confusion. [color=#000000]It seems like a physics formula, what is $i$ the intensity? Wolframalpha may interpret it as the complex $i^2=-1$.[/color]
August 11th, 2011, 04:21 PM   #3
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Re: Confusion.

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 Originally Posted by ZardoZ [color=#000000]It seems like a physics formula, what is $i$ the intensity? Wolframalpha may interpret it as the complex $i^2=-1$.[/color]
Oops. Should've noted that $i$ is the imaginary unit.

 August 11th, 2011, 06:30 PM #4 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Confusion. Intriguing. Seems it agrees with me now: http://www.wolframalpha.com/input/?i=%28%5Cfrac{R%2Bi*w*L}{i*w*C}%29%2F%28%5Cfrac{R*i*w*C%2Bi^2*w ^2*L*C%2B1}{i*w*C}%29 Wolfram must have an issue with complex fractions?
August 12th, 2011, 08:21 AM   #5
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Re: Confusion.

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 Originally Posted by ZardoZ [color=#000000]It seems like a physics formula, what is $i$ the intensity? Wolframalpha may interpret it as the complex $i^2=-1$.[/color]
You are right it's an RCL circuit. $i$ isn't an imaginary number. It's already in the simplest form after truncating a long series. I nonplused what Ms Pi is trying to achieve.

August 12th, 2011, 08:58 AM   #6
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Re: Confusion.

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Originally Posted by Math Dreamer
Quote:
 Originally Posted by ZardoZ [color=#000000]It seems like a physics formula, what is $i$ the intensity? Wolframalpha may interpret it as the complex $i^2=-1$.[/color]
You are right it's an RCL circuit. $i$ isn't an imaginary number. It's already in the simplest form after truncating a long series. I nonplused what Ms Pi is trying to achieve.
The book I took this from states that it's an imaginary number and the actual problem set is in the chapter on Complex Numbers. . .

 August 12th, 2011, 09:14 AM #7 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Confusion. [color=#000000]I think that if the i denotes the imaginary $i^2=-1$, then Z is the impedance of an AC circuit, since electric flow changes direction periodically, so if this formula describes an AC circuit it is correct. I haven't studied physics since high-school though so I am not sure! As far as I can remember R stands for resistance and C for capacity. [/color]
August 12th, 2011, 09:36 AM   #8
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Re: Confusion.

Quote:
 Originally Posted by ZardoZ [color=#000000]I think that if the i denotes the imaginary $i^2=-1$, then Z is the impedance of an AC circuit, since electric flow changes direction periodically, so if this formula describes an AC circuit it is correct. I haven't studied physics since high-school though so I am not sure! As far as I can remember R stands for resistance and C for capacity. [/color]
Thank you. That's what I was asking about. I think that Wolfram Alpha has an issue with complicated fractions--maybe it renders them wrong?

August 12th, 2011, 10:04 AM   #9
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Re: Confusion.

Quote:
 Originally Posted by ZardoZ [color=#000000]I think that if the i denotes the imaginary $i^2=-1$, then Z is the impedance of an AC circuit, since electric flow changes direction periodically, so if this formula describes an AC circuit it is correct. I haven't studied physics since high-school though so I am not sure! As far as I can remember R stands for resistance and C for capacity. [/color]
R is the resistance, L the induction, C the capacitance, and i the current. I found it in library from a physics book, then it leads to a WWII HF radio. Radio at it's crudest stage. It got my friend fired up. We will have a trip to a junk yard for parts. The junk yard is our Walmart.

August 12th, 2011, 12:07 PM   #10
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Re: Confusion.

Quote:
 Originally Posted by ZardoZ [color=#000000]???????? ????????, ???? ??????? ???????? (????? ???????? ?????????? ???? ???????????, ????? ???????? ???????????? ????????????, ???????? ??? ????? ?????????? ????? ?????? ??????? ????????), ???????? ??????????, ???? ????? ????????? ???????? ????? ????? ??? ????????? ????????? ??????, ????????? ?????????? ???????? ??????? ?????????? ?????. [/color]
This is what I see there, I know nothing of physics

I searched for "impedance" and the definition was the same of resistance, but it says the resistance is the real part of the impedance.. how come is a fraction of two real numbers (potential difference and current) a nonreal number in any sense? Physics is just so unearthly sometimes..

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