# Complex Impedance

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## Homework Statement

Question 2 on this paper:
http://www.eng.cam.ac.uk/teaching/past_papers/examples_student/examples_IA/P3_LCD_Q/1ap3lcdq003

Z = V/I
V=L di/dt

## The Attempt at a Solution

for part a of the question:
I began with V = Vo cos(wt)
And I = Io cos (wt + k)

I expressed them as exponentials' real parts using Euler's relation:

Z = V/I
So, Z = [Vo (e^(-jk))]/Io

I don't know what to do further.

I've looked around the internet and apparently that leads to
Z = R + jX

So, the answer for the part a would be 10 + j10
But how?

Thanks

## Answers and Replies

Eloc Jcg
For the formulas that you have posted, I personally don't have any experience dealing with those, unless the ones that I use use different variables. And I'm not really sure what the "complex" part of Impedence means except maybe complex numbers. I would solve the question using the formula Z = R + j(XL - XC) = 10 Ohm + j(10 Ohm) = 10 + 10j, which is the answer that you got. Not entirely sure if my way of doing this problem is correct, but that is how I would go about solving this question with my current Electronics Engineering intelligence.

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For the formulas that you have posted, I personally don't have any experience dealing with those, unless the ones that I use use different variables. And I'm not really sure what the "complex" part of Impedence means except maybe complex numbers. I would solve the question using the formula Z = R + j(XL - XC) = 10 Ohm + j(10 Ohm) = 10 + 10j, which is the answer that you got. Not entirely sure if my way of doing this problem is correct, but that is how I would go about solving this question with my current Electronics Engineering intelligence.

I think you are right.
How did you derive the
Z = R + j(XL - XC)

I think you are using the final general solution (sort of the solution you get after you plough through all the maths).

I think you get Z = R + jXL for inductor and Z = R - jXC for a capacitor. But am not too sure.
I wanted to know the mathematical reasoning behind the equation.

Also, using your method, how would you go about solving part c?
Do they add using the same laws as vector addition (since complex numbers can be added like that).

Eloc Jcg
I am not entirely sure on how it is technically derived, but I know that Impedence (Z) is the total effective resistance to the flow of current by a combination of the elements in the circuit. Resistance being in the real plance, while inductance and capacitance are on the imaginary plane, (hence the multiplication of j by XL - XC). Xc is determined by 1/(wC) & XL = wL. w being the angular velocity. Not sure if that clarifies anything or helps.

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I found another thread that went through some of the derivation:

They talk about phasor representations and so on, which is a little beyond my mathematical knowledge so I think I'll just have to learn the relationship you gave me.

How would you go about solving part c using your method then?

Thanks a lot man.

Eloc Jcg
I don't have any current experience for solving with impedence in a parallel circuit, so you may need to look up the rules for combining elements in a parallel circuit. The 10 and 20 ohm resistors should take the inverses of 10 & 20 then add them together then inverse that answer for the combined ohm resistance. As for the L2/3, those looks like they can be combined somehow. Either by using the same method as the ohms, or by simply adding them. Sorry I can't be of much more help. Here is a list of formulas that could help you out: Things in [brackets] are subscripted

V[RLC] = IZ
Z = R + j(X[L] - X[C])
|Z| = square root of (R^2 + (X[L] - X[C])^2)
angle = inverse tan of (X[L] - X[C])/R

Happen to know anything about Nucleur Physics & Potential Electric Difference btw? I got a post with some help needed. lol.

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Thanks man.
I looked at your question but I don't know how to do it.

I was thinking maybe use the F = E q = Vq/d = ma ?
But that means any non zero potential will be able to bring it to rest (lower potential will simply take long to do it and the nucleus will travel a larger distance while decelerating).