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Impedance (Inductor-Capacitor)

by jeff1evesque
Tags: impedance, inductorcapacitor
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jeff1evesque
#1
Aug16-09, 01:00 PM
P: 312
Statement:
If we have a circuit consisting of an inductor and capacitor in series, the impedance is defined by the following:
[tex]Z = L + C = \sqrt{(X_{L} - X_{C})^{2}} = \sqrt{(X_{C} - X_{L})^{2}}[/tex]

My question:
Howcome the impedance (or reactance) isn't defined by the following equation:
[tex]Z = L + C = \sqrt{X_{L}^{2} + X_{C}^{2}} = \sqrt{X_{C}^{2} + X_{L}^{2}}?[/tex]

Reasoning:
I would think that we would add the impedance, even for inductors and capacitors since from my knowledge, impedance for circuits in series is added together.

Thanks,

JL
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rock.freak667
#2
Aug16-09, 01:24 PM
HW Helper
P: 6,207
Inductor: XL=2πfLj
Capacitor:XC= -j/2πfC

Add them together and you get Z=(XL-XC)j.

so |Z|=√(XL-XC)2
jeff1evesque
#3
Aug16-09, 01:37 PM
P: 312
Quote Quote by rock.freak667 View Post
Inductor: XL=2πfLj
Capacitor:XC= -j/2πfC

Add them together and you get Z=(XL-XC)j.

so |Z|=√(XL-XC)2
That makes sense, except when I looked up the definition of impedance for inductors and capacitors I found the following:

[tex]Z = X_{L} = [(\omega)L]j = [(2 \pi f)L]j[/tex]
[tex]Z = X_{C} = [\frac{1}{(\omega)C}]j = [\frac{1}{(2 \pi f L)C}]j[/tex]

Therefore,
[tex]X_{L} + X_{C} = [(2 \pi f)L]j + [\frac{1}{C(2 \pi f L)}]j = ...?[/tex]

rock.freak667
#4
Aug16-09, 03:14 PM
HW Helper
P: 6,207
Impedance (Inductor-Capacitor)

Wikipedia says it is [itex]X_C=\frac{-j}{2\pi f C}[/itex]. The voltage through the capacitor is 90 degrees out of phase with the voltage through an inductor.
diazona
#5
Aug16-09, 03:29 PM
HW Helper
P: 2,155
Actually I believe it's 180 degrees out of phase. The capacitor voltage is 90 degrees out of phase with the voltage through a resistor (or would be, if there were a resistor), and the inductor voltage is 90 degrees out of phase with the resistor in the other direction. For a net difference of 180 degrees.

That's why the impedances are defined as
[tex]X_L = i \omega L[/tex]
and
[tex]X_C = -\frac{i}{\omega C}[/tex]
Those two quantities are 180 degrees out of phase with each other.
jeff1evesque
#6
Aug16-09, 04:15 PM
P: 312
1.)
Quote Quote by diazona View Post
The capacitor voltage is 90 degrees out of phase with the voltage through a resistor (or would be, if there were a resistor)...
What if there were no resistors in the circuit, and it was only an inductor and capacitor (along with the sinusoid power source)?



2.)
Those two quantities are 180 degrees out of phase with each other.
Could someone elaborate more why the two quantities are 180 degrees out of phase (I'm guessing that's why the inductor has a phase [tex]\frac{\pi}{2}[/tex], while the capacitor has a phase of -[tex]\frac{\pi}{2}[/tex])?



3.)
That's why the impedances are defined as
[tex]X_L = i \omega L[/tex]
and
[tex]X_C = -\frac{i}{\omega C}[/tex]
I looked on the Wikipedia page and found my definition on the "Device Examples" section:
[tex]
Z = X_{L} = [(\omega)L]j = [(2 \pi f)L]j
[/tex]
[tex]
Z = X_{C} = [\frac{1}{(\omega)C}]j = [\frac{1}{(2 \pi f L)C}]j
[/tex]

But further down I also find your definition,
[tex]X_C = -\frac{j}{\omega C}.[/tex]
How does this definition help us to conclude our result,
[tex]
Z = L + C = \sqrt{(X_{L} - X_{C})^{2}} = \sqrt{(X_{C} - X_{L})^{2}} ?
[/tex]



4.) On the same Wikipedia page,
I understand the following line of equation,
[tex]-j = cos(-\frac{\pi}{2}) + jsin(\frac{\pi}{2}) = e^{j(- \frac{\pi}{2})}[/tex]

But I don't understand how [tex]\frac{1}{j} = -j ?[/tex]

Thanks,


JL
vk6kro
#7
Aug18-09, 04:14 AM
Sci Advisor
P: 4,016
But I don't understand how 1 / j = -j ?

A common trick when solving these complex number problems is to multiply by J / J to bring the J to the top line of a fraction.

j = √-1

1 / j = 1 / √-1

Multiply by J / J or √-1 / √-1

Gives you 1 / J = (√-1) / -1 or
1 / J = - (√-1)
1 / J = -J

So, 1 / J = -J


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