Impedance (Inductor-Capacitor)

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In summary, impedance is a measure of the opposition to the flow of alternating current and is represented by the symbol Z. It takes into account both resistance and reactance, which is the opposition to alternating current caused by inductance and capacitance. Inductance and impedance are directly related in an inductor-capacitor (LC) circuit, as inductance increases, so does impedance. On the other hand, a capacitor's reactance decreases with increasing frequency, resulting in a decrease in impedance. In an LC circuit, the total impedance can be calculated using the formula Z = √(R² + (Xl - Xc)²), which considers both resistance and reactance.
  • #1
jeff1evesque
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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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  • #2
Inductor: XL=2πfLj
Capacitor:XC= -j/2πfC

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

so |Z|=√(XL-XC)2
 
  • #3
rock.freak667 said:
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]
 
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  • #5
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.
 
  • #6
1.)
diazona said:
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 http://en.wikipedia.org/wiki/Electrical_impedance#Complex_voltage_and_current" 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 http://en.wikipedia.org/wiki/Electrical_impedance#Complex_voltage_and_current",
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
 
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  • #7
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
 

1. What is impedance?

Impedance is a measure of the opposition that a circuit presents to the flow of alternating current. It is represented by the symbol Z and is measured in ohms (Ω).

2. How is impedance different from resistance?

Resistance is a measure of the opposition to the flow of direct current, while impedance takes into account both resistance and reactance (the opposition to alternating current caused by inductance and capacitance).

3. What is the relationship between inductance and impedance?

Inductance and impedance are directly related in an inductor-capacitor (LC) circuit. As inductance increases, so does impedance, because the inductor's reactance increases with frequency.

4. How does a capacitor affect impedance?

A capacitor's reactance decreases as frequency increases, so as the frequency of the alternating current increases, the capacitor's impedance decreases. This means that capacitors are more effective at filtering out high-frequency signals in a circuit.

5. How is impedance calculated in an LC circuit?

In an LC circuit, the total impedance is calculated using the formula Z = √(R² + (Xl - Xc)²), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. This formula takes into account both the resistance and reactance in the circuit.

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