# Impedance (Inductor-Capacitor)

1. Aug 16, 2009

### jeff1evesque

Statement:
If we have a circuit consisting of an inductor and capacitor in series, the impedance is defined by the following:
$$Z = L + C = \sqrt{(X_{L} - X_{C})^{2}} = \sqrt{(X_{C} - X_{L})^{2}}$$

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

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

Last edited: Aug 16, 2009
2. Aug 16, 2009

### rock.freak667

Inductor: XL=2πfLj
Capacitor:XC= -j/2πfC

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

so |Z|=√(XL-XC)2

3. Aug 16, 2009

### jeff1evesque

That makes sense, except when I looked up the definition of impedance for inductors and capacitors I found the following:

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

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

Last edited: Aug 16, 2009
4. Aug 16, 2009

### rock.freak667

Last edited by a moderator: Apr 24, 2017
5. Aug 16, 2009

### diazona

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
$$X_L = i \omega L$$
and
$$X_C = -\frac{i}{\omega C}$$
Those two quantities are 180 degrees out of phase with each other.

6. Aug 16, 2009

### jeff1evesque

1.)
What if there were no resistors in the circuit, and it was only an inductor and capacitor (along with the sinusoid power source)?

2.)
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 $$\frac{\pi}{2}$$, while the capacitor has a phase of -$$\frac{\pi}{2}$$)?

3.)
I looked on the http://en.wikipedia.org/wiki/Electrical_impedance#Complex_voltage_and_current" and found my definition on the "Device Examples" section:
$$Z = X_{L} = [(\omega)L]j = [(2 \pi f)L]j$$
$$Z = X_{C} = [\frac{1}{(\omega)C}]j = [\frac{1}{(2 \pi f L)C}]j$$

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

4.) On the same http://en.wikipedia.org/wiki/Electrical_impedance#Complex_voltage_and_current",
I understand the following line of equation,
$$-j = cos(-\frac{\pi}{2}) + jsin(\frac{\pi}{2}) = e^{j(- \frac{\pi}{2})}$$

But I don't understand how $$\frac{1}{j} = -j ?$$

Thanks,

JL

Last edited by a moderator: Apr 24, 2017
7. Aug 18, 2009

### vk6kro

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