Register to reply 
Impedance (InductorCapacitor) 
Share this thread: 
#1
Aug1609, 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 


#2
Aug1609, 01:24 PM

HW Helper
P: 6,202

Inductor: X_{L}=2πfLj
Capacitor:X_{C}= j/2πfC Add them together and you get Z=(X_{L}X_{C})j. so Z=√(X_{L}X_{C})^{2} 


#3
Aug1609, 01:37 PM

P: 312

[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] 


#4
Aug1609, 03:14 PM

HW Helper
P: 6,202

Impedance (InductorCapacitor)
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.



#5
Aug1609, 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. 


#6
Aug1609, 04:15 PM

P: 312

1.)
2.) 3.) [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 


#7
Aug1809, 04:14 AM

Sci Advisor
P: 4,032

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 


Register to reply 
Related Discussions  
Inductor (and Capacitor) Discharge  Introductory Physics Homework  23  
Capacitor and Inductor in Series  Engineering, Comp Sci, & Technology Homework  9  
Capacitor & Inductor ?  Introductory Physics Homework  2  
Impedance of a Capacitor  Introductory Physics Homework  4  
Capacitor & Inductor?  General Physics  5 