Register to reply 
Capacitor's Reactance 
Share this thread: 
#1
Dec2908, 10:09 AM

P: 47

How do we derive the opposition to current flow (AC signal) due to capacitance without using complex numbers?
Is the capacitor's reactance a constant or an average? If I were to measure the instantaneous voltage across a capacitor and the instantaneous current through it, shouldn't I get its reactance (according to Ohm's Law)? Why do we get the reactance of the capacitor dividing the peak voltage by the peak current if they are always 90 degrees out of phase? 


#2
Dec2908, 10:46 AM

P: 359

Xc (reactance) = 1/2 x pi x f x C IF the C is essentially lossless.
R in series with C would give impedance Z = Root of the sum of the squares of Xc and R (vector sum) Xc is constant at given frequency f. I = V/Xc. Xc = Vrms/Irms and as Vrms is proportional Vpk and Irms is prop to Ipk. Xc = Vpk/Ipk 


#3
Dec2908, 12:23 PM

P: 47

I don't know if you've noticed but you didn't answer any of my questions. However, thanks for reply anyway :)



#4
Dec2908, 12:42 PM

Mentor
P: 40,717

Capacitor's Reactance



#5
Dec2908, 01:04 PM

P: 47




#6
Dec2908, 02:17 PM

Mentor
P: 40,717

http://en.wikipedia.org/wiki/Reactance_(electronics) . 


#7
Dec3008, 10:11 AM

P: 47

I don't really know how to explain what I mean :S
However, does Ohm's Law only work when both voltage and current are in phase? Imagine an AC circuit comprised of a resistor only. We both know that the voltage and current are both in phase. Now, if we want to know the R value we can just apply Ohm's Law at any given time. Obviously such value will also give you the circuit's opposition to current. If we now replace the resistor with a capacitor how can we manage to get the same current opposition as with the resistor?! Well, we just need to make X equal to R and we are done! Nonetheless, if we measure the instantaneous voltage (across it) and current (through it) at a given time and work out their ratio (Ohm's Law) we won't get the correct value of the opposition to current. Therefore Ohm's Law doesn't work with instantaneous values in AC circuits where there's a phase difference between the voltage and current. Lol... I think I've just answered my question although it seems a bit counter intuitive?! 


#8
Dec3008, 10:45 AM

P: 359

Intantaneous voltage or current is a difficult concept in AC circuits as it suggests a frozen moment in time and DC values which wont exist with a capacitor. It is a circuit element with energy storage and you have to analyse it over one or more AC cycles. The current flow depends on frequency. It has to be analysed dynamically.



#9
Dec3008, 12:24 PM

Mentor
P: 40,717




#10
Dec3008, 01:06 PM

P: 47

Fair enough, but could you tell me why do we get a capacitor's reactance working out the ratio between the Vp and Ip? Why the peak values if they are 90 degrees out of phase? If you apply Ohm's Law directly to instantaneous values and work out their ratio it will fail.
Where does such ratio come from? 


#11
Dec3008, 07:07 PM

P: 1,070




Register to reply 
Related Discussions  
DC and capacitive reactance  Advanced Physics Homework  1  
Phase Delay of capacitor's voltage  Engineering, Comp Sci, & Technology Homework  1  
Reactance is equal  Electrical Engineering  1  
AC waveform and reactance  Electrical Engineering  4  
What the expression CIVIL means  Advanced Physics Homework  1 