Effect of alternating current (a.c.) on a capacitor

[PainterGuy]

the book says:

Effect of alternating current (a.c.) on a capacitor
The size of the current will depend on the capacitance and on the frequency of the supply. If the capacitance is small, or the frequency low, only a small amount of charge needs to flow onto the capacitor plates for the p.d. across the capacitor to equal the supply p.d. In other words, the current will be very low. High capacitance and high frequency give high current.

i do not understand the effect of frequency and why small charge need to flow to equal p.d. across capacitor to that of supply when frequency is low. would you help me with this, please?

many thanks for any help you can give.

cheers

 

[tiny-tim]

hi painterguy!

in a capacitor V is not related to I, instead dV/dt is related to I (CdV/dt = dQ/dt = I)…

if V = V_maxsinωt, then I = CdVdt = ωCV_maxcosωt …

(in other words: the impedance of a capacitor is 1/jωC)

ie I_max = ωCV_max …

high capacitance and high frequency give high current

 

[PainterGuy]

hi painterguy!

in a capacitor V is not related to I, instead dV/dt is related to I (CdV/dt = dQ/dt = I)…

if V = V_maxsinωt, then I = CdVdt = ωCV_maxcosωt …

(in other words: the impedance of a capacitor is 1/jωC)

ie I_max = ωCV_max …

high capacitance and high frequency give high current

hi tiny-tim,

oh, you have used so many math formulas! now i am more confused...is there any simple explanation without that math formula for conceptual understanding...please, please.

 

[tiny-tim]

hi painterguy!

...is there any simple explanation without that math formula for conceptual understanding

not really …

you can't understand capacitors without understanding how the equations work …

the best i can do is to translate the equations into english for you …

the current is proportional to the capacitance and to the rate of change of the voltage, and the rate of change of the voltage is proportional to the voltage and to the frequency

(obviously, the faster the frequency, the faster the voltage is changing )​

 

[sophiecentaur]

If the supply resistance is zero (an ideal voltage source) then there is no "effect of AC". The only time that AC will be relevant will be when there is a resistance involved. When there is a resistance in circuit, the Capacitor will take time to charge or discharge. For a large C and / or large R, this can be a long time. If the AC is at a high frequency then there is not enough time, before polaraty reverses, for the Voltage on the Capacitor to change significantly. This is what gives the frequency selectivity when using a Capacitor. It will follow a slow variation closely but average out a fast variation to zero.

 

[PainterGuy]

the current is proportional to the capacitance and to the rate of change of the voltage, and the rate of change of the voltage is proportional to the voltage and to the frequency

hello tiny-tim,

does this mean if voltage being low or high also affects rate of change of voltage? i thought rate of change of voltage is only proportional to frequency!

many thanks for the help.

cheers

 

[sophiecentaur]

Rate of change of voltage depends both on frequency AND peak voltage aamof. (That is, the volts per second which will be maximum at the zero crossing)

 

[tiny-tim]

the bigger something is, the more it is changing

 

[sophiecentaur]

Rate of change of voltage depends both on frequency AND peak voltage aamof. (That is, the volts per second which will be maximum at the zero crossing)

In OP Amps it's characterised as slew rate.

 

[PainterGuy]

Rate of change of voltage depends both on frequency AND peak voltage aamof. (That is, the volts per second which will be maximum at the zero crossing)

hello sophiecentaur,

i am trying to first understand first relationship between rate of change of voltage and frequency. i think it has something to do with derivates in math. suppose a wave has 2hz frequency - two wave cycles per second. if there are going to 100 waves per second then adjust 100 waves in the same space of 1 second, each wave is going to have sharper slope as compared when there were only two waves in one second. do u see any correctness in this? tell me please. many thanks for your helping me.

cheers

 

[sophiecentaur]

For a sinusoid, the rate of change (the time derivative) is just a cos wave.

d(A sin(wt))/dt =A w cos(wt)
so the maximum slope for a sinewave of amplitude A and angular frequency w is Aw.

 

[tiny-tim]

… each wave is going to have sharper slope as compared when there were only two waves in one second. do u see any correctness in this?

yes, that's exactly correct …

higher frequency (and same amplitude) obviously squeezes the graph, so makes the slope higher, and the slope is the derivative (the rate of change)

 

[PainterGuy]

yes, that's exactly correct …

higher frequency (and same amplitude) obviously squeezes the graph, so makes the slope higher, and the slope is the derivative (the rate of change)

hi tiny-tim,

many thanks for clearing this confusion. so one mystery is solved.

but as sophiecentaur and you said:
Rate of change of voltage depends both on frequency AND peak voltage.

okay, increasing frequency squeezes the graph and make the slope sharper. but how can increasing the peak make the slope sharper?

here it is my "little" understanding:
please check this picture:
http://img600.imageshack.us/img600/1142/sinewavenormalamplitude.png

the slope is almost constant between the two red marks (on the same side of the wave). do you agree? as one starts ascending the top red mark slope starts decreasing and as one starts descending the top red mark slope starts increasing.

now see this picture with doubled amplitude:
http://img22.imageshack.us/img22/5705/sinewaveamplitudedouble.gif

now the section during which the slope is almost constant has enlarged because the amplitude is double than before.

so this implies that as amplitude is increased the values of slope where is has higher value as compared to section where is decreasing spread over larger section. so average of those values during which slope is constant or increasing will be greater than when the amplitude was less.

do you understand and find correct what i am trying to say?

warning: these diagrams were not created by me. i found them on google and modified them only for the purpose of understanding. no claim of ownership.

many thanks for teaching me all this. much grateful.

cheers

 

[sophiecentaur]

A picture is worth a thousand words but an equation can be worth a thousand pictures. The little equation I put in my last post says it all. The maximum slope (rate of change) is A w (or 2pi X frequency X amplitude of the signal). That simple statement avoids any arm waving and people who want to get a grasp of things can help themselves by getting used to as many of the simple maths as possible.

 

[PainterGuy]

A picture is worth a thousand words but an equation can be worth a thousand pictures. The little equation I put in my last post says it all. The maximum slope (rate of change) is A w (or 2pi X frequency X amplitude of the signal). That simple statement avoids any arm waving and people who want to get a grasp of things can help themselves by getting used to as many of the simple maths as possible.

hi sophiecentaur,

yes, you are completely correct where you say an equation can be worth a thousand pictures. i did read your post but couldn't really grasp the hidden meaning behind the math. so was trying to approach the problem from my own angle in steps. i would have definitely returned to your equation once i had decipher the meaning behind the math. i do appreciate your help with many problems. every human learns in different way so i wasn't avoiding your post. i just avoided it for time being in order to understand it afterwards once i have understood some of the basics. i hope it is clear now.

now will you please give you opinion on my last post?

cheers

 

[sophiecentaur]

Hi
The Maths may look scary but I would say that it IS the basics and not something you come back to. Don't put off the evil day but grasp the nettle. There is very little in the Physical Sciences that can be 'understood ' without using some Maths and this is a great example of what I say. Without some Maths, it is sooo easy to extrapolate what one thinks one has understood and to jump to dodgy conclusions.

Your description in graphical terms is fine up to a point. Basically, you can say that, as you increase the frequency, you also increase the rates of change. But, this para_

"so this implies that as amplitude is increased the values of slope where is has higher value as compared to section where is decreasing spread over larger section. so average of those values during which slope is constant or increasing will be greater than when the amplitude was less."

, while it is broadly correct, is a great demo of how involved a purely verbal description can be. I had to read it three times before it meant anything to me. And I knew what you were trying to say.

 

[tiny-tim]

hi painterguy!

… so this implies that as amplitude is increased the values of slope where is has higher value as compared to section where is decreasing spread over larger section. so average of those values during which slope is constant or increasing will be greater than when the amplitude was less.

I had to read it three times before it meant anything to me. And I knew what you were trying to say.

i have to agree with sophiecentaur!

i think what you mean is: you can increase the amplitude (but not the frequency) by putting four fingers of one hand into four adjacent peaks, and four fingers of the other hand into four adjacent troughs, and pulling … obviously, that increases the amplitude and the slope everywhere in the same proportion!