# when capacitance is small, or frequency low, small amount of charge needed...

#### PG1995

Joined Apr 15, 2011
823
Hi

Please have a look on this scanned page from a book (or, you can check the attachment):
http://img683.imageshack.us/img683/2193/imgdec.jpg

Please focus on where it says under the topic "Effect of alternating current (a.c.) on a capacitor":
If the capacitance is small, or the frequency low, only a small amount of charge needs to flow onto the capacitor to equal the supply p.d. In other words, the current will be very low. High capacitance and high frequency gives high current.
I understand that if the capacitance is small, then only a small amount of charge (from the supply) needs to flow into the capacitor before the capacitor's voltage equals the p.d. of the supply. But what I don't understand is frequency part. I know it can be proved easily with some math expressions. I'm unable to create a conceptual picture of this. Why would a small amount of charge be needed to equal the p.d. of the supply when the the frequency of the a.c. current is low?

Regards
PG

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#### colinb

Joined Jun 15, 2011
351
Maybe what they mean is that when frequency is low, the average current is low because the capacitor will be charged and discharged fewer times per second.

#### crutschow

Joined Mar 14, 2008
29,515
As colinb said, the average current depends upon the number of times per second you charge and discharge the capacitor. Another way to look at it is that the faster you charge and discharge the capacitor, the higher the peak current to do the charging. But the total amount of charge needed to charge the capacitor for each cycle is constant, independent of frequency, and only depends upon the peak AC voltage.

#### PG1995

Joined Apr 15, 2011
823
Thank you, Colin, crutschow.

@crutschow: Could you please clarify the bold part?
Another way to look at it is that the faster you charge and discharge the capacitor, the higher the peak current to do the charging.
Thanks.

Regards
PG

#### crutschow

Joined Mar 14, 2008
29,515
@crutschow: Could you please clarify the bold part?

Another way to look at it is that the faster you charge and discharge the capacitor, the higher the peak current to do the charging.
I meant the faster you charge and discharge the capacitor, the higher the peak current required to charge the capacitor to the peak voltage. Obviously if it takes a certain amount of charge to charge a capacitor to a given voltage, the faster you do it, the higher the current.

Is that clear?

#### steveb

Joined Jul 3, 2008
2,436
Could you please clarify the bold part?

Another way to look at it is that the faster you charge and discharge the capacitor, the higher the peak current to do the charging.
This is just a basic property of capacitors. Let's repeat the math first and then talk about the concepts.

Take the the definition of capacitance.

C=Q/V, and rearrange to Q=CV

Then take the derivative of both sides, assuming capacitance is constant

dQ/dt=C dV/dt

and remember the definition of current is I=dQ/dt

So, I=C dV/dt

If the voltage is a sinusoid of the form V=Vo sin(wt) then

I=C Vo w cos(wt)

Now it is clear that frequency is proportional to current.

You probably already know this because you understand the math, but not the concept. So, now on to the concept.

Compare a capacitor with a spring.

The relationship for a spring is F=kX where k is the spring constant, X is the stretching distance and F is the stretching force. The 1/k is analogous to C, X is analogous to Q and F is analogous to V. Now imagine putting a force on the spring by pulling it between both hands. The distance that the spring stretches is independent of frequency, right? This is analogous to the charge on the capacitor being independent of frequency. But, the faster you pull the spring with that same force, the faster the speed is. But, speed is analogous to current since s=dX/dt. Hopefully, you can see the concept now.

The math is the same, note the following where speed is s=dX/dt

F=kX which leads to dF/dt=ks or s=(1/k) dF/dt

For example, a small child may not be able to put much force on a spring, while a large gorilla could put much force. But, the child might be able to move the spring at higher speed, simply by pulling and pushed at a faster rate. Remember, theoretical springs have no mass, just like theoretical capacitors have no inductance.

#### PG1995

Joined Apr 15, 2011
823
Thank you, crustschow, Steve; my special thanks to you, Steve, for your detailed reply.

If the voltage is a sinusoid of the form V=Vo sin(wt) then

I=C Vo w cos(wt)
"Vo" is peak voltage, "t" is period, and "w" is angular velocity. Please correct me if I'm wrong.

Let's talk about the capacitor a little more. I do have made some progress!

When a battery is connected to a capacitor an instantaneous current flows into the the capacitor. Now how much current (which is charge per unit time) flows into the capacitors depends on the capacitance of the capacitor. If capacitance is large, then it would take more current (hence charge) to reach the state where voltage of the capacitor counterbalances the voltage supplied by the battery.

A battery is also limited by a maximum current it can supply. Suppose we have two batteries which can supply the same voltage but battery_A can supply half the maximum current of battery_B. If a capacitor is connected to batter_B (it can supply the double the maximum current of battery_A) then the capacitor will take half the time to reach the stage where it can counterbalance the battery's voltage as compared to when it is connected to the battery_A. Do I make sense?

Now we will talk about AC and frequency - the main point of original discussion.

Just like a battery an AC source also has maximum current limitation, I believe. Let's say a capacitor is connected to an AC source, and needs 10 coulombs of charge to match the peak voltage of the source, and the maximum current the supply can deliver is 1A (1A current is flow of one coulomb of charge per second). If we keep on increasing the frequency of the source and reach a point when time taken for the voltage to change from 0 value to the peak value becomes 5 seconds. In 5 seconds only 5 coulombs of charge would flow but the capacitor needs 10 coulombs to match the supply's voltage. This means that the capacitor will start discharging even before it has been fully charged. Do I make sense? Please keep your reply simple.

Regards
PG

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#### colinb

Joined Jun 15, 2011
351
"Vo" is peak voltage, "t" is period, and "w" is angular velocity. Please correct me if I'm wrong.
I think the slightly more proper term is angular frequency. As you might assume, it is in radians/second.

#### MrChips

Joined Oct 2, 2009
25,930
A resistor has a resistance to DC.
A capacitor has resistance to AC, and we call this reactance.
The reactance of a capacitor decreases with increasing frequency.
Hence, as frequency increases, the current increases in a capacitive load.

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#### kubeek

Joined Sep 20, 2005
5,772
When a battery is connected to a capacitor an instantaneous current flows into the the capacitor. Now how much current (which is charge per unit time) flows into the capacitors depends on the capacitance of the capacitor. If capacitance is large, then it would take more current (hence charge) to reach the state where voltage of the capacitor counterbalances the voltage supplied by the battery.
This is wrong. The peak current that happens in the time of switching on is only set by the power supply's internal resistance - the capactor acts like a short at first. Then it starts charging and the current gradually goes to 0.

The capacitance only changes the time that it takes to charge it with some current.

#### PG1995

Joined Apr 15, 2011
823
Thank you, Colin, MrChips, Kubeek.

I'm sure some of the things I said in my previous posting aren't correct but I hope you get the general idea what I was trying to say. And would request some of you to see if it is makes sense generally. Thanks.

Regards
PG