# capacitor charging

#### paks_paks2000

Joined Nov 24, 2004
5
when current is flowing through a capacitor, say from left to right ,what and how is the potential developed across the capacitor.negative/positive?

#### blackrider

Joined Mar 19, 2005
11
u mean practical or paper work?

#### Brandon

Joined Dec 14, 2004
306
Originally posted by paks_paks2000@Apr 24 2005, 12:15 AM
when current is flowing through a capacitor, say from left to right ,what and how is the potential developed across the capacitor.negative/positive?
[post=7198]Quoted post[/post]​
Current can't flow through a cap. A Cap is an open circuit.

When a voltage is applied to a cap, the cap will charge up to that voltage in a time span related to the amount of capacitence in the cap and the resistance between the cap and the voltage source.

The formula which describes this charging is

Vc(t)=[1-e^-(t/RC)]*Vin

where t is time in seconds. Typically with any e^-x equasions, when x=5, your at about 99% of what the total value is going to be in the cap.

SO, if you wanted to see how long it would take the cap to charge to 99% of Vin,
you would set t/RC = 5 and solve for t.

The same happens when a cap discharges. I.e, if you charged the cap then tied it to ground.

The formula then changes to

Vc(t)=[e^-(t/RC)]*Vin

Hope this helped.

#### dragan733

Joined Dec 12, 2004
152
At the capacitor, in the start when it is supplied with a DC voltage, transient processes happen and in these transient processes the current flows across the capacitor and the formula for the current is:
I=(Vcc/R)*exp(-t/RC), if the capacitor is supplied across a resistor R.
After the transient process, in the stationary regime, the current doesn't flow across the capacitor, when the capacitor is supplied with a DC voltage.
In the TRANSIENT process, the capacitor voltage is expressed with the formula:
V=Vcc(1-exp(-t/RC)), if the capacitor is supplied across a resistor R

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by Brandon@Apr 24 2005, 04:22 PM
Current can't flow through a cap.  A Cap is an open circuit.
To d.c., once it is fully charged.

When a voltage is applied to a cap, the cap will charge up to that voltage in a time span related to the ... value of the cap ... and the resistance between the cap and the voltage source.
And, assuming a step function, the initially high current will reduce to zero over the same time.

On the other hand, if the voltage is a.c , the current will be a.c. too.

#### rukrazy?

Joined Mar 5, 2005
21
a MONKEY WRENCH MAYBE?
If I connect a large value Capacitor in series with a light bulb and connect that circuit to 120 Volts A.C.
Depending on the value of capacitance used ,I can get the bulb to glow or go to full brightness.
If this is the case, where is the current going if not through the capacitor? It goes through the light bulb right? please dont throw a wrench at me. OW that hurt. Oh! come on, It really doesn't flow through the Cap. or does it??

#### Brandon

Joined Dec 14, 2004
306
In a sense, it flows through the Capcitor as displacement current meaning that the field made by the current is what flows. When the field reaches the other plate, it causes the potetial to shift and it appears as if the current went right through the cap. Since DC does not set up a changing displacement current field, it can't pass through the cap.

What is going to happen with the cap and the lightbulb is a typical RC filter. THe resistance in the bulb + the capacitence will make the signal attenuate depending upon the values of both. The attenuation lowers the voltage coming across the lightbulb so you end up getting less current through the bulb.

Vout = Vin * 1/(1 + j fo/F)

This describes the attenuation factor. Its 1st order since its just 1 cap and 1 res.

fo is the frequency of the filter you have. IN this case, you have

fo=1/(2piRC)

F is the frequency of the input signal. 60Hz in this case.

You just need to measure the resistance of the bulb, or do it math like. 100 Watt bulb = (120 volts)^2 / R
R=144 Ohms.

Vout = Vin / sqrt(1 + (fo/60)^2 ) <-- removed the j term and went to magnitude

So now if you want to get the current, divide Vout/R = Iac