# Charging a parallel plate capacitor

1. Apr 25, 2006

### al_201314

Hi guys,

I've been reading the forums and finally I have something to post

I've got a question regarding charging parallel plate capacitors. Take this for example:

Connect a 6 Volt power supply to a resistor of 2Mohm and in series with a 15uF capacitor initially uncharged, with a current of 0.0000018A flowing through.

A simple calculation would show that the P.D across the capacitor is 2.4V. The question then goes on to ask what is the new p.d across the capacitor when it is fully charged and reduced to 5uF without any loss of charges and the answer is 16.8V

I figured in order to get the answer the charges in the capacitor must be that of 6V x 15uF. My question is that how then would the capacitor gets charged up to the maximum of 6V (how would one know the max is 6V?) with the presence of the resistor?

I can't really grasp the concept here hope someone can help me out with the explanation and sorry for being so long winded cant find an easier way to put it across!

Thanks

Last edited: Apr 25, 2006
2. Apr 25, 2006

### Hammie

V(capacitor) = V(source, 6 volts in this instance)*(1-e^-(t/rc)).

Theoretically, the potential difference accross the capacitor will never reach the source voltage, but after a very very long time, it will be close enough that you won't be able to measure the difference anyway.

In electronics, after five or so time constants (r*c) they call it good and say the capacitor is fully charged. Plugging in the numbers, at that time the capacitor is charged up to 99 percent of the source voltage. For this particular circuit the time required would be somewhat less than three minutes.

Leave it hooked up for a day, and the difference between the source voltage and that accross the capacitor is something to the order of 10^-54 or so.

Last edited: Apr 25, 2006
3. Apr 25, 2006

### al_201314

Thanks for the reply Hammie... pardon me but could you elaborate on how the capacitor could charge up close to the source voltage when in this case there's a resistor in series leaving a p.d of only 2.4V across the capacitor?

Thanks!

4. Apr 25, 2006

### Staff: Mentor

The current through the resistor is maximum initially, when the capacitor is uncharged and the voltage across the resistor is maximum. The current decays exponentially as the voltage on the cap increases with a decaying exponential shape. The final current through the resistor after a long time is zero, and the voltage on the capacitor equals the source voltage.

Remember that the equation for the current into a capacitor that is charging is I = C dV/dt. The higher the current, the faster the capacitor charges. Solve this equation to see the exponential equation for the capacitor voltage versus time when charged through a series resistor.

5. Apr 25, 2006

### Hammie

You are correct, the voltage accross the capacitor is 2.4 volts, but at a particular time. The voltage will approach six volts, at a much greater time than this.

The question then goes on:

before the capacitor is reduced to 5 uf, its potential difference is (or is as good as..) six volts, it is said to be fully charged.

This may be more than you want to know about charging capacitors, but look here..

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/RCSeries.html [Broken]

Last edited by a moderator: May 2, 2017
6. Apr 27, 2006

### al_201314

Thanks for the responses much appreciated!