# Questions on Capacitor

Discussion in 'Homework Help' started by zener_16, Feb 20, 2010.

1. ### zener_16 Thread Starter New Member

Feb 1, 2010
9
0
I am not very sure how a capacitor works in a dc circuit.

#1: A circuit consists of a dc supply and a switch and a capacitor.
The capacitor is initially uncharged. When the switch is closed, the source starts to charge the capacitor. Initially the charging current is at maximum because the capacitor momentarily acts as a short circuit.

Qn1) Why is the capacitor's reactance=0 (which make it short circuit) in the first place? What makes it 0?

Ty

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
See if this material on capacitors in the AAC ebook can help you get a better understanding of the behavior of capacitors.

hgmjr

3. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Do not think in terms of reactance. Reactance is a term people invented to deal with AC circuits. Initially, the voltage across the capacitor is zero.

Thus, the current through the circuit is:

i=(Vbattery-Vcapacitor)/Rwires

As time passes, Vcapacitor increases and thus i reduces because Vbatery and Rwires are constant.

It is similar to a water tank with large capacity which fills a small capacity water tank with a hose connected at the base of the tanks. Initially, the small tank is empty and the pressure difference (voltage difference) is high. Thus, the amount of water flowing per second (current) is high. As the water level in the small tank rises the pressure difference reduces and so does the amount of water per minute.

4. ### zener_16 Thread Starter New Member

Feb 1, 2010
9
0
Then what makes the voltage of the capacitor to be 0 in the first place?

5. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
It is so stated in the initial conditions.

6. ### mik3 Senior Member

Feb 4, 2008
4,846
63
The charge difference on the plates of the capacitor is initially zero (i.e the plates of the capacitor have the same amount and kind -negative or positive- of charge). Therefore, the voltage across the capacitor is zero, no charge difference - no electric field. When a voltage is applied, electrons flow from the one plate to the other. Thus, the plate which loses electrons is charged positively (less electrons than protons) and the plate which gains electrons is chraged negatively (more electrons than protons). An electric field is created with this charge difference and hence a voltage difference. When this voltage difference equals the voltage of the battery electrons stop to flow and the capacitor is fully charged.
The amount of charge in each plate needed to develop this voltage difference depends on the area of the plates, the length between them and the permitivity of the dielectric material between them. This is the reason larger capacitance requires more time to charge with the same series resistance. More electrons are required as to build up the needed charge and thus more time.

7. ### zener_16 Thread Starter New Member

Feb 1, 2010
9
0
ty guys. haha finalli got my concept right.