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

 

See if this material on capacitors in the AAC ebook can help you get a better understanding of the behavior of capacitors.

hgmjr

 

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.

 

Then what makes the voltage of the capacitor to be 0 in the first place?

 

The capacitor is initially uncharged.

It is so stated in the initial conditions.

 

Then what makes the voltage of the capacitor to be 0 in the first place?

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.

 

ty guys. haha finalli got my concept right.