# What happens in the RC circuit when being charged?

Say I have this RC circuit and at the start, the switch is off (current going from left to right and the starting point is just above the battery). I understand that then the capacitor acts just like a metal wire as if it isn't there. But when you turn on the switch, then I was taught that the current goes to 0. And since there is no current flowing in the resistor there is no voltage drop in the resistor. My questions are

1. why when you turn on the circuit, does the current go to 0?

2. and why does the capacitor suck in all the voltage drop from the resistor?

3. Also how do charges build on up the capacitor?

4. In my textbook, the positive charges always build up on the upper plate of the capacitor. why is it so? Is this some kind of a rule?

I generally have no idea how the RC circuit works and why the current suddenly goes to 0 when the switch is on please help!

any help is appreciated.

• You ask about what happens when "the switch is off", but your schematic doesn't have any switch in it. – The Photon Apr 17 '16 at 16:33

First of all, you should see some properties of a capacitor. A capacitor is a device capable of storing electrical energy. The capacitor can induce an impedance depending on the capacitance of the capacitor and the frequency of input voltage applied. It's called capacitive reactance and is given by:

$X_C$ = $1/2πfC$

where f is the frequency of input voltage and C is the capacitance of the capacitor.

From the equation, it is clear that the reactance offered by the capacitor decreases as the frequency of the input voltage increases.

You have a DC voltage source. A DC voltage source is characterized by a constant (not time-varying) voltage. That means a DC voltage has zero frequency. From the equation

$X_C$ = $1/2πfC$

the reactance offered by the capacitor to a DC voltage is infinitely high (f=0).

So the capacitor behaves as like an open switch offering infinite resistance. To have a current flow, the circuit should be closed. But the presence of capacitor makes the circuit open. So no current flows through the circuit.

Hence the entire voltage drop occurs across the capacitor. (The applied voltage has to come across capacitor because the energy is not utilized by the circuit to make current flow, but to charge the capacitor)

Current flows from the positive terminal of battery. So positive charges are developed at the capacitor plate connected to the positive terminal of the battery and negative charges develop at the plate connected to the negative of the battery. There is nothing special about upper or lower plate.

This is what happens in an RC circuit. (The assumptions are all based on the fact that the capacitor is not initially charged)

• AC analysis (impedance, reactance, etc.) assumes AC steady state operation. This problem is for a switched DC circuit (according to the text) so I find it puzzling that your answer begins with a few paragraphs on capacitive reactance. For an unswitched DC circuit, all we need is that the current through a capacitor is proportional to the rate of change of voltage across and, since the voltages are constant in a DC circuit, there is zero rate of change and thus, zero current through. – Alfred Centauri Apr 19 '16 at 13:34
• I just thought I could start by talking the reactance offered by a capacitance to a dc signal is infinitely high and will act as an open switch. When the circuit is open, there will be no current flow. To me this statement seems to explain the rest of questions. – UKH Apr 19 '16 at 15:19