Is 2V the same as Vcc in this circuit?

[-EquinoX-]

Homework Statement

[PLAIN]http://img130.imageshack.us/img130/7782/77019121.jpg

Homework Equations

The Attempt at a Solution

The answer is 2V, is it because that vc is basically the same as Vcc?

 

[Feldoh]

Think about it. Steady state means the long-term behavior of the circuit.

What will physically happen with the capacitor after a long time if we supply a DC voltage?

 

[-EquinoX-]

The capacitor will be charged after a long time with the same voltage as the source?

 

[Feldoh]

You are correct.

But perhaps the more interesting thing to think about is this:

Kirchoff's Voltage Rule states that the sum of the voltages across all the components in a circuit adds up to zero.

So we have a +2V from the DC power supply and -2V from the capacitor (since they are in opposite directions following the circuit around in a loop). The sum equals 0 which means that the resistor has a voltage drop of 0. Why is there no voltage drop across a resistor in the steady state of that circuit?

 

[-EquinoX-]

That I can't answer...

 

[Feldoh]

Well from PF's wiki: "A capacitor (or condenser) is something across which charge cannot move, but across which an accumulation of charge on one side can affect charge on the opposite side."

In other words a capacitor is a set of two conductors that are not physically touching. What if we attach a DC voltage source to a capacitor? Since a capacitor is a break in a circuit there should be no current right? WRONG.

-------------| |-----------------

The voltage source will "push" charge in a certain direction right? Well when that charge reaches one of the conductors of a capacitor it can't physically move any further since there is physically nothing attached to the conductor.

-------------|(+charge buildup) |-----------------However the other conductor is made up of electrons and protons, which are just in equal amounts so initially the total charge is zero. But since one side of the capacitor is building up a (call it positive) charge the other side of the capacitor wants to get rid of it's positive charge since positive charges repel each other.

-------------|(+charge buildup) (-charge buildup)|----------------- (+charge moves away)

So this is how a current can exist with a capacitor even though a capacitor is a physical break in a circuit.

But there is a catch. The capacitor's individual sides only have a certain number of charges to give away. When the sides have given up as much charge as possible the capacitors are fully charged

-------------|(all + charge here) (all - charge here)|-----------------

So there are no more like charges to repel each other and the current stops for steady state. Does that make sense?