# Graph of RC circuit

## Homework Statement

Sketch a graph of Vout vs time after the switch is closed
*See attached diagram*

## The Attempt at a Solution

I am not really sure how to do this, because I don't know if it makes a difference whether or not the resistor is before the switch or not. However, if i were to make a guess, Since the switch is closed in an RC circuit, that would mean that it would charge exponentially, and once it is fully charged, it does not have a voltage drop across it, and thus, the entire voltage drop would be across the resistor, so my graph would start out at 12V for time=0, and decrease exponentially to zero at time t when then entire voltage drop is across the resistor.

Am I right? Any help would be great :)

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berkeman
Mentor

## Homework Statement

Sketch a graph of Vout vs time after the switch is closed
*See attached diagram*

## The Attempt at a Solution

I am not really sure how to do this, because I don't know if it makes a difference whether or not the resistor is before the switch or not. However, if i were to make a guess, Since the switch is closed in an RC circuit, that would mean that it would charge exponentially, and once it is fully charged, it does not have a voltage drop across it, and thus, the entire voltage drop would be across the resistor, so my graph would start out at 12V for time=0, and decrease exponentially to zero at time t when then entire voltage drop is across the resistor.

Am I right? Any help would be great :)
You are part right. However, the only way a resistor has a voltage drop across it is when there is current flowing through it, as in V = IR.

Well when the capacitor is full there is no current running through it. I guess i don't really know what you're getting at

berkeman
Mentor
Well when the capacitor is full there is no current running through it. I guess i don't really know what you're getting at
It's all one loop in that circuit, right? So the current is the same everywhere. So when the cap is all charged up and there is no current running through it anymore, what is the current in the resistor? And what is the resulting voltage drop across the resistor in this condition?

gneill
Mentor
Well when the capacitor is full there is no current running through it. I guess i don't really know what you're getting at
When it's "full", what's it full of?

So then the entire voltage drop would be across the capacitor? Cause then there would be no current thus no voltage drop across the resistor? I don't understand how the capacitor creates a voltage drop though...

berkeman
Mentor
So then the entire voltage drop would be across the capacitor? Cause then there would be no current thus no voltage drop across the resistor? I don't understand how the capacitor creates a voltage drop though...
Yes.

The capacitor does not "create" a voltage drop. It has a voltage across it when it is charged up (your "full" reference).

Q = C * V

Charge (on the cap) is equal to the capacitance multiplied by the voltage.

ahhh it makes sense!

Thanks so much for your help!