Graph Vout vs Time for RC Circuit Switch Closure

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Homework Help Overview

The discussion revolves around sketching a graph of Vout versus time for an RC circuit after a switch closure. Participants are exploring the behavior of voltage across the capacitor and resistor in the circuit setup.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the impact of the resistor's position relative to the switch and the implications for voltage behavior over time. There is uncertainty about the relationship between current flow, voltage drops, and the state of the capacitor.

Discussion Status

Some participants are questioning the assumptions regarding current flow when the capacitor is fully charged and its effect on the voltage across the resistor. Guidance has been offered regarding the nature of voltage drops in relation to current flow and the charging of the capacitor.

Contextual Notes

There is a lack of clarity regarding the definitions of "full" in relation to the capacitor and the conditions under which voltage drops occur across circuit components. Participants are navigating these concepts without a definitive resolution.

dancergirlie
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Homework Statement



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


Homework Equations





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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dancergirlie said:

Homework Statement



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


Homework Equations





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.

Try again with your answer?
 
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
 
dancergirlie said:
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?
 
dancergirlie said:
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...
 
dancergirlie said:
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!
 

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