Graph Vout vs Time for RC Circuit Switch Closure

AI Thread Summary
The discussion focuses on understanding the behavior of an RC circuit after a switch is closed. The initial assumption is that the output voltage (Vout) starts at 12V and decreases exponentially to zero as the capacitor charges. Clarification is provided that the voltage drop across the resistor occurs only when current flows through it, which ceases once the capacitor is fully charged. It is emphasized that when the capacitor is fully charged, the voltage drop is across the capacitor, not the resistor. The conversation concludes with a better understanding of how voltage and charge relate in the circuit.
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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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