Final Charge of a Capacitor in a Circuit with Capacitors and Resistors

  • #1
Tofuwu6
4
1
Homework Statement
What is the final charge on the capacitor?
Relevant Equations
Q=CΔV

q(t)=Cε(1 - e^t/RC)
YF-26-72.jpg
I had two trains of thought. One is that the capacitor will fully charge when t = infinity, so when you plug t = infinity into the equation of charge as a function of time you get 1.68E-4, which you also could've gotten from Q = CΔV where ΔV = 42V. My other train of thought was that when t = infinity, R = infinity so all the current will be diverted into the inner loop as the capacitor will serve as a roadblock; this would make the potential difference across the capacitor 0 and the resulting charge also 0.

However, both these answers were wrong, and it keeps telling me to "consider the circuit when the capacitor is fully charged". Can anyone help?
 
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  • #2
Tofuwu6 said:
R = infinity so all the current will be diverted into the inner loop as the capacitor will serve as a roadblock; this would make the potential difference across the capacitor 0 and the resulting charge also 0.
Uh ... you may want to rethink that.
 
  • #3
phinds said:
Uh ... you may want to rethink that.
Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?
 
  • #4
If the capacitor isn't charged at all, the potential difference across the capacitor is zero. When it is fully charged then it has some nonzero potential difference. If you were to 'pluck' the capacitor out of the circuit after a very long time and measure the voltage across it, you would find a nonzero voltage.

Tofuwu6 said:
Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?
Let's say you have a simple circuit with a voltage source of 10V that runs through a resistor of 10 ohms and then through a broken part of wire, creating an open in the circuit. What is the voltage across the open?
 
  • #5
Tofuwu6 said:
Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?
Consider the entire 3 ohm, 4 mu-f leg to be an open circuit (which it IS, since as you point out, the cap is a blockage to current at t = infinity). What is the voltage across that open circuit?
 
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  • #6
phinds said:
Consider the entire 3 ohm, 4 mu-f leg to be an open circuit (which it IS, since as you point out, the cap is a blockage to current at t = infinity). What is the voltage across that open circuit?
Ohhhhhhhhhhhhh, so the current of the circuit after the 3 ohm, 4 mu-f leg is blocked off is 3 because I = ΔV/R = 42/(8+6) = 3 then you find the voltage across the 6 ohm resistor which is 6 * 3 = 18V and since this leg is parallel to the leg with the capacitor it would take the same amount of voltage to hold Kirchoff's voltage loop rule true. Then CΔV = 18(4E-6) = 7.2E-5C. Thanks so much!
 
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  • #7
I know you already solved the problem, but I thought I'd point out the reasons why what you tried didn't work.

Tofuwu6 said:
I had two trains of thought. One is that the capacitor will fully charge when t = infinity, so when you plug t = infinity into the equation of charge as a function of time you get 1.68E-4, which you also could've gotten from Q = CΔV where ΔV = 42V.
The equation you are referring to doesn't apply to the given circuit. It only applies to the basic RC circuit—a battery, a resistor R, and a capacitor C connected in series.

Tofuwu6 said:
Wouldn't the current be 0 and ΔV = IR so ΔV is also 0?
That's the voltage across the resistor, not the capacitor. Because there's no potential drop across the 3-ohm resistor, the voltage across the capacitor is the same as the voltage across the 6-ohm resistor.
 
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  • #8
I mean, the first thing to do here is to have the OP clarify what the question actually is rather than encouraging sloppy questions.

”What is the final charge on the capacitor?” and the attached image is not a complete question. At the very least OP needs to specify if S is open or closed. Ideally OP should quote the full problem statement as required by the homework guidelines.
 
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  • #9
Orodruin said:
I mean, the first thing to do here is to have the OP clarify what the question actually is rather than encouraging sloppy questions.

”What is the final charge on the capacitor?” and the attached image is not a complete question. At the very least OP needs to specify if S is open or closed. Ideally OP should quote the full problem statement as required by the homework guidelines.
My apologies, this was my first time posting anything and this is all the question gave me
Screenshot 2024-03-09 130321.png

above it states "The capacitor in (Figure 1) is initially uncharged. The switch is closed at t=0" which I guess I should have included.
 
  • #10
Tofuwu6 said:
My apologies, this was my first time posting anything and this is all the question gave me View attachment 341529
above it states "The capacitor in (Figure 1) is initially uncharged. The switch is closed at t=0" which I guess I should have included.
Thanks for clearing up that one bit of information.

However, there is no way to read what is posted in that screen shot.

So rather than repeating what @Orodruin posted, I'll just repost it as a quote.

Orodruin said:
I mean, the first thing to do here is to have the OP clarify what the question actually is.
. . .
”What is the final charge on the capacitor?” and the attached image is not a complete question.
. . .
Ideally OP should quote the full problem statement as required by the homework guidelines.
 

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