RC Circuits, time to charge a capacitor

AI Thread Summary
The discussion revolves around calculating the time it takes for the electric potential across a capacitor to equal that across a resistor in an RC circuit. The relevant equations include q = CE(1-e^-t/RC) and V = Vmax(1 - e^-t/RC), where E is the EMF of the battery. The solution involves setting the voltage across the capacitor to E/2 and solving for time t, leading to the result of approximately 0.216 milliseconds. Participants clarify the relationship between the capacitor's charge, voltage, and the total voltage supplied by the battery, referencing Kirchhoff's law. The conversation emphasizes understanding the exponential nature of charging in RC circuits.
iiiiaann
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Homework Statement


Switch S in the figure below is closed at time t = 0, to begin charging an initially uncharged capacitor of capacitance C = 13.0 µF through a resistor of resistance R = 24.0 . At what time is the electric potential across the capacitor equal to that across the resistor?

hrw7_27-52.gif



Homework Equations



i = dq/dt
q = CE(1-e^-t/RC)

The Attempt at a Solution



I really don't even know where to start with this one
 
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I assume that E is the EMF of the battery.

Is q the charge on the capacitor at time, t ?

How much charge would be on this capacitor of the potential difference across it was E ?
 
the charge on the capacitor would be

Q = CV = (13e-6)(E)

if the potential difference were E
 
iiiiaann said:
... At what time is the electric potential across the capacitor equal to that across the resistor? ...

At the above time, how do these two potentials compare with E , the electric potential provided by the battery?
 
SammyS said:
At the above time, how do these two potentials compare with E , the electric potential provided by the battery?

I'm not sure i follow what you are saying? Would the combination of the 2 be equal to E?
 
iiiiaann said:
I'm not sure i follow what you are saying? Would the combination of the 2 be equal to E?
Yes, according to Kirchhoff.

Since the two are equal, what is the electric potential across the capacitor ?
 
Would it just be 1/2 E?
 
Last edited:
where do i go from here?
 
If the electric potential on the capacitor is E/2, then how much charge is on the capacitor?
 
  • #10
the charge is Q = CV which would be 13uF * E / 2
 
  • #11
So, solve this equation for t when q = C(E/2)

q = CE(1-e^-t/RC)
 
  • #12
C(E/2) = CE(1-e^-t/RC)
1/2 = 1 - e^-t/RC
e^-t/RC = 1/2
-t/RC = ln(1/2)
-t = ln(1/2) * RC
t = -1 * ln(1/2) * RC

t = -1 * -.6931 * 24 * 13e-6 = 0.000216 s = 0.216 ms

Thanks again for the help, these forums (you especially) are fantastic
 
  • #13
Hello, may I ask what kind of equation q = CE(1-e^-t/RC) is? Does the problem give it as the equation which determines the time to charge of the capacitor through this RC circuit? How do they get to such an equation?
 
  • #14
If you check your capacitor equations you will find a more convenient expression for the voltage across a capacitor during charging.
V = Vmax(1 - e^-t/RC) so you can calculate the voltage across the capacitor t sec after switch on.
The charge equation is the same exponential form
Q = Qmax(1-e^-t/RC)
hope this helps
 
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