Potential across a capacitor in an RC circuit

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 5K views
Elbobo
Messages
145
Reaction score
0

Homework Statement


http://img14.imageshack.us/img14/4822/2116b.png

The circuit has been connected as shown in the figure for a “long” time.
What is the magnitude of the electric potential across the capacitor?
Answer in units of V.

Homework Equations


Kirchoff's rules
[tex]V = IR[/tex]
[tex]V(t) = V_{0} e^{-t/\tau}[/tex]
[tex]C = \frac{Q}{C}[/tex]
[tex]Q(t) = Q_{f}(1-e^{-t/\tau)[/tex]


The Attempt at a Solution


http://img215.imageshack.us/img215/7758/2116.png

That's how I redrew the circuit. But then I got confused. Does the capacitor short circuit the rest of the circuit (i.e., does no current flow in the 15 ohm and 48 ohm resistors)?

If not, I don't get what happens to (I - i1) and i1 once they get to the junction with the capacitor.

Help please?
 
Last edited by a moderator:
Physics news on Phys.org
Elbobo said:
Does the capacitor short circuit the rest of the circuit (i.e., does no current flow in the 15 ohm and 48 ohm resistors)?

No. The capacitor might act as a short the very instant the switch is closed, if the capacitor's initial charge was 0. But then when the capacitor gains charge, the potential across its terminals changes until it reaches steady-state. Once the charge in the capacitor reaches steady-state, the potential across its terminals does not change either.

If not, I don't get what happens to (I - i1) and i1 once they get to the junction with the capacitor.

Without giving you the answer, here is a big hint. If the potential (i.e. voltage) across a capacitor's terminals does not change, meaning the capacitor's charge is not changing either, what does that tell you about the current flowing through the capacitor? :wink:
 
Ah! Thank you, I get it :D