Potential across a capacitor in an RC circuit

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SUMMARY

The discussion focuses on the behavior of a capacitor in an RC circuit after a long time has passed since the circuit was connected. It establishes that once the capacitor reaches steady-state, the potential across its terminals remains constant, indicating that no current flows through the capacitor. The relevant equations discussed include Kirchhoff's rules, Ohm's Law (V = IR), and the exponential decay function for voltage across a capacitor (V(t) = V_{0} e^{-t/\tau}). The confusion regarding current flow in resistors connected in parallel with the capacitor is clarified, emphasizing that the capacitor does not short circuit the resistors in steady-state.

PREREQUISITES
  • Understanding of RC circuit dynamics
  • Familiarity with Kirchhoff's rules
  • Knowledge of Ohm's Law (V = IR)
  • Concept of exponential decay in capacitors (V(t) = V_{0} e^{-t/\tau})
NEXT STEPS
  • Study the behavior of capacitors in transient analysis
  • Learn about steady-state conditions in RC circuits
  • Explore the implications of Kirchhoff's rules in complex circuits
  • Investigate the mathematical modeling of capacitor charging and discharging
USEFUL FOR

Students studying electrical engineering, educators teaching circuit analysis, and anyone interested in understanding the behavior of capacitors in RC circuits.

Elbobo
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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:
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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
 

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