Another RC circuit problem, don't understand why my answer is wrong

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Homework Help Overview

The problem involves an RC circuit with two resistors and a capacitor, where the switch is closed to reach a steady state before being opened. The parameters include resistances R1 and R2, capacitance C, and an emf from an ideal battery. The original poster is trying to determine the current in resistor R2 at a specific time after the switch is opened.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the current through R2 after the switch is opened, using the charging and discharging equations for the capacitor. They express confusion over their calculated answer being incorrect.
  • Some participants question the validity of the equation used for the capacitor's potential during the discharging phase, suggesting a need to reconsider the approach.
  • There is a discussion about the correct form of the equation for a discharging capacitor, with participants exploring the implications of initial conditions.

Discussion Status

The discussion is ongoing, with participants providing guidance on the correct equations to use for the discharging capacitor scenario. There is a recognition of the need to clarify the initial conditions and the proper mathematical representation for the situation.

Contextual Notes

Participants are addressing potential misunderstandings regarding the behavior of the capacitor when the switch is opened, particularly focusing on the transition from charging to discharging states. The original poster's calculations are based on assumptions that are being questioned and refined.

iiiiaann
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Homework Statement


In the figure below, R1 = 13.0 kΩ, R2 = 18.0 kΩ, C = 0.600 µF, and the ideal battery has emf = 20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

27-55.gif


Homework Equations



V = IR

V_{Capacitor}(t) = V_{0}(1-e^{-t/\tau})

The Attempt at a Solution



Okay so i put a decent amount into this one but the answer I'm coming up with is wrong and I can't figure out why. Here's what I've done:

When the capacitor is fully charged, the current through the battery is \frac{\xi}{R_{1} + R_{2}} which is \frac{20}{13 * 10^{3} + 18 * 10^{3}} = 6.45e-4

The current through R2 would be the same as this, 6.45e-4, so the potential difference over R2 = IR = 6.45e-4 * 18e3 = 11.61 V

When the switch is first opened, the potential difference across the capacitor would be the same as this, 11.61.

The time constant for the capacitor would be RC = 18e3 * .6e-6 = 0.0108

so to calculate the potential difference at t = 4 ms, you would use the following equation:
V_{capacitor}(0.004) = 11.61 * (1 - e^{-0.004 / 0.0108}) which comes out to 3.59

This is the same as the potential difference in R2, and so current through R2 should be 3.59 / 18e3 = 1.99e-4, but that is not right. Can someone let me know where I went wrong?
 
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The potential across the capacitor will be falling, not rising. Check your equation for the potential on the capacitor!
 
This equation

\displaystyle V_\text{Capacitor}(t) = V_{0}(1-e^{-t/\tau})

is for the case of charging a capacitor which initially has a charge (and thus potential) of zero. How can I tell ? Look at V when t = 0 .

What's the correct equation for discharging a capacitor which initially has a potential of V0 across its plates?
 
is it just the same equation except without the "1 -" part? so V0(e^-t/TAU)?
 
got it, 0.000445358666. Thanks for putting me on the right track guys
 

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