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

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The discussion revolves around solving an RC circuit problem involving a capacitor and two resistors after a switch is opened. The user initially misapplies the charging equation for the capacitor instead of using the correct discharging equation. The correct formula for the voltage across a discharging capacitor is V(t) = V0 * e^(-t/τ), where V0 is the initial voltage. After correcting the equation, the user finds the current through resistor R2 to be approximately 0.000445 A. The key takeaway is the importance of using the appropriate equations for charging versus discharging scenarios in RC circuits.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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