(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

V = IR

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

3. 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 [itex]\frac{\xi}{R_{1} + R_{2}}[/itex] which is [itex]\frac{20}{13 * 10^{3} + 18 * 10^{3}}[/itex] = 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:

[itex]V_{capacitor}(0.004) = 11.61 * (1 - e^{-0.004 / 0.0108})[/itex] 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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# Homework Help: Another RC circuit problem, don't understand why my answer is wrong

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