I have a question for 2. a. Why is voltage across C0 equal to V0? I understand how this be the case if the resistor R1 was not there (since voltages are the same for components connected in parallel); however, since the resistor is connected to the capacitor in series, wouldn't the voltage V0 be split across C0 and R1 (which means C0 would have a voltage less than V0)?

Also, for 2. b - I originally thought that I2 would increase, because when the switch is first closed, the capacitor acts like a wire and allows current to flow freely, so the current will be split between both the R1 and R2 parallel branches. Then, after a long time, the capacitor acts like an open switch, so no current will flow through R1 and it will all flow through R2 - therefore, the current in R2 would increase. After watching this video, I understand Mr. Fullerton's reasoning as to why I2 remains constant, but I'm not sure what was wrong with my original thinking.

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## kd2eom

I just watched this APlusPhysics video: https://www.youtube.com/watch?v=1wAwkyzxB0k

I have a question for 2. a. Why is voltage across C0 equal to V0? I understand how this be the case if the resistor R1 was not there (since voltages are the same for components connected in parallel); however, since the resistor is connected to the capacitor in series, wouldn't the voltage V0 be split across C0 and R1 (which means C0 would have a voltage less than V0)?

Also, for 2. b - I originally thought that I2 would increase, because when the switch is first closed, the capacitor acts like a wire and allows current to flow freely, so the current will be split between both the R1 and R2 parallel branches. Then, after a long time, the capacitor acts like an open switch, so no current will flow through R1 and it will all flow through R2 - therefore, the current in R2 would increase. After watching this video, I understand Mr. Fullerton's reasoning as to why I2 remains constant, but I'm not sure what was wrong with my original thinking.

Thank you for any help!

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