What is the Relationship Between Voltage and Capacitors in a Series Circuit?

AI Thread Summary
In a series circuit with a capacitor and resistor, the voltage across the capacitor (C0) equals the source voltage (V0) when the circuit reaches steady state, which occurs after a sufficient time (typically around 5 to 6 time constants, RC). Initially, while the capacitor is charging, the voltage is not equal to V0, as it is split between the capacitor and the resistor. The charge on the capacitor increases over time according to the equation Q = CV(1-e^-t/RC), indicating that at t=0, the charge is zero. Understanding that steady state means the current through the capacitor is zero clarifies why C0 eventually equals V0. This concept is crucial for analyzing capacitor behavior in circuits.
kd2amc
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Homework Statement


I was going through this PDF on my own to review for my AP Physics C E&M exam: https://apcentral.collegeboard.org/...course=ap-physics-c-electricity-and-magnetism

I was also watching this video (which discusses the solutions):

The part I am stuck on is part 2(a) (it is hard for me to copy and paste the contents of the PDF into here).

(The solution to part 2(a) is at about 16 minutes into the video).

Homework Equations


Q=CV

The Attempt at a Solution


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, I thought that the voltage V0 would be split across C0 and R1 (which means C0 would have a voltage less than V0). Why is this not the case?

Thank you so much for any help!
 
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kd2amc said:
I thought that the voltage V0 would be split across C0 and R1 (which means C0 would have a voltage less than V0).
Yes, that will be the case when the capacitor is 'charging' i.e. the charge on the capacitor will be increasing with time. You need to find the equation for charge as a function of time.
 
cnh1995 said:
Yes, that will be the case when the capacitor is 'charging' i.e. the charge on the capacitor will be increasing with time. You need to find the equation for charge as a function of time.

Thank you for your help. According to my notes, the charge as a function of time is Q = CV(1-e^-t/RC). I'm not sure how to use this equation to conclude that the voltage across C0 = V0, though. Doesn't this equation actually imply that Q = 0 at time t=0?
 
kd2amc said:
Doesn't this equation actually imply that Q = 0 at time t=0?
Right.
kd2amc said:
that the voltage across C0 = V0
You get this condition at t=∞(mathematically) and that represents the steady state. In steady state, the current through the capacitor is zero, meaning that it is fully charged.

Practically, you don't need infinite time to reach steady state. It is assumed that the steady state is reached at t=5RC (or 6RC)(where RC is the time constant) since the capacitor will have more than 99% of its steady state voltage by that time.
 
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Oh! I see now! Thank you!
 
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