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

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SUMMARY

The relationship between voltage and capacitors in a series circuit is clarified through the equation Q = CV, where Q represents charge, C is capacitance, and V is voltage. In steady state, which is reached approximately at t=5RC, the voltage across the capacitor C0 equals the source voltage V0, as the capacitor becomes fully charged and the current through it becomes zero. This understanding is crucial for solving problems related to capacitor charging in series circuits, particularly in the context of AP Physics C E&M exams.

PREREQUISITES
  • Understanding of basic circuit components: resistors and capacitors
  • Familiarity with the concept of voltage and charge in electrical circuits
  • Knowledge of the time constant in RC circuits
  • Ability to interpret and apply the equation Q = CV
NEXT STEPS
  • Study the concept of time constants in RC circuits
  • Learn about the charging and discharging equations for capacitors
  • Explore the implications of steady state in electrical circuits
  • Review AP Physics C E&M exam problems related to capacitors in series
USEFUL FOR

Students preparing for the AP Physics C E&M exam, educators teaching circuit theory, and anyone interested in understanding the behavior of capacitors in series 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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