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How do capacitors combine in circuits, and how does an RC circuit charge and discharge over time?

Topic 11.8 Capacitors in Circuits and RC Circuits: combine capacitors in series and parallel and describe charging and discharging through a resistor.

A focused answer to AP Physics 2 Topic 11.8, covering capacitors in series and parallel, the equivalent capacitance rules, the behavior of a capacitor in a circuit at the first instant and after a long time, and the charging and discharging of an RC circuit with its time constant, with full worked examples.

Generated by Claude Opus 4.812 min answer

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  1. What this topic is asking
  2. Combining capacitors
  3. A capacitor in a circuit: first instant and long time
  4. RC circuits and the time constant
  5. Try this

What this topic is asking

The College Board (Topic 11.8) wants you to combine capacitors in series and parallel, describe how a capacitor behaves in a circuit at the first instant and after a long time, and analyze the charging and discharging of an RC circuit using the time constant.

Combining capacitors

The capacitor combination rules are the mirror image of the resistor rules, which is a frequent source of error. The reasoning is in the shared quantity: parallel capacitors share the voltage, so their stored charges (each Q=CVQ = CV) simply add, making a larger effective capacitor; series capacitors share the charge, so their voltages add and the combination stores less charge per volt. A useful check: parallel adds (bigger), series gives less than the smallest.

A capacitor in a circuit: first instant and long time

These two limiting cases answer most capacitor-circuit questions without any time-dependent math. When a switch first closes, the empty capacitor offers no opposition (no charge means no voltage), so it is effectively a wire and the current is largest. As it fills, the voltage across it grows and opposes the source, choking off the current until, fully charged, it blocks the branch entirely like an open switch. Reasoning with these limits is a core exam skill.

RC circuits and the time constant

Between the two limits, the charge builds up (or drains away) exponentially over time, governed by the time constant τ=RC\tau = RC. In charging, the charge rises toward its final value; after one time constant it has reached about 63%63\%, and after about five it is essentially complete. In discharging, the charge falls to about 37%37\% of its value after one time constant. A larger resistance or capacitance lengthens τ\tau, slowing the process, because more charge must move (larger CC) or it moves more slowly (larger RR). The strategic role of this topic is that it adds the dimension of time to circuits: capacitors do not respond instantly but smooth and delay changes, which is why RC circuits are used for timing and filtering. Combined with the capacitor energy 12CV2\tfrac{1}{2}CV^2 from Topic 10.6 and the resistor rules of Topic 11.5, this completes the picture of how charge, energy and time interact in a circuit.

Try this

Q1. Find the equivalent capacitance of two 6.06.0 microfarad capacitors in parallel. [1 point]

  • Cue. Cp=6.0+6.0=12C_p = 6.0 + 6.0 = 12 microfarads (parallel capacitances add).

Q2. State how a fully charged capacitor behaves in a DC circuit after a long time. [1 point]

  • Cue. Like an open switch: no current flows through that branch.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). A 1212 V battery charges a 4.04.0 microfarad capacitor through a 2.0×1052.0 \times 10^5-ohm resistor. (a) State the current through the capacitor at the first instant the switch closes and after a long time, with justification. (b) Calculate the time constant of the circuit. (c) Calculate the final charge on the capacitor. (d) If instead two 4.04.0 microfarad capacitors were placed in parallel, calculate their equivalent capacitance.
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A 7-point FRQ on an RC circuit.

(a) Currents (2 points): at the first instant the uncharged capacitor acts like a wire (no voltage across it), so the current is a maximum, I0=E/R=12/(2.0×105)=6.0×10−5I_0 = \mathcal{E}/R = 12/(2.0 \times 10^5) = 6.0 \times 10^{-5} A. After a long time the capacitor is fully charged, no more charge flows, so the current is zero.
(b) Time constant (2 points): τ=RC=(2.0×105)(4.0×10−6)=0.80\tau = RC = (2.0 \times 10^5)(4.0 \times 10^{-6}) = 0.80 s.
(c) Final charge (2 points): Q=C E=(4.0×10−6)(12)=4.8×10−5Q = C\,\mathcal{E} = (4.0 \times 10^{-6})(12) = 4.8 \times 10^{-5} C.
(d) Parallel (1 point): capacitors in parallel add, Cp=4.0+4.0=8.0C_p = 4.0 + 4.0 = 8.0 microfarads.

Markers reward the initial and final currents, the time constant, the final charge, and the parallel sum.

AP 2023 (style)1 marksSection I (multiple choice). Two capacitors are connected in series. How does their equivalent capacitance compare with either one alone? (A) larger than both (B) smaller than both (C) equal to their sum (D) equal to the larger one. Justify your reasoning.
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A 1-point MCQ on series capacitance. The answer is (B).

Capacitors in series combine through reciprocals, 1Cs=1C1+1C2\dfrac{1}{C_s} = \dfrac{1}{C_1} + \dfrac{1}{C_2}, so the equivalent capacitance is smaller than the smallest one (the opposite of resistors). The trap is (C): series capacitors do not add; that is the parallel rule.

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