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Physics 2 · 14 min read · Updated 2026-05-11

Steady-State DC Circuits with Resistors and Capacitors — AP Physics 2

AP Physics 2 · AP Physics 2 CED Unit 4: Electric Circuits · 14 min read

1. Core Behavior of a Fully Charged Capacitor ★★☆☆☆ ⏱ 4 min

Steady-state DC means the circuit has been connected to a constant voltage source long enough for all transients (charging/discharging) to stop, so no circuit quantities change over time. The entire analysis of mixed steady-state RC circuits rests on one core rule, derived directly from the definition of electric current.

I_C = 0

Since current is the rate of change of charge, zero current means a fully charged capacitor in steady state acts as an open circuit (infinite effective resistance). This does **not** mean the voltage across the capacitor is zero: the capacitor stores charge on its plates, so it has a non-zero potential difference equal to the difference between the two nodes it connects.

Exam tip: On any steady-state DC problem, first replace all capacitors with open circuits before you start any analysis. This simplifies the circuit immediately and eliminates wrong assumptions about current flow.

2. Equivalent Capacitance for Series and Parallel Networks ★★★☆☆ ⏱ 5 min

When multiple capacitors form a network, you can combine them into a single equivalent capacitor to simplify analysis, just like resistors. However, the equivalent capacitance formulas are reversed from equivalent resistance, which is a very common source of error on exams.

For capacitors in parallel: all capacitors are connected across the same two nodes, so all have the same voltage $V$. Total stored charge is the sum of individual charges, so substituting $Q=CV$ gives:

C_{eq,\text{parallel}} = \sum_{i=1}^n C_i

Intuitively, adding capacitors in parallel increases total plate area, so total capacitance increases. The equivalent capacitance is always larger than the largest individual capacitor in the combination.

For capacitors in series: capacitors are connected end-to-end, so all have the same stored charge $Q$. Total voltage across the combination is the sum of individual voltages, so substituting $V=Q/C$ gives:

\frac{1}{C_{eq,\text{series}}} = \sum_{i=1}^n \frac{1}{C_i}

The equivalent capacitance for a series combination is always smaller than the smallest individual capacitor in the group.

Exam tip: Always use the intuitive consistency check after calculating equivalent capacitance: parallel $C$ > largest individual $C$, series $C$ < smallest individual $C$. This lets you quickly catch formula-swapping errors on MCQs.

3. Full Analysis of Mixed Steady-State RC Circuits ★★★☆☆ ⏱ 5 min

Once you understand the core capacitor rule and equivalent capacitance, full analysis of any mixed steady-state RC circuit follows a straightforward step-by-step process:

  1. Replace all capacitors with open circuits to remove them from the conducting network
  2. Analyze the remaining resistive network using standard tools: Ohm's law, Kirchhoff's rules, equivalent resistance to find currents and node voltages
  3. To find the voltage across any capacitor, calculate the potential difference between the two nodes the capacitor connects
  4. To find stored charge on a capacitor, use $Q = C V_C$. Capacitors always dissipate zero power in steady state, since $I_C = 0$.

Exam tip: Set the negative terminal of the battery as 0 V when calculating node potentials. This eliminates 90% of sign errors when calculating capacitor voltage.

4. Concept Check (AP Style) ★★☆☆☆ ⏱ 2 min

Common Pitfalls

Why: Students confuse steady-state capacitor behavior with fully discharged capacitors at the start of charging, or with inductors in steady DC, which act as short circuits.

Why: Muscle memory from working with resistor networks leads students to automatically apply resistor formulas by mistake.

Why: Students assume all resistors have current, so they calculate a non-zero voltage drop for the resistor connected in series with an open capacitor.

Why: Students confuse Ohm's law for resistors ($V=IR$) with capacitor behavior, incorrectly assuming zero current means zero voltage.

Why: Students use $P=V I$ and plug in $V_C$, forgetting that $I_C$ is zero.

Quick Reference Cheatsheet

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