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

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Behavior of fully charged capacitors in steady-state DC, equivalent capacitance for series/parallel networks, Kirchhoff’s rules applied to mixed RC circuits, and voltage/current/power analysis of steady mixed resistor-capacitor DC circuits.

You should already know: Ohm's law for ohmic resistors. Kirchhoff's junction and loop rules for DC circuits. Equivalent resistance calculations for series and parallel resistor networks.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Steady-State DC Circuits with Resistors and Capacitors?

Steady-state DC means the circuit has been connected to a constant voltage source for a sufficiently long time that all transient changes (the charging or discharging of capacitors) have stopped, and no quantities in the circuit change over time. Per the AP Physics 2 Course and Exam Description (CED), this topic makes up approximately 1-2% of total exam weight, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) questions, most often as a conceptual check or the first step of a longer circuit analysis problem.

Standard notation for this topic: $C$ for capacitance (units: farads, F), $V_C$ for voltage across a capacitor, $I_C$ for current through a capacitor, $R$ for resistance (ohms, $\Omega$), $\epsilon$ for source emf, and $Q$ for stored charge on a capacitor. Unlike transient RC circuits (which focus on changing charge over time), this topic exclusively considers long-term behavior after capacitors have reached their maximum stored charge. Most exam questions test whether you correctly apply the unique rules for steady-state capacitors instead of incorrectly treating them like resistors or transient capacitors.

2. Core Behavior of a Fully Charged Capacitor in Steady-State DC

The entire analysis of steady-state mixed RC circuits rests on one simple core rule, derived directly from the definition of steady state. In steady state, the charge on a capacitor’s plates is no longer changing: $dQ/dt=0$. By definition, electric current is the rate of flow of charge, so $I_C=dQ/dt=0$. This gives us the core rule: $$I_C=0$$ A capacitor with zero current acts as an open circuit: it is equivalent to a break in the wire at its location, with 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 potential difference between the two nodes it connects.

Intuition: When you first connect a DC source to an uncharged capacitor, current flows to add charge to the plates, increasing $V_C$ until $V_C$ matches the potential difference between the two nodes the capacitor is connected to. At that point, there is no longer a driving force to add more charge, so current stops, and we have reached steady state.

Worked Example

A 9 V DC battery is connected in a single series loop with a 1.5 $\Omega$ resistor, a 100 $\mu$F capacitor, and a 3 $\Omega$ resistor. What is the current through the 3 $\Omega$ resistor in steady state?

1. Apply the core steady-state rule: current through the capacitor is $I_C=0$.
2. For a series loop, the same current flows through every component in the loop.
3. If current through the capacitor is zero, the current through all other components (including the 3 $\Omega$ resistor) must also be zero.
4. The current through the 3 $\Omega$ resistor is 0 A.

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.

3. Equivalent Capacitance for Series and Parallel Capacitor Networks

When multiple capacitors are connected together in a network, you can combine them into a single equivalent capacitor to simplify analysis, just like you do for resistors. However, the formulas for equivalent capacitance are reversed from equivalent resistance, which is a common source of error.

For capacitors in parallel (all connected across the same two nodes, so all have the same voltage $V$), total stored charge is the sum of charge on each individual capacitor: $Q_{eq}=Q_1+Q_2+...+Q_n$. Substituting $Q=CV$ gives: $$C_{eq,\text{parallel}}=\sum _{i=1}^n{C}_i$$ This makes intuitive sense: adding capacitors in parallel increases total plate area, so total capacitance increases, and the equivalent capacitance is always larger than the largest individual capacitor in the combination.

For capacitors in series (connected end-to-end, so all have the same stored charge $Q$), total voltage across the combination is the sum of individual voltages: $V_{eq}=V_1+V_2+...+V_n$. Substituting $V=Q/C$ gives: $$\frac{1}{C_{eq,\text{series}}}=\sum _{i=1}^n\frac{1}{C_i}$$ Here, the equivalent capacitance is always smaller than the smallest individual capacitor in the series combination.

Worked Example

Find the equivalent capacitance of a network with a 4 $\mu$F capacitor, a 12 $\mu$F capacitor, and a 6 $\mu$F capacitor, where the 4 $\mu$F and 12 $\mu$F are in parallel with each other, and that parallel combination is in series with the 6 $\mu$F capacitor.

1. First combine the parallel capacitors: $C_{parallel}=C_1+C_2=4\mu \text{F}+12\mu \text{F}=16\mu \text{F}$.
2. Now combine this 16 $\mu$F in series with the 6 $\mu$F capacitor: $\frac{1}{C_{eq}}=\frac{1}{16\mu \text{F}}+\frac{1}{6\mu \text{F}}=\frac{3+8}{48\mu \text{F}}=\frac{11}{48\mu \text{F}}$.
3. Invert to solve for $C_{eq}$: $C_{eq}=\frac{48}{11}\mu \text{F}\approx 4.36\mu \text{F}$.
4. Check for consistency: the series equivalent is smaller than the smallest capacitor in the series combination (6 $\mu$F), which matches the expected behavior.

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.

4. Full Voltage and Power Analysis of Mixed Steady-State RC Circuits

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 voltages at all nodes.
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 $Q=C{V}_C$. Power dissipated by any component is $P=VI$; since $I_C=0$, capacitors always dissipate zero power in steady state.

The most common mistake here is forgetting that any resistor in series with a capacitor in an open branch has zero current, so zero voltage drop across the resistor.

Worked Example

A 15 V DC battery with negligible internal resistance is connected to a circuit where the positive terminal splits into two parallel branches. Branch 1: 2 $\mu$F capacitor in series with a 5 $\Omega$ resistor. Branch 2: 10 $\Omega$ resistor in series with a 5 $\Omega$ resistor. What is the voltage across the capacitor in steady state?

1. Replace the capacitor with an open circuit, so Branch 1 has zero current. This means the voltage drop across the 5 $\Omega$ resistor in Branch 1 is $V=IR=0\times 5=0$.
2. Analyze Branch 2, the only conducting branch: total resistance is $10\Omega +5\Omega =15\Omega$, so total current in Branch 2 is $I=15\text{V}/15\Omega =1\text{A}$.
3. The top plate of the capacitor connects directly to the positive battery terminal (15 V relative to the negative terminal at 0 V). The bottom plate of the capacitor connects directly to the node between the 10 $\Omega$ and 5 $\Omega$ resistors in Branch 2. The voltage drop across the 10 $\Omega$ resistor is $1\text{A}\times 10\Omega =10\text{V}$, so the potential at the middle node is $15\text{V}-10\text{V}=5\text{V}$.
4. The voltage across the capacitor is $V_C=15\text{V}-5\text{V}=10\text{V}$.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Treating a capacitor in steady-state DC as a short circuit (zero resistance, zero voltage) instead of an open circuit. 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. Correct move: Always write $I_C=0$ next to every capacitor at the start of a steady-state problem, and explicitly replace it with an open circuit before analyzing current.
• Wrong move: Swapping the equivalent capacitance formulas for series and parallel, using the same formulas as equivalent resistance. Why: Muscle memory from working with resistor networks leads students to automatically apply resistor formulas by mistake. Correct move: After calculating equivalent capacitance, check the intuitive rule: parallel $C$ > largest individual $C$, series $C$ < smallest individual $C$ to confirm you used the right formula.
• Wrong move: Forgetting that a capacitor in series with a resistor in the same branch has zero voltage drop across the resistor. 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. Correct move: Any branch that contains a capacitor (and no other parallel conducting path) has zero current, so all resistors in that branch have $V=IR=0$.
• Wrong move: Claiming that the voltage across a capacitor in steady-state DC is zero because current is zero. Why: Students confuse Ohm's law for resistors ($V=IR$) with capacitor behavior, incorrectly assuming zero current means zero voltage. Correct move: Remember that capacitor voltage comes from stored charge ($V_C=Q/C$), not from Ohm's law; zero current only means $Q$ is not changing, not that $Q$ is zero.
• Wrong move: Calculating power dissipated by a capacitor in steady-state DC as non-zero. Why: Students use $P=VI$ and plug in $V_C$, forgetting that $I_C$ is zero. Correct move: Any power calculation for a capacitor in steady DC will always give $P=V_C{I}_C=0$, so capacitors dissipate no power in steady state.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A capacitor connected to a 9 V DC battery in a mixed steady-state circuit has a potential difference of 6 V across it and is in series with a 100 $\Omega$ resistor in its branch. What is the current through the 100 $\Omega$ resistor in this branch, and what is the power dissipated by the capacitor in steady state? A) Current = 0 A, Power = 0 W B) Current = 6 mA, Power = 36 mW C) Current = 90 mA, Power = 0 W D) Current = 0 A, Power = 360 mW

Worked Solution: By the core steady-state rule, any capacitor in steady DC has zero current through it. The 100 $\Omega$ resistor is in series with the capacitor, so it shares the same zero current. Power dissipated by the capacitor is $P=V_C{I}_C=(6\text{V})(0\text{A})=0\text{W}$. All other options incorrectly assume non-zero current through the capacitor branch or non-zero power dissipation by the capacitor. The correct answer is A.

Question 2 (Free Response)

A 24 V DC battery with negligible internal resistance is connected to a circuit with three parallel branches connected across the battery:

1. Branch 1: 6 $\mu$F capacitor in series with an 8 $\Omega$ resistor
2. Branch 2: 4 $\mu$F capacitor in parallel with a 12 $\Omega$ resistor
3. Branch 3: 4 $\Omega$ resistor in series with an 8 $\Omega$ resistor

(a) What is the total current drawn from the battery in steady state? (b) What is the voltage across the 4 $\mu$F capacitor? (c) What is the total charge stored on all capacitors combined in steady state?

Worked Solution: (a) Replace all capacitors with open circuits: Branch 1 is open (no current), Branch 2 conducts through the 12 $\Omega$ resistor, Branch 3 conducts through its two resistors. Conducting branches are 12 $\Omega$ (Branch 2) and 12 $\Omega$ (Branch 3, $4+8=12\Omega$) in parallel. Equivalent resistance: $\frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$, so $R_{eq}=6\Omega$. Total current: $I_{total}=24\text{V}/6\Omega =\boxed{4\text{A}}$. (b) The 4 $\mu$F capacitor is in parallel with the 12 $\Omega$ resistor, which is connected directly across the 24 V battery. The voltage across the capacitor equals the voltage across the resistor, so $V_C=\boxed{24\text{V}}$. (c) Charge on 6 $\mu$F capacitor: the 8 $\Omega$ resistor in its branch has zero current, so $V_{C1}=24\text{V}$, $Q_1=6\mu \text{F}\times 24\text{V}=144\mu \text{C}$. Charge on 4 $\mu$F capacitor: $Q_2=4\mu \text{F}\times 24\text{V}=96\mu \text{C}$. Total charge: $Q_{total}=144+96=\boxed{240\mu \text{C}}$.

Question 3 (Application / Real-World Style)

A low-power touch-sensor circuit uses a steady-state DC setup to detect touches. The baseline circuit has a 1.5 V coin cell connected to two parallel branches: one branch has a 10 k$\Omega$ resistor, the other branch has a 1 nF capacitor in series with a 10 M$\Omega$ resistor. What is the total power drawn from the coin cell in steady state before a finger touches the sensor? Interpret your result in terms of battery life.

Worked Solution:

1. In steady state, the capacitor creates an open branch, so the second branch has zero current. Only the 10 k$\Omega$ branch conducts.
2. Current through the conducting branch: $I=V/R=1.5\text{V}/(10\times 10^3\Omega )=1.5\times 10^{-4}\text{A}$.
3. Total power drawn: $P=VI=(1.5\text{V})(1.5\times 10^{-4}\text{A})=2.25\times 10^{-4}\text{W}=0.225\mu \text{W}$.
4. In context: This extremely low steady-state power draw means the touch sensor will have a very long battery life from the small coin cell, since the capacitor branch draws no power when no touch is detected.

7. Quick Reference Cheatsheet

8. What's Next

This chapter gives you the foundational rules for analyzing capacitors in DC circuits, which is a prerequisite for the next core topic in Unit 4: transient RC circuits, where you study how capacitors charge and discharge over time. Without mastering the steady-state behavior of capacitors, you will not be able to correctly identify the boundary conditions (fully charged final state, fully discharged initial state) for transient problems, which are a commonly tested topic on the AP Physics 2 exam. This topic also reinforces your understanding of node potential analysis, which is used for all complex circuit problems across the course, and sets up the contrast between capacitor behavior in steady DC vs alternating current, which you will explore later in the syllabus.

Follow-on topics for further study:

• Transient RC Charging and Discharging
• Kirchhoff's Rules for DC Circuits
• Alternating Current Circuits