# Steady-State DC Circuits with Resistors and Capacitors

> AP Physics 2 · AP Physics 2 CED Unit 4: Electric Circuits
> Source: https://www.owlsprep.com/study/ap-physics-2-u4-steady-state-dc-circuits-with/

This subtopic covers behavior of fully charged capacitors in steady-state DC, equivalent capacitance for series/parallel networks, and full voltage/current/power analysis of mixed RC circuits for AP Physics 2.

**Prerequisites:** Ohm's law for ohmic resistors; Kirchhoff's junction and loop rules for DC circuits; Equivalent resistance calculations for series/parallel resistor networks

## Learning objectives

- Identify steady-state behavior of fully charged capacitors in DC circuits
- Calculate equivalent capacitance for series and parallel capacitor networks
- Analyze voltage, current, and power in mixed resistor-capacitor steady-state DC circuits
- Avoid common exam traps in steady-state RC circuit problems

## Core Behavior of a Fully Charged Capacitor

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.

**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 final 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.

## Equivalent Capacitance for Series and Parallel Networks

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.

**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. Next combine the 16 $\mu$F parallel combination 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 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.

## Full 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 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 $I_C = 0$.

**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

   $$V = IR = 0 \times 5 = 0\ \text{V}$$
2. Analyze Branch 2, the only conducting branch: total resistance is $10\ \Omega + 5\ \Omega = 15\ \Omega$, so total current in Branch 2 is

   $$I = \frac{15\ \text{V}}{15\ \Omega} = 1\ \text{A}$$
3. Set the negative battery terminal to 0 V. The top plate of the capacitor connects directly to the positive terminal at 15 V. The bottom plate connects 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 the difference between the two node potentials:

   $$V_C = 15\ \text{V} - 5\ \text{V} = 10\ \text{V}$$

> **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.

## Concept Check (AP Style)

**Check your understanding**

Test your understanding of core steady-state rules with this AP-style multiple choice question:

1. 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, and what is the power dissipated by the capacitor in steady state?

   - Current = 0 A, Power = 0 W
   - Current = 6 mA, Power = 36 mW
   - Current = 90 mA, Power = 0 W
   - Current = 0 A, Power = 360 mW

   *Answer:* Current = 0 A, Power = 0 W

   *Why:* Correct. By the core steady-state rule, $I_C = 0$, so all components in series with the capacitor have zero current, and power $P = V_C I_C = 0$.

## Common pitfalls

- **Wrong:** Treating a capacitor in steady-state DC as a short circuit (zero resistance, zero voltage) instead of an open circuit.
  - Why it fails: 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: 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:** Swapping the equivalent capacitance formulas for series and parallel, using the same formulas as equivalent resistance.
  - Why it fails: Muscle memory from working with resistor networks leads students to automatically apply resistor formulas by mistake.
  - Correct: 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:** Forgetting that a capacitor in series with a resistor in the same branch has zero voltage drop across the resistor.
  - Why it fails: 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: 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:** Claiming that the voltage across a capacitor in steady-state DC is zero because current is zero.
  - Why it fails: Students confuse Ohm's law for resistors ($V=IR$) with capacitor behavior, incorrectly assuming zero current means zero voltage.
  - Correct: 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:** Calculating power dissipated by a capacitor in steady-state DC as non-zero.
  - Why it fails: Students use $P=V I$ and plug in $V_C$, forgetting that $I_C$ is zero.
  - Correct: 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.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Steady-state capacitor current | $I_C = 0$ | Always true for fully charged capacitors; capacitor acts as an open circuit |
| Power dissipated by steady capacitor | $P_C = 0$ | No energy is lost as heat in a capacitor in steady state |
| Equivalent capacitance (parallel) | $C_{eq} = C_1 + C_2 + ... + C_n$ | All capacitors share the same voltage across two common nodes |
| Equivalent capacitance (series) | $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$ | All capacitors share the same stored charge; $C_{eq}$ < smallest individual $C$ |
| Capacitor stored charge | $Q = C V_C$ | Relates stored charge to capacitance and voltage across the capacitor |
| Power for resistors | $P = I^2 R = V^2/R = V I$ | Applies only to resistors in the conducting network, not capacitors |
| Capacitor voltage | $V_C = \Delta V_{\text{two nodes}}$ | Equal to the potential difference between the two nodes the capacitor connects |

## What's next

This subtopic 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.

- [Unit 4 Electric Circuits Overview](https://www.owlsprep.com/study/ap-physics-2-u4-overview/)
- [Magnetism and Electromagnetic Induction Overview](https://www.owlsprep.com/study/ap-physics-2-u5-overview/)
- [Magnetic Systems](https://www.owlsprep.com/study/ap-physics-2-u5-magnetic-systems/)

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