# Steady-State Direct Current Circuits

> AP Physics C: Electricity and Magnetism · AP Physics C: E&M Unit 3: Electric Circuits
> Source: https://www.owlsprep.com/study/ap-physics-c-em-u3-steady-state-direct-current-circuits/

This study guide covers core concepts of steady-state direct current (DC) circuits for AP Physics C: E&M, including equivalent resistance, Kirchhoff's rules, emf/terminal voltage, and power calculations, with exam-aligned worked examples.

**Prerequisites:** Electric potential and potential difference; Definitions of current, resistance, and conservation of charge/energy

## Learning objectives

- Calculate equivalent resistance for series and parallel resistor combinations
- Apply Kirchhoff's junction and loop rules to solve multi-loop DC circuits
- Relate emf, internal resistance, and terminal voltage for real voltage sources
- Calculate power dissipated and supplied in DC circuits, apply maximum power transfer theorem
- Identify steady-state behavior of capacitors and inductors in DC circuits

## Fundamentals of Steady-State DC

Steady-state direct current (DC) circuits are circuits where the magnitude and direction of current in every branch is constant over time, with no build-up or depletion of charge at any point. By convention, we use conventional current (flow of positive charge) from high to low potential.

**Steady-State Behavior for Reactive Components** — In steady-state DC, capacitors act as open circuits (no current flow) and inductors act as short circuits (zero resistance). This simplifies all steady-state analysis to only resistors and voltage sources.

> **info**
>
> This topic makes up 20% of the total AP Physics C: E&M exam score, and regularly appears in both multiple-choice and free-response sections.

## Equivalent Resistance for Series & Parallel

Equivalent resistance simplifies a complex network of resistors into a single equivalent value that behaves the same way as the original network when connected to a voltage source. The rules for series and parallel combinations are:

- **Series**: Same current flows through each resistor, connected end-to-end. Equivalent resistance: $R_{\text{eq}} = \sum_{i=1}^n R_i$. Adding series resistors increases total resistance, like increasing conductor length.
- **Parallel**: Each resistor connected across the same potential difference, forming separate branches. Equivalent resistance: $\frac{1}{R_{\text{eq}}} = \sum_{i=1}^n \frac{1}{R_i}$. Adding parallel resistors decreases total resistance, by adding more current paths.

**Worked example:** Find the equivalent resistance between terminals A and B for the following network: a 2 Ω resistor connected in series with a parallel combination of 3 Ω and 6 Ω. This entire block is connected in parallel with a 4 Ω resistor between A and B.

1. First solve for the innermost parallel combination (3 Ω and 6 Ω):
2. $$\frac{1}{R_1} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \implies R_1 = 2\ \Omega$$
3. Add the series 2 Ω resistor to get the equivalent resistance of the full series-parallel block:
4. $R_2 = 2 + 2 = 4\ \Omega$
5. This 4 Ω block is parallel with the final 4 Ω resistor, so total equivalent resistance is:
6. $$\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \implies R_{\text{eq}} = 2\ \Omega$$

> **Exam tip:** Always simplify circuits starting from the innermost (furthest from the source terminals) combination and work back toward the terminals; starting from the source end often leads to misidentifying series vs parallel combinations.

## Kirchhoff's Rules for Multi-Loop Circuits

For circuits that cannot be reduced to a single equivalent resistance (e.g., multiple batteries in different branches), we use Kirchhoff's two rules, derived from fundamental conservation laws:

**Kirchhoff's Junction Rule** — Conservation of charge: sum of currents entering a junction equals sum of currents leaving. For $n$ junctions, only $n-1$ independent equations are needed.

*Notation:* \sum I_{\text{in}} = \sum I_{\text{out}}

**Kirchhoff's Loop Rule** — Conservation of energy: sum of all potential changes around any closed loop equals zero. The standard sign convention is: (1) moving with current through resistor: $\Delta V = -IR$, (2) moving against current through resistor: $\Delta V = +IR$, (3) moving negative to positive through battery: $\Delta V = +\varepsilon - Ir$, (4) moving positive to negative through battery: $\Delta V = -\varepsilon + Ir$.

*Notation:* \sum \Delta V = 0

**Worked example:** A two-loop circuit has $\varepsilon_1 = 12\ \text{V}$ ($r_1 = 0.5\ \Omega$) and $\varepsilon_2 = 6\ \text{V}$ ($r_2 = 0.5\ \Omega$). Both positive terminals connect at junction A, both negatives at common junction C. A is connected to B via 4 Ω, B connected to C via 3 Ω. Find the current through the 4 Ω resistor.

1. Assign currents: $I_1$ (C to A through $\varepsilon_1$), $I_2$ (C to A through $\varepsilon_2$), $I_3$ (A to B to C through resistors). Junction rule at A:
2. $I_1 + I_2 = I_3$
3. Apply loop rule to left loop ($\varepsilon_1$, 4 Ω, 3 Ω):
4. $$12 - 0.5I_1 - 4I_3 - 3I_3 = 0 \implies 12 = 0.5I_1 + 7I_3$$
5. Apply loop rule to right loop ($\varepsilon_2$, 4 Ω, 3 Ω):
6. $$6 - 0.5I_2 - 7I_3 = 0 \implies 6 = 0.5I_2 + 7I_3$$
7. Substitute $I_1 = I_3 - I_2$ into first equation, add to second equation:
8. $$12 = 0.5(I_3 - I_2) + 7I_3 \implies 18 = 14I_3 \implies I_3 \approx 1.29\ \text{A}$$

> **Exam tip:** Never change your assumed current direction if you get a negative value. The negative sign already indicates direction opposite your assumption; changing directions mid-calculation almost always causes sign errors.

## Emf, Terminal Voltage, and Power

All real voltage sources have internal resistance from their constituent materials. Emf ($\varepsilon$) is the open-circuit potential difference across the source when no current is drawn. When current $I$ is drawn from a discharging source, terminal voltage (potential across the source terminals) is:

$$V = \varepsilon - Ir$$

Power dissipated by a resistor can be written three equivalent ways, and power supplied by a source is $P = \varepsilon I$. The maximum power transfer theorem states that power delivered to an external load resistor $R$ from a source with emf $\varepsilon$ and internal resistance $r$ is maximized when $R = r$, with maximum power:

$$P_{\text{max}} = \frac{\varepsilon^2}{4r}$$

**Worked example:** A 9 V battery with internal resistance 0.8 Ω is connected to a variable external resistor $R$. (a) Find terminal voltage when $R = 3.2\ \Omega$. (b) Find the maximum power delivered to the external resistor.

1. Part (a): Total resistance = $R + r = 3.2 + 0.8 = 4.0\ \Omega$, so circuit current is:
2. $$I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9}{4} = 2.25\ \text{A}$$
3. Calculate terminal voltage:
4. $V = \varepsilon - Ir = 9 - (2.25)(0.8) = 7.2\ \text{V}$, which matches $V = IR = 2.25 \times 3.2 = 7.2\ \text{V}$.
5. Part (b): Maximum power occurs when $R = r = 0.8\ \Omega$. Substitute into the formula:
6. $$P_{\text{max}} = \frac{9^2}{4(0.8)} = \frac{81}{3.2} \approx 25.3\ \text{W}$$

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. Three identical resistors $R$ are connected: two in parallel, this combination in series with the third, connected to a battery of emf $\varepsilon$ with negligible internal resistance. What is the ratio of power dissipated in one parallel resistor to power dissipated in the series resistor?

   - $\frac{1}{4}$
   - $\frac{1}{2}$
   - $\frac{1}{\sqrt{2}}$
   - $2$

   *Why:* Correct. Total current $I_{\text{total}} = \frac{2\varepsilon}{3R}$, power in series resistor $P_{\text{series}} = \frac{4\varepsilon^2}{9R}$. Each parallel resistor gets half the current, so $P_{\text{par}} = \frac{\varepsilon^2}{9R}$, ratio = $\frac{1}{4}$.

> **Exam tip:** When asked for power from a battery, clarify if the question asks for total power supplied by the source (includes power lost to internal resistance) or power delivered to the external load.

## Common pitfalls

- **Wrong:** After adding reciprocals of parallel resistors, forget to take the reciprocal of the sum, leaving $R_{\text{eq}} = \sum 1/R_i$.
  - Why it fails: Rushing the final step; AP MCQ distractors are specifically designed to match this common error.
  - Correct: Explicitly write $R_{\text{eq}} = 1/\sum(1/R_i)$ and compute this final step before moving on.
- **Wrong:** Getting the sign of potential change wrong in the loop rule, writing $+IR$ when moving in the direction of current.
  - Why it fails: Confusion between potential rise and drop, mixing up conventional and electron current direction.
  - Correct: Remind yourself: current flows from high to low potential, so moving with current is a drop (negative), moving against is a rise (positive).
- **Wrong:** Treating capacitors as short circuits in steady-state DC analysis.
  - Why it fails: Confusing steady-state with transient charging/discharging where capacitors carry current.
  - Correct: Always open-circuit any capacitor in steady-state DC; remove its branch entirely when calculating current or equivalent resistance.
- **Wrong:** Adding emfs of parallel batteries to get total emf, instead of applying Kirchhoff's rules.
  - Why it fails: Confusing parallel and series battery combinations, where series emfs do add.
  - Correct: Only add emfs for series batteries with aligned polarities; always use Kirchhoff's rules for parallel-connected batteries with different emfs.
- **Wrong:** Calculating terminal voltage as $V = \varepsilon + Ir$ when a battery is discharging.
  - Why it fails: Confusing potential drop direction when discharging vs charging.
  - Correct: Internal resistance always drops voltage when discharging, so $V = \varepsilon - Ir$ for discharging; only use $V = \varepsilon + Ir$ for charging a battery.
- **Wrong:** Writing one junction equation for every junction, leading to a dependent system of equations.
  - Why it fails: Not realizing charge conservation at the last junction is automatically implied by previous equations.
  - Correct: For $n$ junctions, write exactly $n-1$ independent junction equations, then use loop equations to match the number of unknown currents.

## Cheatsheet

| Concept | Formula/Rule | Key Notes |
| --- | --- | --- |
| Series Resistance | $R_{\text{eq}} = \sum R_i$ | Same current through all resistors |
| Parallel Resistance | $\frac{1}{R_{\text{eq}}} = \sum \frac{1}{R_i}$ | Same voltage across all resistors |
| Junction Rule | $\sum I_{\text{in}} = \sum I_{\text{out}}$ | Use $n-1$ equations for $n$ junctions |
| Loop Rule | $\sum \Delta V = 0$ | $-IR$ moving with current, $+\varepsilon$ for - to + through battery |
| Terminal Voltage (discharging) | $V = \varepsilon - Ir$ | $V = \varepsilon + Ir$ when charging |
| Power (resistor) | $P = VI = I^2R = \frac{V^2}{R}$ | Always positive |
| Max Power Transfer | $P_{\text{max}} = \frac{\varepsilon^2}{4r}$ at $R=r$ | For load connected to source with internal resistance $r$ |
| Steady-State Behavior | Capacitors: open; Inductors: short | No current through capacitor branches |

## What's next

Mastering steady-state DC circuits is the foundation for all further circuit analysis in AP Physics C: E&M, including transient behavior of RC, RL, and RLC circuits, which are also tested in Unit 3. The problem-solving skills you developed here — assigning variables, applying conservation laws, solving systems of equations, and checking energy consistency — transfer directly to more complex dynamic circuit problems. You will reuse these same core concepts when analyzing AC circuits later in your study, though AC adds time-dependent voltage and current that requires additional tools. Next, build fluency with practice problems, then move on to study transient dynamic circuits, a common free-response topic on the AP exam.

- [Unit 3 Electric Circuits Overview](https://www.owlsprep.com/study/ap-physics-c-em-u3-overview/)
- [Capacitors in Circuits](https://www.owlsprep.com/study/ap-physics-c-em-u3-capacitors-in-circuits/)
- [Magnetic Fields Overview](https://www.owlsprep.com/study/ap-physics-c-em-u4-overview/)

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