Steady-State Direct Current Circuits

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Electric Circuits· 14 min read

1. Fundamentals of Steady-State DC★★☆☆☆⏱ 3 min

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.

📘 Definition

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.

2. Equivalent Resistance for Series & Parallel★★☆☆☆⏱ 4 min

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: $R_{eq} = \sum_{i = 1}^{n} R_{i}$ . Adding series resistors increases total resistance, like increasing conductor length.
Parallel: Each resistor connected across the same potential difference, forming separate branches. Equivalent resistance: $\frac{1}{R _{eq}} = \sum_{i = 1}^{n} \frac{1}{R _{i}}$ . 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
$R_{2} = 2 + 2 = 4 Ω$
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.

3. Kirchhoff's Rules for Multi-Loop Circuits★★★☆☆⏱ 5 min

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:

📘 Definition

Kirchhoff's Junction Rule

\sum I_{\text{in}} = \sum I_{\text{out}}

Conservation of charge: sum of currents entering a junction equals sum of currents leaving. For $n$ junctions, only $n - 1$ independent equations are needed.

📘 Definition

Kirchhoff's Loop Rule

\sum \Delta V = 0

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: $Δ V = - I R$ , (2) moving against current through resistor: $Δ V = + I R$ , (3) moving negative to positive through battery: $Δ V = + ε - I r$ , (4) moving positive to negative through battery: $Δ V = - ε + I r$ .

📐 Worked Example

A two-loop circuit has $ε_{1} = 12 V$ ( $r_{1} = 0.5 Ω$ ) and $ε_{2} = 6 V$ ( $r_{2} = 0.5 Ω$ ). 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 $ε_{1}$ ), $I_{2}$ (C to A through $ε_{2}$ ), $I_{3}$ (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 ( $ε_{1}$ , 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 ( $ε_{2}$ , 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.

4. Emf, Terminal Voltage, and Power★★★☆☆⏱ 4 min

All real voltage sources have internal resistance from their constituent materials. Emf ( $ε$ ) 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 $P = ε I$ . The maximum power transfer theorem states that power delivered to an external load resistor $R$ from a source with emf $ε$ 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 $R = 3.2 Ω$ . (b) Find the maximum power delivered to the external resistor.

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

✓ Quick check

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

Three identical resistors $R$ are connected: two in parallel, this combination in series with the third, connected to a battery of emf $ε$ with negligible internal resistance. What is the ratio of power dissipated in one parallel resistor to power dissipated in the series resistor?
- $\frac{1}{4}$
- $\frac{1}{2}$
- $\frac{1}{2}$
- $2$
Reveal answer
A —
Correct. Total current $I_{total} = \frac{2 ε}{3 R}$ , power in series resistor $P_{series} = \frac{4 ε ^{2}}{9 R}$ . Each parallel resistor gets half the current, so $P_{par} = \frac{ε ^{2}}{9 R}$ , 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.

5. Common Pitfalls

Wrong move:

After adding reciprocals of parallel resistors, forget to take the reciprocal of the sum, leaving $R_{eq} = \sum 1/ R_{i}$ .

Why:

Rushing the final step; AP MCQ distractors are specifically designed to match this common error.

Correct move:

Explicitly write $R_{eq} = 1/ \sum (1/ R_{i})$ and compute this final step before moving on.

Wrong move:

Getting the sign of potential change wrong in the loop rule, writing $+ I R$ when moving in the direction of current.

Why:

Confusion between potential rise and drop, mixing up conventional and electron current direction.

Correct move:

Remind yourself: current flows from high to low potential, so moving with current is a drop (negative), moving against is a rise (positive).

Wrong move:

Treating capacitors as short circuits in steady-state DC analysis.

Why:

Confusing steady-state with transient charging/discharging where capacitors carry current.

Correct move:

Always open-circuit any capacitor in steady-state DC; remove its branch entirely when calculating current or equivalent resistance.

Wrong move:

Adding emfs of parallel batteries to get total emf, instead of applying Kirchhoff's rules.

Why:

Confusing parallel and series battery combinations, where series emfs do add.

Correct move:

Only add emfs for series batteries with aligned polarities; always use Kirchhoff's rules for parallel-connected batteries with different emfs.

Wrong move:

Calculating terminal voltage as $V = ε + I r$ when a battery is discharging.

Why:

Confusing potential drop direction when discharging vs charging.

Correct move:

Internal resistance always drops voltage when discharging, so $V = ε - I r$ for discharging; only use $V = ε + I r$ for charging a battery.

Wrong move:

Writing one junction equation for every junction, leading to a dependent system of equations.

Why:

Not realizing charge conservation at the last junction is automatically implied by previous equations.

Correct move:

For $n$ junctions, write exactly $n - 1$ independent junction equations, then use loop equations to match the number of unknown currents.

6. Quick Reference Cheatsheet

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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · FRQ
Multi-loop circuit current and power calculation
2022 · MCQ
Equivalent resistance of mixed network
2021 · FRQ
Terminal voltage and maximum power transfer

Going deeper

Official SyllabusAP Physics C: E&M Course and Exam DescriptionOfficial unit 3 learning objectives

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.