Steady-State Direct Current Circuits — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Ohm’s law, equivalent resistance for series/parallel combinations, Kirchhoff’s junction and loop rules, emf and terminal voltage, power calculation, and solution methods for single and multi-loop steady-state DC circuits.

You should already know: Electric potential and potential difference, definitions of electric current and resistance, conservation of charge and energy.

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 C: E&M 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 Direct Current Circuits?

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 in the circuit. By convention, current is treated as conventional current (flow of positive charge) from high to low potential, with notation $I$ for current, $R$ for resistance, $\epsilon$ for battery emf, $r$ for internal resistance, and $V$ for potential difference. This topic makes up the core of Unit 3 (Electric Circuits) on the AP Physics C: E&M exam, which accounts for 20% of the total exam score. Steady-state DC circuit problems regularly appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a full 10-15 point FRQ or a set of 2-3 MCQ questions. Synonyms for this topic include constant-current DC circuits and DC steady-state analysis. In steady-state DC, capacitors act as open circuits (no current flow) and inductors act as short circuits (zero resistance), simplifying analysis to only resistor and source components.

2. Equivalent Resistance for Series and Parallel Combinations

Equivalent resistance is a method to simplify a complex network of resistors into a single resistance value that behaves the same way as the full network when connected to a voltage source. For resistors connected in series (same current flows through each resistor, connected end-to-end in a single branch), total equivalent resistance is the sum of individual resistances: $$R_{\text{eq}}=R_1+R_2+\cdots +R_n=\sum _{i=1}^n{R}_i$$ This makes intuitive sense: resistance of a conductor is proportional to its length, so adding resistors in series increases total length, hence increases total resistance. For resistors connected in parallel (each resistor is connected across the same potential difference, forming multiple separate branches between the same two nodes), equivalent resistance is given by: $$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots +\frac{1}{R_n}=\sum _{i=1}^n\frac{1}{R_i}$$ This also makes intuitive sense: adding parallel resistors adds more current paths between the two nodes, so total current increases for the same applied voltage, meaning total resistance decreases. For complex circuits, you simplify step-by-step, starting from the innermost (furthest from the source terminals) combination and working outward to the source terminals.

Worked Example

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

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

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 which combinations are series vs parallel.

3. Kirchhoff's Rules for Multi-Loop Circuits

For circuits that cannot be reduced to a single equivalent resistance (e.g., circuits with multiple batteries in different branches, non-reducible resistor networks), we use Kirchhoff's two rules to solve for branch currents directly, based on fundamental conservation laws. The first rule, Kirchhoff's Junction Rule (Current Rule), is a statement of conservation of charge: the sum of currents entering any junction must equal the sum of currents leaving the junction: $$\sum I_{\text{in}}=\sum I_{\text{out}}$$ For a circuit with $n$ distinct junctions, only $n-1$ independent junction equations are needed; the final equation will always be redundant. The second rule, Kirchhoff's Loop Rule (Voltage Rule), is a statement of conservation of energy: the sum of all potential changes around any closed loop in a circuit is zero: $$\sum \Delta V=0$$ The standard sign convention for the loop rule is: (1) moving through a resistor in the direction of the assumed current: $\Delta V=-IR$ (potential drop); (2) moving through a resistor opposite the assumed current: $\Delta V=+IR$ (potential rise); (3) moving through a battery from the negative terminal to the positive terminal: $\Delta V=+\epsilon -Ir$ (net rise, accounting for internal resistance); (4) moving from positive to negative terminal: $\Delta V=-\epsilon +Ir$ (net drop). To solve, assign currents to every branch, write $n-1$ junction equations and the required number of independent loop equations, then solve the system of linear equations.

Worked Example

A two-loop circuit has two batteries: ${\epsilon }_1=12\text{V}$ (internal resistance $r_1=0.5\Omega$) and ${\epsilon }_2=6\text{V}$ (internal resistance $r_2=0.5\Omega$). Both batteries have their positive terminals connected at junction A, and negative terminals connected at common junction C. Junction A is connected to junction B through a 4 Ω resistor, and junction B is connected to junction C through a 3 Ω resistor. Find the current through the 4 Ω resistor.

1. Assign currents: $I_1$ from C to A through ${\epsilon }_1$, $I_2$ from C to A through ${\epsilon }_2$, and $I_3$ from A to B to C through the 4 Ω and 3 Ω resistors. Junction rule at A: $I_1+I_2=I_3$.
2. Apply loop rule to the left loop (${\epsilon }_1$, 4 Ω, 3 Ω): moving from C to A through ${\epsilon }_1$ gives $+12-0.5I_1$, then through 4 Ω and 3 Ω with current $I_3$ gives $-4I_3-3I_3$, back to C. Sum to zero: $12-0.5I_1-7I_3=0⟹12=0.5I_1+7I_3$.
3. Apply loop rule to the right loop (${\epsilon }_2$, 4 Ω, 3 Ω): $6-0.5I_2-7I_3=0⟹6=0.5I_2+7I_3$.
4. Substitute $I_1=I_3-I_2$ into the first loop equation: $12=0.5(I_3-I_2)+7I_3=7.5I_3-0.5I_2$. Add this to the second loop equation: $18=14I_3⟹I_3=\frac{18}{14}\approx 1.29\text{A}$. The current through the 4 Ω resistor is approximately 1.29 A.

Exam tip: Never change your assumed current direction if you get a negative current value; the negative sign itself tells you the actual direction is opposite your assumption, and changing directions mid-calculation almost always introduces unnecessary sign errors.

4. Emf, Terminal Voltage, and Power in DC Circuits

A real battery (or any voltage source) does not provide a constant potential difference regardless of current. The electromotive force (emf) $\epsilon$ of a source is the work per unit charge done by the source to separate charge, equal to the open-circuit potential difference across the source when no current is drawn. All real sources have internal resistance $r$ from the materials they are made of, so when current $I$ is drawn from the source (discharging), the terminal potential difference $V$ (the potential difference across the source's terminals) is: $$V=\epsilon -Ir$$ The term $Ir$ is the potential drop across the internal resistance. For power in DC circuits, power dissipated by a resistor is always given by any of the equivalent forms: $$P=VI=I^{2}R=\frac{V^2}{R}$$ Power supplied by a source is $P=\epsilon I$. A useful result for many exam problems is the maximum power transfer theorem: when an external load resistor $R$ is connected to a source of emf $\epsilon$ and internal resistance $r$, the power delivered to the load is maximized when $R=r$, and the maximum power is $P_{\text{max}}=\frac{{\epsilon }^2}{4r}$.

Worked Example

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

1. For part (a): Total circuit resistance is $R_{\text{total}}=R+r=3.2+0.8=4.0\Omega$. Circuit current is $I=\frac{\epsilon }{R_{\text{total}}}=\frac{9}{4}=2.25\text{A}$.
2. Terminal voltage is $V=\epsilon -Ir=9-(2.25)(0.8)=9-1.8=7.2\text{V}$. We can confirm this with $V=IR=(2.25)(3.2)=7.2\text{V}$, which matches.
3. For part (b): By the maximum power transfer theorem, maximum power is delivered when $R=r=0.8\Omega$.
4. Use the maximum power formula: $P_{\text{max}}=\frac{{\epsilon }^2}{4r}=\frac{9^2}{4(0.8)}=\frac{81}{3.2}\approx 25.3\text{W}$.

Exam tip: When asked for total power from a battery, remember that some power is dissipated in the internal resistance; always clarify whether the question asks for power delivered to the load or total power supplied by the source.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After adding the reciprocals of resistors in parallel, forget to take the reciprocal of the sum, leaving $R_{\text{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$ instead of $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$. Why: Students rush the final step, and AP MCQ distractors are often designed to match this common error. Correct move: Explicitly write $R_{\text{eq}}=1/\sum (1/R_i)$ after adding the reciprocals, and compute that final step before moving on.
• Wrong move: Getting the sign of the potential change wrong when applying the loop rule, writing $+IR$ 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 before each loop: "current flows from high to low potential, so moving with current is a drop (negative), moving against is a rise (positive)".
• Wrong move: Treating a capacitor as a short circuit (zero resistance) in steady-state DC analysis. Why: Confusing steady-state DC 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 equivalent resistance or current.
• Wrong move: Adding the emf of parallel batteries to get total emf, instead of applying Kirchhoff's rules to find the net effect. Why: Confusing parallel and series battery combinations, where series emfs do add. Correct move: Only add emfs for batteries connected in series with aligned polarities; always use Kirchhoff's rules for parallel-connected batteries with different emfs.
• Wrong move: Calculating terminal voltage as $V=\epsilon +Ir$ when a battery is discharging. Why: Confusing the direction of the potential drop across internal resistance when discharging vs charging. Correct move: Remember that internal resistance always drops voltage when discharging, so $V=\epsilon -Ir$ for discharging, only use $V=\epsilon +Ir$ for charging a battery.
• Wrong move: Writing one junction equation for every junction in the circuit, leading to a dependent, contradictory system of equations. Why: Students do not realize that charge conservation at the last junction is automatically implied by the other equations. Correct move: For a circuit with $n$ junctions, write exactly $n-1$ independent junction equations, then use loop equations to make up the required number of equations for the number of unknown currents.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

Three identical resistors of resistance $R$ are connected to a battery of emf $\epsilon$ with negligible internal resistance. The circuit configuration is: two resistors connected in parallel with each other, and this parallel combination connected in series with the third resistor and the battery. What is the ratio of the power dissipated in one of the parallel resistors to the power dissipated in the series resistor? A) $\frac{1}{4}$ B) $\frac{1}{2}$ C) $\frac{1}{\sqrt{2}}$ D) $2$

Worked Solution: First, calculate the equivalent resistance of the circuit: the parallel combination of two $R$ resistors has equivalent resistance $R_{\text{par}}=\frac{R\cdot R}{R+R}=\frac{R}{2}$. Adding the series resistor gives total equivalent resistance $R_{\text{eq}}=R+\frac{R}{2}=\frac{3R}{2}$. Total current from the battery (equal to the current through the series resistor) is $I_{\text{total}}=\frac{\epsilon }{R_{\text{eq}}}=\frac{2\epsilon }{3R}$. Power dissipated in the series resistor is $P_{\text{series}}=I_{\text{total}}^{2}R=\frac{4{\epsilon }^2}{9R}$. By junction rule, current splits equally between the two identical parallel resistors, so each carries $I_{\text{par}}=\frac{I_{\text{total}}}{2}=\frac{\epsilon }{3R}$, giving $P_{\text{par}}=I_{\text{par}}^{2}R=\frac{{\epsilon }^2}{9R}$. The ratio is $\frac{1}{4}$. The correct answer is A.

Question 2 (Free Response)

A three-branch DC circuit is constructed as follows: Branch 1 contains a $10\text{V}$ battery with internal resistance $1\Omega$, connected between nodes X and Y. Branch 2 contains a $5\text{V}$ battery with internal resistance $0.5\Omega$, also connected between X and Y, with its positive terminal at X aligned with the larger battery. Branch 3 contains a $10\Omega$ resistor connected between X and Y. (a) State the direction of currents you assign and write the system of equations from Kirchhoff's rules needed to solve for the three branch currents. (b) Solve for the magnitude and direction of each branch current. (c) Calculate the total power supplied by the batteries, and the total power dissipated by all resistors, and confirm energy is conserved.

Worked Solution: (a) Assign currents: Let $I_1$ be the current from X to Y through branch 1 (the 10V battery), $I_2$ current from X to Y through branch 2 (the 5V battery), $I_3$ current from X to Y through the 10Ω resistor. Junction rule at X: $I_1+I_2=I_3$. Loop rule for the loop containing branch 1 and branch 2: $10-I_1(1)+I_2(0.5)-5=0⟹5=I_1-0.5I_2$. Loop rule for the loop containing branch 1 and branch 3: $10-I_1(1)-I_3(10)=0⟹10=I_1+10I_3$. (b) Substitute $I_3=I_1+I_2$ into the second loop equation: $10=I_1+10(I_1+I_2)=11I_1+10I_2$. From the first loop equation, $I_1=5+0.5I_2$. Substitute: $10=11(5+0.5I_2)+10I_2=55+15.5I_2⟹I_2=\frac{-45}{15.5}\approx -2.90\text{A}$. The negative sign means $I_2$ flows opposite to the assumed direction: Y to X through branch 2 (the 5V battery is being charged). Then $I_1=5+0.5(-2.90)\approx 3.55\text{A}$ (direction X to Y matches assumption). Then $I_3=3.55-2.90=0.65\text{A}$ (direction X to Y matches assumption). (c) Total power supplied: Only the 10V battery supplies net power: $P_{\text{supplied}}={\epsilon }_1{I}_1+{\epsilon }_2(-I_2)=10(3.55)-5(2.90)=35.5-14.5=21.0\text{W}$. Total power dissipated: $P_{\text{dissipated}}=I_1^2{r}_1+I_2^2{r}_2+I_3^2{R}_3=(3.55{)}^2(1)+(2.90{)}^2(0.5)+(0.65{)}^2(10)\approx 12.6+4.2+4.2=21.0\text{W}$. Power supplied equals power dissipated, so energy is conserved, as expected.

Question 3 (Application / Real-World Style)

A portable LED flashlight uses three 1.5 V AA batteries connected in series to power a 20 Ω LED (modeled as a constant ohmic resistor for steady state operation). New AA batteries have an internal resistance of 0.2 Ω each, while near end-of-life, the internal resistance of each battery rises to 5.0 Ω. Calculate the power output of the LED for both new and end-of-life batteries, and explain the observed gradual dimming behavior of flashlights as batteries die.

Worked Solution: For new batteries: total emf ${\epsilon }_{\text{total}}=3\times 1.5=4.5\text{V}$. Total internal resistance $r_{\text{total}}=3\times 0.2=0.6\Omega$. Total circuit resistance $R_{\text{total}}=0.6+20=20.6\Omega$. Circuit current $I=\frac{4.5}{20.6}\approx 0.218\text{A}$. LED power $P_{\text{new}}=I^2{R}_{\text{LED}}=(0.218{)}^2(20)\approx 0.95\text{W}$. For end-of-life batteries: $r_{\text{total}}=3\times 5.0=15.0\Omega$, $R_{\text{total}}=15+20=35\Omega$, $I=\frac{4.5}{35}\approx 0.129\text{A}$, $P_{\text{end}}=(0.129{)}^2(20)\approx 0.33\text{W}$. The LED output power drops by more than 65% from new to end-of-life batteries, which explains the gradual dimming: as internal resistance increases over the battery's lifespan, more of the battery's emf is dropped across the internal resistance, leaving steadily less power for the LED, rather than causing an abrupt cutout.

7. Quick Reference Cheatsheet

8. What's Next

Immediately after mastering steady-state DC circuits, you will move on to study time-varying RC circuits, the next core topic in Unit 3. Steady-state DC analysis is the non-negotiable foundation for RC circuits: you need Kirchhoff's rules to set up the governing differential equation for charging and discharging, and you already understand the steady-state open-circuit behavior of capacitors that gives the final boundary condition for these problems. Beyond Unit 3, the circuit analysis skills you learned here are required for analyzing induced emfs from electromagnetic induction, where you will apply the same rules to find currents from changing magnetic flux. Without a solid command of equivalent resistance and Kirchhoff's sign conventions, solving any circuit-based problem in later topics will be extremely difficult.

RC Circuits and Time-Varying Current Capacitance and Capacitor Combinations Power and Energy in Electric Circuits Faraday's Law of Electromagnetic Induction