AP Physics C: E&M · Topic 11.6

Topic 11.6: Kirchhoff's Loop Rule

Unit 11: Electric Circuits15-25% of the multiple-choice section

Kirchhoff's loop rule says the sum of the potential differences around any single closed loop equals zero. That is conservation of energy: carry a charge once around a loop and it returns to the same potential. The rule is not printed on the AP Physics C E&M sheet, so write it from the principle.

AP Physics: Unit 11 (topics 11.6 Kirchhoff's Loop Rule). Topic 11.6 of the current AP Physics C: Electricity and Magnetism course and exam description, in Unit 11 Electric Circuits (15 to 25% of the multiple-choice section, about 12 to 24 class periods). One learning objective, 11.6.A, describe a circuit or elements of a circuit by applying Kirchhoff's loop rule, with four essential knowledge statements: 11.6.A.1 with the relevant equation for change in electric potential energy equals q times the potential difference, 11.6.A.2 that the loop rule is a consequence of the conservation of energy, 11.6.A.2.i stating the rule with the relevant equation that the sum of the potential differences equals zero, and 11.6.A.2.ii on representing potential as a function of position within a loop. The topic prints no boundary statement. The loop rule itself is not printed on the equation sheet, and neither is the junction rule; the electric potential energy relation is printed. Suggested skills 1.C, 2.A, 2.C and 3.B. The AP Physics 2 framework contains the same four statements with the same wording and the same four skills, listing the last two flat as 11.6.A.3 and 11.6.A.4 rather than nested. In AP Physics C the loop rule additionally generates differential equations, at 11.8.B.1 for RC circuits and 13.5.A.2 for LR circuits, and the scoring guidelines for sample Question 4 award a point for a derivation beginning with a correct application of either Kirchhoff rule.

What Topic 11.6 requires

Topic 11.6 of AP Physics C: Electricity and Magnetism Unit 11 has one learning objective and four essential-knowledge statements. It prints no boundary statement.

11.6.A, describe a circuit or elements of a circuit by applying Kirchhoff's loop rule.

11.6.A.1 Energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements. Relevant equation $\Delta U_E = q \Delta V$ .
11.6.A.2 Kirchhoff's loop rule is a consequence of the conservation of energy.
11.6.A.2.i Kirchhoff's loop rule states that the sum of potential differences across all circuit elements in a single closed loop must equal zero. Relevant equation $\sum \Delta V = 0$ .
11.6.A.2.ii The values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop.

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Notice the order the CED puts these in. The energy statement comes first, at 11.6.A.1, and the rule itself arrives at 11.6.A.2 as a consequence of conservation of energy. The framework never presents the loop rule as a formula to memorise, and that ordering is why it is not printed on the equation sheet.

The same four statements as AP Physics 2, renumbered

This topic is an honest case of two courses saying the same thing. Every statement above appears in the AP Physics 2 framework in identical wording, and the four suggested skills are the same four. The only difference is bookkeeping: AP Physics C nests the last two as 11.6.A.2.i and 11.6.A.2.ii under the conservation statement, while AP Physics 2 lists them flat as 11.6.A.3 and 11.6.A.4.

That nesting is not entirely cosmetic. Putting the statement of the rule underneath "Kirchhoff's loop rule is a consequence of the conservation of energy" makes the derivation the parent and the rule the child, which matches how the C exam asks about it.

So if you want the rule explained from scratch, the AP Physics 2 Topic 11.6 page covers this content for the algebra-based exam. That page is for AP Physics 2 students; this one is for AP Physics C students and spends its space on three things that are genuinely different on the C side.

The C sheet gives you the line-integral version. It prints $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$ and $E_x = -dV/dx$ , where the AP Physics 2 sheet prints only the finite-difference relation between field magnitude and potential difference. The loop rule becomes a statement about a closed line integral, and that matters in Unit 13.

The loop rule is the engine that produces the course's differential equations. Statement 11.8.B.1 says the charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchhoff's loop rule, and statement 13.5.A.2 says the same for a series LR circuit. Neither has an analogue in AP Physics 2, whose Topic 11.8 boundary statement explicitly excludes modelling those behaviours with respect to time.

A loop-rule start earns a point on the free-response section. The scoring guidelines for the CED's own sample Question 4 award a point for a multi-step derivation starting with a correct application of either Kirchhoff's junction rule or loop rule, and give $\Delta V_C - \Delta V_R = 0$ as an acceptable opening.

Why the sum has to be zero

Start where the CED starts, with 11.6.A.1: energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements, and

\Delta U_E = q\, \Delta V

That equation is printed on the C: E&M sheet. It says that moving a charge $q$ through a potential difference $\Delta V$ changes its electric potential energy by $q\Delta V$ .

Now carry a test charge once around a closed loop and return it to its starting point. Electric potential is a property of position, so the charge is back at the same potential, so its potential energy is unchanged: $\Delta U_E = 0$ for the trip. Since $q$ is not zero, the sum of the potential differences it passed through must be zero:

\sum \Delta V = 0

That is 11.6.A.2.i, and 11.6.A.2 names the principle behind it as conservation of energy. Every joule per coulomb the sources put in gets taken out again by the elements before the charge gets home.

Two things this derivation makes obvious that a memorised formula hides:

The rule is per loop, and every closed loop qualifies. Statement 11.2.A.2 defines a closed electrical loop as a closed path through which charges may flow, with no requirement that it contain a source. A path out through one parallel branch and back through another is a loop, and the loop rule applies to it.
The rule holds whether or not you know the currents. It is a statement about potentials, so it is true before you have solved anything. That is what makes it usable as an equation-generating tool.

The practical form most students actually use is the rearranged one: the sum of the potential rises equals the sum of the potential drops. Same statement, fewer sign errors, and it is what a loop-rule table amounts to.

The line-integral version, which only the C sheet gives you

The C: E&M sheet prints two relations between field and potential that the AP Physics 2 sheet does not:

\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}, \qquad E_x = -\frac{dV}{dx}

Put the first of those into the loop rule and $\sum \Delta V = 0$ becomes the statement that the closed line integral of the electrostatic field around any loop is zero. That is the same conservative-field property that lets electric potential exist at all, which is what Unit 9 establishes.

Three places this pays off in AP Physics C and nowhere in the algebra-based course.

Inside a resistive element the potential drops continuously. Statement 11.1.A.2.iii gives the field inside a current-carrying conductor as $\vec{E} = \rho \vec{J}$ . Integrating that along the element recovers $\Delta V = I R$ . So the "drop across a resistor" is not a discontinuous jump: it is a field integrated over a length, and a graph of potential against position falls along a slope rather than a step. That is precisely the representation 11.6.A.2.ii asks for.

A resistor with varying resistivity still obeys the loop rule. If $\rho$ varies along the length, as in 11.3.A.2.iii, the potential falls at a varying rate, but the total drop is still $IR$ with $R = \int \rho(\ell)\, d\ell / A$ , and the loop still sums to zero.

When a magnetic flux is changing, the loop equation gains a term. The C: E&M sheet prints Faraday's law as $\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\dfrac{d\Phi_B}{dt}$ , so the closed line integral of the field around a loop threaded by a changing flux is not zero. The CED's approach in that case is to carry the induced emf as a source term in the loop equation: statement 13.5.A.2 says Kirchhoff's loop rule can be applied to a series LR circuit with a battery, resulting in a differential equation, and the inductor's contribution enters as $L\, dI/dt$ . So the bookkeeping survives; what changes is that an inductor is a source of emf rather than a passive drop.

None of that appears in AP Physics 2, which does not print the line integral and whose Unit 12 stops before inductance.

Signs, declared once and held

The loop rule is easy to state and easy to get wrong, and the failures are almost always sign failures. Two conventions have to be fixed before the first term is written, and then never changed:

A direction for each unknown current. Guess freely. If the guess is backwards, the algebra returns a negative number and that is the answer, correctly reported as a magnitude in the opposite direction. It costs nothing.
A direction to traverse each loop. Clockwise, say. This is independent of the current directions, and it does not have to agree with them.

With those two fixed, every term follows mechanically:

Element	Traversing	Contribution
Resistor	with the assumed current	$-IR$
Resistor	against the assumed current	$+IR$
Battery	from the short line to the long line	$+\mathcal{E}$
Battery	from the long line to the short line	$-\mathcal{E}$
Capacitor	from the negative plate to the positive plate	$+q/C$
Capacitor	from the positive plate to the negative plate	$-q/C$

The battery rows are worth reading twice. A battery's sign depends only on which way you walk through it, not on which way the current runs. A battery being charged, with current forced into its positive terminal, still contributes $+\mathcal{E}$ when you traverse from its short line to its long line. That is why a charging circuit works out correctly with no special rules.

The resistor rows are the reverse: a resistor's sign depends on the current direction relative to your traversal, because a resistor is not a source and the potential always falls in the direction of conventional current through it.

Two habits that catch errors before they propagate:

Set the sum to zero and check the units. Every term must be in volts. A stray $I^2$ or a missing $R$ shows up instantly.
Traverse a second, different loop and check consistency. If a network has three loops, only two of the loop equations are independent, so the third is a free check on your arithmetic. Topic 11.2 covers counting the independent loops.

The graph of potential against position (11.6.A.2.ii)

Statement 11.6.A.2.ii names a representation that is easy to skip and is directly assessable under skill 1.C, creating qualitative sketches of graphs that represent features of a model or the behavior of a physical system: the values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop.

How to build one:

Pick a reference point and call it zero. Usually the negative terminal of the battery. Only differences matter, so the choice is free, and the exam's reference information already uses the same idea in electrostatics by setting the potential to zero at infinite distance from an isolated point charge.
Walk the loop in one direction, plotting position along the horizontal axis in the order the elements come.
Rise at each source, by its emf.
Fall at each resistor, by $IR$ .
Return to the reference value when you get back to the start. The graph must close. If it does not, the loop rule is telling you a term is wrong.

What the shape communicates:

The vertical extent is the emf. A single-source loop rises once and comes back down in stages.
Each fall is proportional to that element's resistance, since the current is shared in a series loop. Two resistors in series produce two drops in the ratio of their resistances, which is the visual form of a potential divider.
An ideal wire is horizontal. A resistive wire is not, which is the 11.5.B.1 distinction drawn as a picture.
A nonideal battery has an internal drop inside the source itself, so the graph rises by $\mathcal{E}$ and immediately falls by $Ir$ . The height you land on is the terminal potential difference $\mathcal{E} - Ir$ from 11.5.B.3, read straight off the axis.
A capacitor at steady state carries no current in its branch, by 11.8.B.3.iv, so there is no drop across any resistor in that branch and the capacitor's own potential difference is whatever the graph says it is at that point.

The CED works exactly this representation in its Instructional Strategies section, under Changing Representations: for a situation involving analyzing circuits, have students create a circuit diagram, a set of Kirchhoff's rules equations, and a graph of the potential around the circuit as a function of the position in the circuit. Three representations of one circuit is the task, and the graph is the one students practise least.

The loop rule as an equation generator

In the algebra-based course the loop rule produces linear equations. In AP Physics C it also produces differential equations, and that is the largest single thing this topic does for the rest of the course.

The mechanism is one substitution. Statement 11.1.A.1 defines current as $I = dq/dt$ . Put a capacitor in a loop, write the loop rule, and the capacitor contributes $q/C$ while the resistor contributes $IR = R\, dq/dt$ . Both the function and its derivative are now in the same equation:

\mathcal{E} = \frac{dq}{dt}R + \frac{q}{C}

That is statement 11.8.B.1 verbatim, and the CED describes it as "a fundamental differential equation derived from Kirchhoff's loop rule". It carries the Derived equation label, which the Required Equations page defines as a final result of a derivation expected of students on the exam. So the loop rule is not just a way to solve circuits in this course; it is the step you are graded on producing.

The same move runs in Unit 13. Statement 13.5.A.2 applies the loop rule to a series LR circuit and gets $\mathcal{E} = IR + L\,dI/dt$ , another derived equation, another differential equation, same technique. Topic 13.5 develops it.

The scoring guidelines for the CED's sample Question 4 show what earns the point. The first mark of part B is for a multi-step derivation starting with a correct application of either Kirchhoff's junction rule or loop rule, and the rubric gives two acceptable openings, $I_{\text{const}} = I_R + I_C$ for the junction route and $\Delta V_C - \Delta V_R = 0$ for the loop route. The question stem itself says: begin your derivation by writing a fundamental physics principle or an equation from the reference information.

So the habit to build is: when a question says derive, the first line is a rule, not an answer. Write $\sum \Delta V = 0$ for the named loop, then substitute the element relations one at a time. That sequence is worth marks even when the algebra afterwards goes wrong.

What is printed, and how Topic 11.6 is tested

Checked against the Table of Information appendix directly:

Equation	CED statement	On the sheet
$\Delta U_E = q \Delta V$	11.6.A.1, relevant	yes
$\sum \Delta V = 0$	11.6.A.2.i, relevant	no

Kirchhoff's loop rule is not printed on the AP Physics C: Electricity and Magnetism equation sheet. Neither is the junction rule. Two of Unit 11's eight topics are named after these rules and neither appears anywhere in the Table of Information. That is not an oversight: the CED introduces them as consequences of conservation laws, at 11.6.A.2 for energy and 11.7.A.1 for charge, so the expectation is that you write them from the principle. A student who plans to look them up during the exam has a problem; a student who can derive them in one sentence has none.

What is printed and is useful here: $\Delta U_E = q\Delta V$ , $I = \Delta V / R$ , $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$ , $C = Q/\Delta V$ and $I = dq/dt$ . Those five are enough to build any loop equation the course can pose.

On assessment, Topic 11.6 behaves like Topic 11.2: it is rarely the subject of a question and constantly the tool inside one. The skill mix says the same. 2.A derivation and 1.C qualitative graph sketching are both listed, and on the free-response section Science Practice 1 carries 20 to 35% and Science Practice 2 carries 40 to 45%, while Science Practice 1 is not assessed at all on the multiple-choice section.

The CED's own Topic 11.6 sample activity is a Changing Representations task: have students solve a typical multi-loop circuit problem with batteries and resistors, then construct a representation for each possible loop that visually shows Kirchhoff's loop rule and a representation for each junction that visually shows Kirchhoff's junction rule. Practising the drawing, not just the algebra, is the point of it.

A two-loop network with two batteries

Two branches run upward from node B to node A. Branch 1 contains a battery of emf $\mathcal{E}_1 = 12$ V in series with $R_1 = 3.0\ \Omega$ ; branch 2 contains a battery of emf $\mathcal{E}_2 = 10$ V in series with $R_2 = 4.0\ \Omega$ . Both batteries drive charge from B toward A. A third branch returns from A to B through $R_3 = 2.0\ \Omega$ . All batteries are ideal. Find the three branch currents, then audit the energy.

Declare the conventions first. Take conventional current, assume $I_1$ and $I_2$ flow upward from B to A in their branches and $I_3$ flows down from A to B through $R_3$ , and traverse each loop in the direction of its own assumed current. Hold all of that to the end.
Junction rule at A, from 11.7.A.2: everything arriving must leave, so $I_1 + I_2 = I_3$ .
Loop containing branch 1 and branch 3, starting at B and going up: rise $\mathcal{E}_1$ at the battery, fall $I_1 R_1$ across $R_1$ , then fall $I_3 R_3$ coming back down through $R_3$ . Setting the sum to zero: $12 - 3.0 I_1 - 2.0 I_3 = 0$ .
Loop containing branch 2 and branch 3, same way: $10 - 4.0 I_2 - 2.0 I_3 = 0$ .
Solve. From the first loop, $I_1 = 4.0 - \dfrac{2}{3} I_3$ . From the second, $I_2 = 2.5 - 0.5 I_3$ . Substituting both into the junction equation: $I_3 = 6.5 - \dfrac{7}{6} I_3$ , so $\dfrac{13}{6} I_3 = 6.5$ and $I_3 = 3.0$ A.
Back-substitute: $I_1 = 4.0 - 2.0 = 2.0$ A and $I_2 = 2.5 - 1.5 = 1.0$ A. All three are positive, so every guessed direction was right, and $2.0 + 1.0 = 3.0$ checks the junction equation.
Free check from the third loop. There is a closed path up branch 1 and back down branch 2 that touches no source-free element in common with the others: it requires $\mathcal{E}_1 - I_1 R_1 = \mathcal{E}_2 - I_2 R_2$ , that is $12 - 6.0 = 10 - 4.0$ , giving $6.0 = 6.0$ . Node A is 6.0 V above node B, which also equals $I_3 R_3 = (3.0)(2.0) = 6.0$ V.
Energy audit, which is the loop rule's parent principle at work. The sources deliver $\mathcal{E}_1 I_1 + \mathcal{E}_2 I_2 = (12)(2.0) + (10)(1.0) = 34$ W. The resistors dissipate $I_1^2 R_1 + I_2^2 R_2 + I_3^2 R_3 = (4.0)(3.0) + (1.0)(4.0) + (9.0)(2.0) = 12 + 4 + 18 = 34$ W. It closes exactly, so the solution is right.

$I_1 = 2.0$ A, $I_2 = 1.0$ A and $I_3 = 3.0$ A, with node A sitting 6.0 V above node B. The energy audit closes at 34 W delivered and 34 W dissipated, which is conservation of energy, the principle essential knowledge 11.6.A.2 names as the source of the loop rule.

Graphing electric potential against position around a loop

A nonideal battery of emf $\mathcal{E} = 15$ V and internal resistance $r = 1.0\ \Omega$ is in series with $R_1 = 4.0\ \Omega$ and $R_2 = 10\ \Omega$ . A capacitor $C = 20\ \mu\mathrm{F}$ is connected in parallel with $R_2$ , and the circuit has been closed for a long time. Find the current, then give the electric potential at each point around the loop taking the negative terminal as zero, and find the capacitor's charge and stored energy.

Steady state first. Statement 11.8.B.3.iv says that after a long time a charging capacitor reaches a maximum potential difference at which there is zero current in the circuit branch containing it. So no current flows into the capacitor branch, and the loop is simply $r$ , $R_1$ and $R_2$ in series.
Total resistance $= 1.0 + 4.0 + 10 = 15\ \Omega$ , so $I = \dfrac{\mathcal{E}}{R_{\text{total}}} = \dfrac{15}{15} = 1.0$ A.
Now walk the loop, setting $V = 0$ at the negative terminal and moving in the direction of conventional current. Rise through the ideal source: $V = +15$ V. Fall across the internal resistance, $Ir = (1.0)(1.0) = 1.0$ V, giving $V = 14$ V. That value is the terminal potential difference, and it matches the derived 11.5.B.3 result $\mathcal{E} - Ir = 15 - 1.0 = 14$ V.
Fall across $R_1$ : $IR_1 = (1.0)(4.0) = 4.0$ V, giving $V = 10$ V. Fall across $R_2$ : $IR_2 = (1.0)(10) = 10$ V, giving $V = 0$ . The graph returns to its starting value, which is the loop rule closing: $+15 - 1.0 - 4.0 - 10 = 0$ .
Sketch it as 11.6.A.2.ii asks. Position runs along the horizontal axis in the order the elements come. A vertical rise of 15 V at the source, then three falls in the ratio $1 : 4 : 10$ , and a return to the reference line. Ideal wires are horizontal segments; the drops inside the resistive elements slope rather than jump, because 11.1.A.2.iii puts a field inside each one and $\Delta V = -\int \vec{E} \cdot d\vec{r}$ accumulates it along the length.
The capacitor sits across $R_2$ , so it holds whatever potential difference $R_2$ has: $\Delta V_C = 10$ V. From the printed $C = Q/\Delta V$ , $Q = (20 \times 10^{-6})(10) = 2.0 \times 10^{-4}$ C, or $200\ \mu\mathrm{C}$ .
Stored energy from the printed $U_C = \tfrac{1}{2} Q \Delta V = \tfrac{1}{2}(2.0 \times 10^{-4})(10) = 1.0 \times 10^{-3}$ J, that is 1.0 mJ.
One reading to carry away: the capacitor's potential difference is 10 V, not the 15 V emf and not the 14 V terminal value. A capacitor takes the potential difference of whatever it is in parallel with, and the graph makes that obvious at a glance.

$I = 1.0$ A. Taking the negative terminal as zero, the potential runs 0 to 15 V at the source, down to 14 V after the internal resistance, down to 10 V after $R_1$ , and back to 0 after $R_2$ . The capacitor holds $\Delta V_C = 10$ V, $Q = 200\ \mu\mathrm{C}$ and $U_C = 1.0$ mJ.

A battery being charged, where the current runs backwards through a source

A charger of emf $\mathcal{E}_1 = 12.0$ V and internal resistance $r_1 = 0.50\ \Omega$ is connected in series with a resistor $R = 2.2\ \Omega$ and a battery of emf $\mathcal{E}_2 = 9.0$ V and internal resistance $r_2 = 0.30\ \Omega$ , with the two sources connected so that they oppose one another. Find the current, the terminal potential difference of each source, and account for all the power.

Set the conventions. Assume conventional current flows in the direction the charger drives it, and traverse the single loop in that same direction. Because the sources oppose, that direction carries current into the positive terminal of the second battery, which is what charging means.
Apply 11.6.A.2.i term by term, using the sign table: rise $+\mathcal{E}_1$ through the charger, fall $-I r_1$ across its internal resistance, fall $-I R$ across the resistor, fall $-\mathcal{E}_2$ through the second source because you traverse it from its long line to its short line, and fall $-I r_2$ across its internal resistance.
$12.0 - I(0.50) - I(2.2) - 9.0 - I(0.30) = 0$ , so $3.0 = I(3.0)$ and $I = 1.0$ A. The positive result confirms the assumed direction.
Charger's terminal potential difference, from 11.5.B.3: $\Delta V_1 = \mathcal{E}_1 - I r_1 = 12.0 - 0.50 = 11.5$ V. Current leaves its positive terminal, so its terminals read below its emf, as usual.
Charged battery's terminal potential difference: here the current enters the positive terminal, so the internal drop adds instead of subtracting, $\Delta V_2 = \mathcal{E}_2 + I r_2 = 9.0 + 0.30 = 9.3$ V. A battery on charge reads above its emf, which is the reverse of the familiar case and is the whole point of the example.
Consistency check across the external part: $\Delta V_1$ must equal $\Delta V_2$ plus the drop across $R$ . $9.3 + (1.0)(2.2) = 11.5$ V, which matches.
Power audit. The charger delivers $\mathcal{E}_1 I = (12.0)(1.0) = 12.0$ W. Dissipation: $I^2 r_1 = 0.50$ W, $I^2 R = 2.2$ W, $I^2 r_2 = 0.30$ W, totalling $3.0$ W. The remaining $\mathcal{E}_2 I = 9.0$ W is stored in the second battery rather than dissipated, and $9.0 + 3.0 = 12.0$ W closes the audit.
The energy split is the useful result: 75 percent of the charger's output is stored and 25 percent is wasted as dissipation, and lowering $R$ would raise the current, raise the stored fraction's rate and raise the wasted fraction faster, since dissipation goes as $I^2$ .

$I = 1.0$ A. The charger's terminals read $11.5$ V, below its 12.0 V emf, while the battery on charge reads $9.3$ V, above its 9.0 V emf, because current enters its positive terminal. Of the charger's 12.0 W, 9.0 W is stored in the second battery and 3.0 W is dissipated.

Frequently asked questions

What is Kirchhoff's loop rule?

Kirchhoff's loop rule states that the sum of the potential differences across all circuit elements in a single closed loop must equal zero. The AP Physics C course description gives it at essential knowledge 11.6.A.2.i and says at 11.6.A.2 that it is a consequence of the conservation of energy. The reason is short: electric potential depends only on position, so a charge carried once around a loop returns to the same potential and its potential energy is unchanged. Since the change in electric potential energy is the charge times the potential difference, and the charge is not zero, the potential differences must sum to zero. The rule applies to every closed loop, including loops that contain no battery.

Is Kirchhoff's loop rule on the AP Physics C E&M equation sheet?

No. Neither Kirchhoff rule is printed anywhere in the Table of Information for AP Physics C: Electricity and Magnetism, even though two of Unit 11's eight topics are named after them. The course description introduces them as consequences of conservation laws rather than as formulas, at essential knowledge 11.6.A.2 for the loop rule and conservation of energy, and 11.7.A.1 for the junction rule and conservation of electric charge, so the expectation is that students write them from the principle. What is printed and useful for building a loop equation is the change in electric potential energy equals q times the potential difference, Ohm's law, the definition of capacitance, and current as dq/dt.

How do you assign signs when applying the loop rule?

Fix two things before writing any term, then never change them: a direction for each unknown current, guessed freely, and a direction to traverse each loop. After that the terms are mechanical. A resistor contributes minus I R when you traverse it with the assumed current and plus I R when you traverse against it. A battery contributes plus its emf when you traverse from its short line to its long line and minus its emf the other way, regardless of which way the current runs through it. A capacitor contributes plus q over C going from the negative plate to the positive plate. If a current comes out negative, the guess was backwards and the magnitude is still correct.

How does the loop rule give you a differential equation?

By putting a capacitor and a resistor in the same loop and using the calculus definition of current. Essential knowledge 11.1.A.1 defines current as dq/dt, so a resistor in the branch contributes R times dq/dt while the capacitor contributes q over C. The loop rule then relates a function and its own derivative, giving the emf equals R dq/dt plus q over C. That is essential knowledge 11.8.B.1 in the AP Physics C course description, which describes it as a fundamental differential equation derived from Kirchhoff's loop rule and gives it the Derived equation label. The same technique in Unit 13 produces the LR circuit equation at essential knowledge 13.5.A.2.

What does a graph of electric potential against position in a circuit look like?

It rises at each source by that source's emf, falls at each resistive element by the current times its resistance, and returns to its starting value when you get back to where you began. Essential knowledge 11.6.A.2.ii names this representation: the values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop. Choose any point as the zero reference, usually the negative terminal of the battery. A nonideal battery rises by its emf and immediately falls by the current times its internal resistance, so the height it settles at is the terminal potential difference. The requirement that the graph close is the loop rule drawn as a picture.

Is AP Physics C Topic 11.6 different from AP Physics 2 Topic 11.6?

The content is the same. All four essential knowledge statements appear in both frameworks in identical wording, and both list the same four suggested skills. The only structural difference is numbering: AP Physics C nests the statement of the rule and the graph representation as 11.6.A.2.i and 11.6.A.2.ii under the conservation of energy statement, where AP Physics 2 lists them flat as 11.6.A.3 and 11.6.A.4. What differs in practice is use. Only the AP Physics C sheet prints the potential difference as a line integral of the field, and only in AP Physics C does the loop rule go on to generate differential equations, in Topic 11.8 for RC circuits and Topic 13.5 for LR circuits.

How many loop equations do you need for a multi-loop circuit?

As many as there are loops that enclose no other loop, with the junction rule supplying the rest. In a network with three closed paths, only two of the three loop equations are independent, because the third is a combination of the other two, so it adds no information but does make a free check on your arithmetic. Together the independent loop equations and the junction equations give one equation per unknown branch current. Essential knowledge 11.2.A.3 is the warning behind the counting: a single circuit element may be part of multiple electrical loops, so the same resistor appears in more than one equation.