Kirchhoff's Loop vs Junction Rule: Difference

The loop rule follows from conservation of energy: the potential differences around any closed loop sum to zero. The junction rule follows from conservation of charge: the current entering a node equals the current leaving it. Apply one around a closed path and the other at a node.

AP Physics: Unit 11 (topics 11.6 Kirchhoff's Loop Rule, 11.7 Kirchhoff's Junction Rule). The two rules are consecutive topics in AP Physics 2 Unit 11, Electric Circuits, which carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods. Topic 11.6 holds the loop rule: 11.6.A.1 gives the energy link delta U_E = q delta V, 11.6.A.2 states that the rule is a consequence of the conservation of energy, 11.6.A.3 states that the sum of potential differences across all circuit elements in a single closed loop must equal zero, and 11.6.A.4 adds that the electric potential at points in a circuit can be represented by a graph of potential as a function of position within a loop. Its suggested skills are 1.C, 2.A, 2.C and 3.B. Topic 11.7 holds the junction rule in two statements: 11.7.A.1 that it is a consequence of the conservation of electric charge, and 11.7.A.2 that the total charge entering a junction per unit time must equal the total exiting, with the equation sum of I in equals sum of I out. Its suggested skills are 1.A, 2.B, 2.C and 3.C. AP Physics C: Electricity and Magnetism states both rules identically, filing the loop-rule statement and the potential graph as 11.6.A.2.i and 11.6.A.2.ii. Neither rule's equation is printed on either equation sheet. A boundary statement under Topic 11.5 in both CEDs excludes circuits with batteries of different potential differences connected in parallel.

Two conservation laws, so two rules

There are two rules rather than one because there are two separate things a circuit conserves, and the CED files them as separate topics for exactly that reason.

[The loop rule](/glossary/kirchhoffs-loop-rule) is conservation of energy. Essential knowledge 11.6.A.2 says so in one line: Kirchhoff's loop rule is a consequence of the conservation of energy. The statement itself is 11.6.A.3, that the sum of potential differences across all circuit elements in a single closed loop must equal zero,

ΔV=0\sum \Delta V = 0

and 11.6.A.1 supplies the link to energy, ΔUE=qΔV\Delta U_E = q\Delta V. Carry a charge once around a loop and it returns to where it started, so the electric potential energy it gained has to equal the energy it gave up. A rule about volts is a rule about joules per coulomb.

[The junction rule](/glossary/kirchhoffs-junction-rule) is conservation of charge. Essential knowledge 11.7.A.1 is the parallel sentence: Kirchhoff's junction rule is a consequence of the conservation of electric charge. The statement, 11.7.A.2, is that the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time,

Iin=Iout\sum I_{in} = \sum I_{out}

Charge is not created at a node and does not pile up there in a steady circuit, so the rates in and out match.

Everything that follows is a consequence of that split. The loop rule constrains potential differences and needs a journey. The junction rule constrains currents and needs a point. Neither can do the other's job, and a problem that has both unknown currents and unknown potential differences needs both.

Side by side

Loop ruleJunction rule
Conservation law behind itEnergy (11.6.A.2)Electric charge (11.7.A.1)
What it statesPotential differences around a single closed loop sum to zero (11.6.A.3)Charge entering a junction per unit time equals charge exiting it (11.7.A.2)
EquationΔV=0\sum \Delta V = 0Iin=Iout\sum I_{in} = \sum I_{out}
Quantity it constrainsPotential difference, in voltsCurrent, in amperes
Where you apply itAround a closed pathAt one node, where three or more conductors meet
What it needs from youA chosen walking direction and a sign conventionA chosen in and out for each branch
Independent equations availableOne per independent loopOne fewer than the number of junctions
What it forbidsA charge arriving back at its start with different potential energyCharge accumulating at a point in a steady circuit
In a series connectionThe drops add up to the source's potential differenceNothing to say: no junction exists between series elements
In a parallel connectionEvery branch spans the same potential differenceBranch currents add to the trunk current
Printed on the AP Physics 2 sheet?NoNo
CED topic11.611.7
Suggested skills in AP Physics 21.C, 2.A, 2.C, 3.B1.A, 2.B, 2.C, 3.C

Neither equation appears on either equation sheet. Render the AP Physics 2 Table of Information and the Electricity box carries I=Δq/ΔtI = \Delta q/\Delta t, R=ρ/AR = \rho \ell / A, P=IΔVP = I\Delta V, I=ΔV/RI = \Delta V/R, the series and parallel combination rules for resistors and capacitors and the time constant, with no ΔV=0\sum \Delta V = 0 and no I\sum I. The AP Physics C: Electricity and Magnetism sheet is the same in this respect. Both rules are known rather than looked up, which is the practical reason the sign convention below is worth memorising rather than reconstructing under pressure.

The sign convention, stated once and kept to the end

The junction rule barely has a sign problem. Label each branch as carrying current into or out of the node, add the ins, add the outs, set them equal. That is the whole procedure.

The loop rule is where marks are lost, and the fix is to fix a convention before writing anything and then not to think about it again.

Step one: choose a direction to walk the loop. Clockwise or anticlockwise, it makes no difference, but choose before you start.

Step two: guess a direction for each unknown current and draw the arrow. A guess that turns out wrong produces a negative number and nothing else. Nothing needs redrawing.

Step three: apply these four cases as you walk.

What you cross, walking in your chosen directionContribution to ΔV\sum \Delta V
A source from its negative terminal to its positive terminal+E+\mathcal{E}
A source from its positive terminal to its negative terminalE-\mathcal{E}
A resistor in the same direction as the current arrow in itIR-IR
A resistor against the current arrow in it+IR+IR

Step four: set the total to zero and solve.

The resistor rows are the ones to internalise. Walking with the current means walking downhill in potential, because charge loses energy in a resistor, so the term is negative. Walking against the current means climbing, so the term is positive. The source rows are the opposite way round from what people expect on a first read: crossing from minus to plus is a rise, which is why it is +E+\mathcal{E} regardless of which way the current happens to be going through that source.

A negative current in your answer means the real current runs opposite to your arrow. That is information, not an error, and rewriting the whole solution to remove the minus sign wastes time you do not have.

Where each one applies: a point against a path

The two rules attach to different features of a schematic, and identifying the feature is most of the work.

A junction is a point where three or more conductors meet. That is where charge has a choice, and where a split or a merge can occur. Two things that look like junctions and are not:

  • A bend in a wire. Nothing branches, so no equation is available.
  • Two components joined end to end. Charge that enters the first must leave through the second, which is exactly the definition of a series connection at 11.5.A.1.i, and it is why series elements carry identical currents.

A loop is any closed path you can trace through the circuit and back to your starting point. It does not have to be a physically obvious rectangle, and a single element may belong to several loops: 11.2.A.3 states that a single circuit element may be part of multiple electrical loops. What you need is a set of independent loops, meaning each new loop includes at least one element the earlier ones did not.

Counting equations before you start is the habit that prevents thrashing:

  • A circuit with nn junctions yields n1n - 1 independent junction equations. The last one is arithmetically implied by the others and adds nothing.
  • Each independent loop yields one loop equation.
  • Together those must number as many as the unknown currents, or the system will not close.

A two-junction, two-loop network therefore gives one junction equation and two loop equations for three branch currents, and it solves. A single-loop circuit has no junctions at all, so the junction rule contributes nothing and the loop rule does all the work by itself.

The case that separates them: one circuit, one rule at a time

Take a 12 V12 \ \mathrm{V} ideal battery in series with a 2.0 Ω2.0 \ \Omega resistor, feeding a parallel pair of 6.0 Ω6.0 \ \Omega and 3.0 Ω3.0 \ \Omega. Call the three branch currents I1I_1 through the series resistor, I2I_2 through the 6.0 Ω6.0 \ \Omega and I3I_3 through the 3.0 Ω3.0 \ \Omega.

Now try to solve it with only one rule and watch it fail.

Loop rule alone. Walk the loop through the battery, the 2.0 Ω2.0 \ \Omega and the 6.0 Ω6.0 \ \Omega: 122.0I16.0I2=012 - 2.0 I_1 - 6.0 I_2 = 0. One equation, two unknowns. Add the second loop, the one containing only the two parallel resistors: 6.0I23.0I3=06.0 I_2 - 3.0 I_3 = 0. Two equations, three unknowns. The loop rule has run out of loops and the circuit is not solved, because nothing has yet said how the current splits.

Junction rule alone. At the node where the branch divides, I1=I2+I3I_1 = I_2 + I_3. One equation, three unknowns, and no way to bring the 12 V12 \ \mathrm{V} into the problem at all. Charge conservation knows nothing about batteries.

Both together. Three equations, three unknowns, and it closes: I2=1.0 AI_2 = 1.0 \ \mathrm{A}, I3=2.0 AI_3 = 2.0 \ \mathrm{A}, I1=3.0 AI_1 = 3.0 \ \mathrm{A}. The full working is the second worked example below.

That is the argument for the pairing in one paragraph. The loop rule fixes how much potential each branch has to spend; the junction rule fixes how the charge divides between them. They constrain different variables, so neither is redundant, and this is the smallest circuit in which you can watch each one fail on its own. The series and parallel guide reduces the same network faster using ReqR_{eq}, and the reason that shortcut works is that the combination rules are themselves derived from these two rules.

When it costs a mark

Flipping the sign of a resistor term halfway round. The single most common loss. Once you have chosen a walking direction, every resistor crossed with the current is IR-IR and every one crossed against it is +IR+IR, for the whole loop and every loop after it.

Reversing a battery term to match the current. A source's contribution depends on which terminal you enter, not on the current direction. Cross from minus to plus and it is +E+\mathcal{E} even if the current is being pushed backwards through that source by a stronger one elsewhere.

Redrawing after a negative current. A negative answer already means the current runs the other way. Restarting costs minutes and introduces new sign errors.

Using the junction rule around a loop, or the loop rule at a node. The clearest symptom is a written equation with a mix of amperes and volts in it. Check the units of every term before solving: a loop equation is entirely volts, a junction equation is entirely amperes.

Counting the same loop twice. Two loops that share every element are one loop. If two of your equations are multiples of each other, you have not gained an equation, and the system will refuse to solve.

Writing all the junction equations. With nn junctions only n1n - 1 are independent. The extra equation is not wrong, it is just empty, and it can disguise the fact that you are short of a loop equation.

"The current is used up." The junction rule forbids it. Whatever enters a series element leaves it. What a bulb consumes is energy, which is the loop rule's business, not charge.

When one rule is enough, and why that hides the pairing

For most of the circuits you meet, one rule sits idle, and that is why students often finish the unit thinking of them as one idea with two names.

A single-loop series circuit has no junctions. No branch points, no choice for the charge, and therefore no junction equation to write. The loop rule alone determines everything, and the statement "the current is the same everywhere" arrives from the geometry rather than from a rule you applied.

A simple parallel pair across an ideal battery needs almost no loop work. Every branch spans the same two nodes, so every branch has the same potential difference by 11.5.A.1.ii, and each branch current comes straight from I=ΔV/RI = \Delta V/R. The junction rule then just totals them. The loop rule is doing something here, but the something is so immediate that it looks like a definition rather than a law.

Reduction by ReqR_{eq} hides both. Collapsing a network to one equivalent resistance and then unwinding it is faster and is the right method in an exam. It is also why the two rules can feel optional: the combination rules already have them baked in.

The pairing becomes unavoidable in exactly one situation, and it is worth recognising it on sight: a circuit with more than one loop and unknown currents in more than one branch. There, neither rule alone produces enough equations, and the CED's own sample activity for Topic 11.6 in AP Physics C asks students to solve a multi-loop problem and then construct a representation for each loop showing the loop rule and one for each junction showing the junction rule. The two are being taught as a pair because they are used as a pair.

Where this sits on the AP exam

The two rules are consecutive topics in Unit 11, Electric Circuits, which carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods.

Topic 11.6, Kirchhoff's Loop Rule, carries suggested skills 1.C, 2.A, 2.C and 3.B. Skill 1.C is creating qualitative sketches of graphs, which is a direct pointer at 11.6.A.4: 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. That graph is the loop rule made visible, and reading or drawing one is a live exam task.

Topic 11.7, Kirchhoff's Junction Rule, carries 1.A, 2.B, 2.C and 3.C. It is the shorter topic, with two essential knowledge statements against the loop rule's four, and the skills lean toward calculating and justifying rather than sketching.

AP Physics C: Electricity and Magnetism states both rules in the same words at its Topic 11.6 and Topic 11.7, with the loop-rule statement and the potential graph filed as 11.6.A.2.i and 11.6.A.2.ii rather than as 11.6.A.3 and 11.6.A.4. The physics does not change; only the numbering does, so cite the numbers for the course you are sitting.

One boundary statement shapes what these rules will be asked about. Under Topic 11.5 both CEDs state that circuits with batteries of different potential differences connected in parallel will not be assessed, which rules out the hardest classical multi-loop exercises. Expect multi-loop networks with a single source, or with sources in series, rather than two competing batteries in separate branches.

For the quantities the two rules constrain, see voltage vs current, and for what the source itself supplies before any of this begins, emf vs voltage.

One loop, two opposing batteries, done with the sign convention in full

A single loop contains a 12 V12 \ \mathrm{V} ideal battery and a 4.0 V4.0 \ \mathrm{V} ideal battery connected so that they drive current in opposite directions, along with a 3.0 Ω3.0 \ \Omega resistor and a 5.0 Ω5.0 \ \Omega resistor. The 12 V12 \ \mathrm{V} battery wins. (a) State the convention you will use. (b) Write the loop equation and find the current. (c) Verify that the potential differences sum to zero. (d) What does the junction rule contribute here?

  1. (a) Declare it before writing anything. Walk the loop in the direction the 12 V12 \ \mathrm{V} battery drives the current, and take the current arrow in that same direction. Crossing a source from minus to plus counts +E+\mathcal{E}, from plus to minus counts E-\mathcal{E}; crossing a resistor with the current counts IR-IR, against it counts +IR+IR.

  2. (b) Walking round: up through the 12 V12 \ \mathrm{V} source, through the 3.0 Ω3.0 \ \Omega with the current, backwards through the 4.0 V4.0 \ \mathrm{V} source (entering its positive terminal), then through the 5.0 Ω5.0 \ \Omega with the current:

  3. 123.0I4.05.0I=012 - 3.0I - 4.0 - 5.0I = 0

  4. Collect: 8.0=8.0I8.0 = 8.0 I, so I=1.0 AI = 1.0 \ \mathrm{A}.

  5. (c) List the four terms with the current substituted: +12 V+12 \ \mathrm{V}, (1.0)(3.0)=3.0 V-(1.0)(3.0) = -3.0 \ \mathrm{V}, 4.0 V-4.0 \ \mathrm{V}, (1.0)(5.0)=5.0 V-(1.0)(5.0) = -5.0 \ \mathrm{V}.

  6. Sum: 123.04.05.0=012 - 3.0 - 4.0 - 5.0 = 0. The loop closes, so the convention was applied consistently.

  7. (d) Nothing. This circuit has no junction: no point in it has three or more conductors meeting, so there is no branch for charge to choose between. The loop rule alone is sufficient, and that is what a single-loop circuit means.

I=1.0 AI = 1.0 \ \mathrm{A}. The loop equation is 123.0I4.05.0I=012 - 3.0I - 4.0 - 5.0I = 0, and the four potential differences +12+12, 3.0-3.0, 4.0-4.0 and 5.0-5.0 volts sum to zero. The junction rule has no work to do in a single-loop circuit.

A two-loop network solved with both rules, then checked against the shortcut

A 12 V12 \ \mathrm{V} ideal battery is in series with a 2.0 Ω2.0 \ \Omega resistor, which feeds a parallel combination of a 6.0 Ω6.0 \ \Omega resistor and a 3.0 Ω3.0 \ \Omega resistor. Let I1I_1 be the current in the 2.0 Ω2.0 \ \Omega, I2I_2 in the 6.0 Ω6.0 \ \Omega and I3I_3 in the 3.0 Ω3.0 \ \Omega. (a) Write the junction equation. (b) Write two independent loop equations. (c) Solve for all three currents. (d) Check the result by reducing the network instead.

  1. (a) The circuit has two junctions, so 21=12 - 1 = 1 independent junction equation. At the node where the branch splits: I1=I2+I3I_1 = I_2 + I_3.

  2. (b) First loop, through the battery, the 2.0 Ω2.0 \ \Omega and the 6.0 Ω6.0 \ \Omega, walking with the currents: 122.0I16.0I2=012 - 2.0 I_1 - 6.0 I_2 = 0.

  3. Second loop, the closed path containing only the two parallel resistors. Walk down through the 6.0 Ω6.0 \ \Omega with I2I_2 and back up through the 3.0 Ω3.0 \ \Omega against I3I_3: 6.0I2+3.0I3=0-6.0 I_2 + 3.0 I_3 = 0, so 3.0I3=6.0I23.0 I_3 = 6.0 I_2 and I3=2I2I_3 = 2 I_2.

  4. (c) Substitute into the junction equation: I1=I2+2I2=3I2I_1 = I_2 + 2I_2 = 3I_2.

  5. Now substitute into the first loop equation: 122.0(3I2)6.0I2=126.0I26.0I2=1212I2=012 - 2.0(3I_2) - 6.0 I_2 = 12 - 6.0I_2 - 6.0I_2 = 12 - 12 I_2 = 0.

  6. So I2=1.0 AI_2 = 1.0 \ \mathrm{A}, giving I3=2.0 AI_3 = 2.0 \ \mathrm{A} and I1=3.0 AI_1 = 3.0 \ \mathrm{A}.

  7. (d) Reduce instead. Parallel pair: 1Rp=16.0+13.0=16.0+26.0=36.0\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}, so Rp=2.0 ΩR_p = 2.0 \ \Omega. Total: 2.0+2.0=4.0 Ω2.0 + 2.0 = 4.0 \ \Omega, and I1=12/4.0=3.0 AI_1 = 12/4.0 = 3.0 \ \mathrm{A}.

  8. Potential difference across the pair: 12(3.0)(2.0)=6.0 V12 - (3.0)(2.0) = 6.0 \ \mathrm{V}. Branch currents: 6.0/6.0=1.0 A6.0/6.0 = 1.0 \ \mathrm{A} and 6.0/3.0=2.0 A6.0/3.0 = 2.0 \ \mathrm{A}. Both routes agree, and the junction check holds: 1.0+2.0=3.0 A1.0 + 2.0 = 3.0 \ \mathrm{A}.

I1=3.0 AI_1 = 3.0 \ \mathrm{A}, I2=1.0 AI_2 = 1.0 \ \mathrm{A}, I3=2.0 AI_3 = 2.0 \ \mathrm{A}. Three unknowns needed three equations: one junction equation and two loop equations. Reduction by equivalent resistance gives the same numbers because the combination rules are derived from these two rules.

The loop rule as a graph of potential against position

A 9.0 V9.0 \ \mathrm{V} ideal battery drives 20 Ω20 \ \Omega, 40 Ω40 \ \Omega and 30 Ω30 \ \Omega resistors in series through ideal wires. Label the negative terminal A, the positive terminal B, and the points after each resistor C, D and back to A. (a) Find the current. (b) Find the potential at each labelled point, taking the potential at A to be zero. (c) Describe the graph of electric potential against position around the loop. (d) Which rule does the graph express, and what would the other rule's graph look like?

  1. (a) Series resistances add: Req=20+40+30=90 ΩR_{eq} = 20 + 40 + 30 = 90 \ \Omega, and I=ΔV/R=9.0/90=0.10 AI = \Delta V/R = 9.0/90 = 0.10 \ \mathrm{A}. The same 0.10 A0.10 \ \mathrm{A} is in all three, since no junction exists between them.

  2. (b) Start at A with VA=0V_A = 0. Crossing the battery from minus to plus is a rise of 9.0 V9.0 \ \mathrm{V}, so VB=9.0 VV_B = 9.0 \ \mathrm{V}.

  3. Across the 20 Ω20 \ \Omega: a drop of IR=(0.10)(20)=2.0 VIR = (0.10)(20) = 2.0 \ \mathrm{V}, so VC=7.0 VV_C = 7.0 \ \mathrm{V}.

  4. Across the 40 Ω40 \ \Omega: a drop of (0.10)(40)=4.0 V(0.10)(40) = 4.0 \ \mathrm{V}, so VD=3.0 VV_D = 3.0 \ \mathrm{V}.

  5. Across the 30 Ω30 \ \Omega: a drop of (0.10)(30)=3.0 V(0.10)(30) = 3.0 \ \mathrm{V}, returning to VA=0V_A = 0. The three drops are 2.0+4.0+3.0=9.0 V2.0 + 4.0 + 3.0 = 9.0 \ \mathrm{V}, matching the source exactly.

  6. (c) The graph steps up by 9.0 V9.0 \ \mathrm{V} at the battery, holds flat along each ideal wire because a wire of negligible resistance has no potential difference across it, then falls by 2.02.0, 4.04.0 and 3.0 V3.0 \ \mathrm{V} in turn, and finishes at the value it started from.

  7. (d) The loop rule, exactly as 11.6.A.4 describes it. The graph closing on its starting value is ΔV=0\sum \Delta V = 0 drawn rather than written. The junction rule has no such graph: it is a statement about currents meeting at a single point, and a point is not a journey.

I=0.10 AI = 0.10 \ \mathrm{A}, with potentials VA=0V_A = 0, VB=9.0 VV_B = 9.0 \ \mathrm{V}, VC=7.0 VV_C = 7.0 \ \mathrm{V}, VD=3.0 VV_D = 3.0 \ \mathrm{V} and back to 00. The graph rises once and falls three times, returning to its start, which is the loop rule in picture form.

Frequently asked questions

What is the difference between Kirchhoff's loop rule and junction rule?

They come from two different conservation laws and constrain two different quantities. The loop rule is a consequence of conservation of energy, per essential knowledge 11.6.A.2, and states that the sum of potential differences across all circuit elements in a single closed loop must equal zero. The junction rule is a consequence of conservation of electric charge, per 11.7.A.1, and states that the total charge entering a junction per unit time equals the total exiting it. Apply the loop rule around a closed path and the junction rule at a node where three or more conductors meet.

What is the sign convention for Kirchhoff's loop rule?

Choose a direction to walk the loop and a direction for each unknown current, then apply four cases consistently. Crossing a source from its negative terminal to its positive terminal contributes plus the emf; crossing it the other way contributes minus the emf. Crossing a resistor in the same direction as the current in it contributes minus I R; crossing against the current contributes plus I R. Set the total to zero. If a current comes out negative, it simply runs opposite to your assumed arrow, and nothing needs redrawing.

Which rule is conservation of energy and which is conservation of charge?

The loop rule is conservation of energy and the junction rule is conservation of charge. The AP Physics 2 CED states both explicitly: 11.6.A.2 says Kirchhoff's loop rule is a consequence of the conservation of energy, and 11.7.A.1 says Kirchhoff's junction rule is a consequence of the conservation of electric charge. A useful memory hook is the unit of the quantity each one constrains: the loop rule adds volts, which are joules per coulomb, and the junction rule adds amperes, which are coulombs per second.

Are Kirchhoff's rules on the AP equation sheet?

No. Neither the sum of potential differences equals zero nor the sum of currents in equals the sum out appears on the AP Physics 2 or the AP Physics C: Electricity and Magnetism equation sheet. Those sheets print the current definition, Ohm's law, electric power, the series and parallel combination rules and the RC time constant, but the two Kirchhoff statements are known rather than looked up. That is a reason to memorise the loop-rule sign convention rather than expect a prompt on exam day.

How many equations do Kirchhoff's rules give you?

A circuit with n junctions yields n minus 1 independent junction equations, because the last one repeats information already contained in the others. Each independent loop, meaning one that includes at least one element no earlier loop used, yields one loop equation. Add the two counts and you need as many equations as you have unknown branch currents. A two-junction, two-loop network gives one junction equation and two loop equations, which is enough for three branch currents.

When do you need both rules instead of just one?

Whenever a circuit has more than one loop and unknown currents in more than one branch. A single-loop circuit has no junctions at all, so the loop rule does everything. A simple parallel pair across a battery is almost entirely handled by the equal-potential-difference rule plus a sum of currents. But once the current splits and you also need the source's potential difference to reach the branches, neither rule produces enough equations on its own: the loop rule fixes how much potential each branch has to spend and the junction rule fixes how the charge divides.