AP® Physics C: Electricity and Magnetism review sheet from Aim for Five (aimforfive.com/physics-c-em/units/11/11-7)
Unit 11 · Topic 11.7
11.7 Kirchhoff’s Junction Rule
Kirchhoff's junction rule says the total current into any junction equals the total current out. It comes from conservation of charge, and paired with the loop rule it lets you solve any multi-loop circuit as a set of equations.
Key terms
- Kirchhoff’s junction rule
- junction
- conservation of charge
- branch current
The junction rule
A junction is a point where three or more wires meet. Charge can't pile up at a junction or vanish there, so whatever flows in each second must flow out: .
Think of water pipes meeting. If 5 liters per second flow into a joint and one outlet carries 2 liters per second, the other must carry 3. Current works the same way, because charge is conserved.
Each section of wire between two junctions is a branch, and every element in one branch carries the same branch current. That's why series elements share a current: they're in the same branch.
Solving a multi-loop circuit
This method works for any circuit, including ones that can't be reduced to series and parallel groups, and it's also a good check on a series-parallel answer.
- Label a current in every branch with a guessed direction, like I₁, I₂ and I₃.
- Write junction equations. A circuit with j junctions gives j − 1 independent ones.
- Write loop equations (11.6) for enough loops that each branch is included at least once.
- You need as many independent equations as unknown currents. Solve them by substitution or elimination.
- A negative current means that branch's current flows opposite to your guess.
Connecting the two rules
The junction rule is conservation of charge; the loop rule is conservation of energy. Together they're the whole toolbox for steady circuits. The series and parallel formulas from 11.5 are shortcuts built from these two rules: parallel branches add currents because of the junction rule, and they share a potential difference because of the loop rule.
Ideal ammeters fit naturally here: an ammeter in a branch reads that branch's current, so a junction equation tells you what a missing ammeter would read.
The junction rule still holds in a branch with a charging capacitor. Charge piles up on one plate, but the same amount leaves the other plate, so the current into the capacitor's branch equals the current out of it at every moment.
Checking your answers
Once you have every current, you have two quick checks. First, walk a loop you didn't use in your equations, such as the outer edge of the circuit. The potential changes should still add to zero.
Second, check energy. The power the batteries supply, , should equal the power all the resistors use, . If they don't match, a sign is wrong somewhere, often a resistor crossed against its current.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Setting up and solving Kirchhoff's equations
A 12 V ideal battery is connected in series with a 2.0 Ω resistor R₁. The current then splits at junction P between R₂ = 3.0 Ω and R₃ = 6.0 Ω, which rejoin before returning to the battery. Use Kirchhoff's rules to find each current.
Show the solutionHide the solution
- Step 1: Label I₁ through the battery and R₁, toward P. Label I₂ through R₂ and I₃ through R₃, both away from P.
- Step 2: Junction rule at P: .
- Step 3: Loop through the battery, R₁ and R₂: .
- Step 4: Loop through R₂ and R₃ only (walking down through R₂ with its current and up through R₃ against its current): , so .
- Step 5: Substitute: , so , giving . So A, A and A.
- Step 6: Check with 11.5: R₂ and R₃ in parallel make 2.0 Ω, plus 2.0 Ω gives 4.0 Ω, and 12/4.0 = 3.0 A. It matches.
Answer: I₁ = 3.0 A, I₂ = 2.0 A, I₃ = 1.0 A
- Example 2Calculator allowed
A negative current (classic trap)
Three wires meet at junction P. Wire 1 carries 3.0 A into P and wire 2 carries 5.0 A out of P. A student labels the current in wire 3 as I₃, flowing out of P. Find the actual current in wire 3.
Show the solutionHide the solution
- Step 1: Junction rule with the student's labels: in = out, so .
- Step 2: .
- Step 3: The negative sign means the guess was backward: wire 3 carries 2.0 A into P. Check: 3.0 + 2.0 = 5.0 A in, and 5.0 A out.
- Step 4: The trap is reporting "−2.0 A out of P" without interpreting it, or quietly dropping the sign and getting other branches wrong later.
Answer: 2.0 A flowing into P
Common mistakes
- Writing a junction equation for every junction. With j junctions, only j − 1 equations are independent.
- Giving two different labels to the current in one branch. Every element in a single branch carries the same current.
- Changing a current's assumed direction between equations. Keep each label's direction fixed throughout.
- Forgetting that a negative result is a direction, not an error.
On the exam
- Expect to write junction and loop equations in terms of labeled currents and to justify a rule by naming what it conserves: charge for junctions, energy for loops.
- Multiple-choice questions often show a junction with some currents labeled and ask for the missing one. Add up everything in and everything out.
Connected topics
Videos
Check yourself
3 questions on 11.7 Kirchhoff’s Junction Rule. Pick an answer to see if you got it, and why.
Four wires meet at a junction. Wire 1 carries 5.0 A into the junction, wire 2 carries 1.5 A into the junction, and wire 3 carries 2.0 A out of the junction. What is the current in wire 4?
Which statement best explains why Kirchhoff’s junction rule holds in a circuit with steady currents?
Students test the junction rule using a junction where wire 1 carries current in and wires 2 and 3 carry current out. Their ammeters are accurate to about ±0.01 A. Which set of readings, in amperes, best supports the rule?
0 of 3 answered