AP® Physics 2: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics-2/units/11/11-7)
Unit 11 · Topic 11.7
11.7 Kirchhoff’s Junction Rule
Kirchhoff's junction rule says the total current into a junction equals the total current out. It's conservation of charge: charge can't pile up or vanish at a point where wires meet. Combined with the loop rule, it lets you solve circuits with several branches.
Key terms
- Kirchhoff's junction rule
- junction
- branch
- conservation of charge
The junction rule
A junction is a point where three or more wires meet, where the current splits or combines. Charge is conserved and doesn't build up at the junction, so in any time interval the charge flowing in equals the charge flowing out:
Think of water pipes: if 5 liters per second flows into a T-joint, 5 liters per second must flow out of the other two pipes combined. This is why the branch currents in a parallel circuit add up to the total current.
Using it with the loop rule
Many circuits can be solved by collapsing series and parallel groups (11.5). When that's hard, or when a question asks you to reason about specific branches, use Kirchhoff's rules directly.
- Label a current in each branch, with an arrow for its direction. If you guess a direction wrong, the answer just comes out negative.
- Write a junction equation for each junction (one fewer than the number of junctions is enough).
- Write a loop equation for each independent loop, using the sign rules from 11.6.
- Solve the equations together. Then check that the battery's power equals the total power used by the resistors.
What the junction rule tells you without math
The rule alone settles many conceptual questions. The current entering a set of parallel branches equals the current leaving them, so a bulb right next to the battery carries the total current, while bulbs in branches share it. An ammeter placed anywhere on the main line, before or after the branches, reads the same value.
When a new branch is added in parallel, the total current from an ideal battery goes up by exactly the new branch's current, and the currents in the existing branches don't change, because each still has the battery's potential difference across it. That's why plugging in another lamp at home doesn't dim the others.
In two parallel branches, the currents split in inverse proportion to the resistances. A branch with half the resistance carries twice the current, because both branches have the same potential difference and I = ΔV/R. Together with the junction rule, that's often enough to find every branch current in your head.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
A busy junction
Four wires meet at a junction. 4.0 A flows in through wire 1, 1.5 A flows in through wire 2, and 2.5 A flows out through wire 3. What is the current in wire 4, and which way does it flow?
Show the solutionHide the solution
- Step 1: Total in so far: 4.0 + 1.5 = 5.5 A. Total out so far: 2.5 A.
- Step 2: To balance, wire 4 must carry 5.5 − 2.5 = 3.0 A out of the junction.
Answer: 3.0 A, flowing out of the junction
- Example 2Calculator allowed
Solving with both rules
A 12 V ideal battery drives a current of 3.0 A through a 2.0 Ω resistor R₁. The current then splits between a 6.0 Ω resistor R₂ and an unknown resistor R₃ in parallel, which rejoin and return to the battery. Find the current in R₂, the current in R₃, and R₃.
Show the solutionHide the solution
- Step 1: Loop rule for the loop through the battery, R₁ and R₂: 12 − (3.0)(2.0) − I₂(6.0) = 0, so 6.0 I₂ = 6.0 and I₂ = 1.0 A.
- Step 2: Junction rule where the current splits: 3.0 = I₂ + I₃, so I₃ = 2.0 A.
- Step 3: Loop rule for the loop through the battery, R₁ and R₃: 12 − 6.0 − (2.0)R₃ = 0, so R₃ = 3.0 Ω.
- Step 4: Check: 6.0 Ω and 3.0 Ω in parallel give 2.0 Ω; plus R₁ makes 4.0 Ω; 12 V / 4.0 Ω = 3.0 A. ✓
Answer: I₂ = 1.0 A, I₃ = 2.0 A, R₃ = 3.0 Ω
Common mistakes
- Thinking current splits equally at every junction. It splits so that each branch's current fits its resistance; less resistance gets more current.
- Treating a bend in a wire as a junction. A junction needs at least three wires meeting.
- Giving up when a current comes out negative. That just means it flows opposite to the arrow you drew.
On the exam
- Expect to justify current rankings with conservation of charge: the current into a group of branches equals the current out.
- Multi-part circuit problems often need one junction equation and one or two loop equations; write them clearly with labeled currents.
Connected topics
Videos
Check yourself
4 questions on 11.7 Kirchhoff’s Junction Rule. Pick an answer to see if you got it, and why.
Currents of 3.0 A and 2.0 A flow into a junction. One wire carries 4.0 A out of the junction. What is the current in the only other wire at the junction?
Why must the total current into a junction equal the total current out of it?
A battery is connected in a single loop with one lightbulb. An ammeter reads the current in the wire just before the bulb, and a second ammeter reads it just after the bulb. How do the readings compare?
A total current of 4.0 A enters a pair of parallel resistors, 2.0 Ω and 6.0 Ω. What is the current in the 2.0 Ω resistor?
0 of 4 answered