AP Physics 2- 11.7 Kirchhoff's Junction Rule- Study Notes- New Syllabus
AP Physics 2- 11.7 Kirchhoff’s Junction Rule – Study Notes
AP Physics 2- 11.7 Kirchhoff’s Junction Rule – Study Notes – per latest Syllabus.
Key Concepts:
- Kirchhoff’s Junction Rule
Kirchhoff’s Junction Rule
Kirchhoff’s junction rule states that the algebraic sum of currents at a junction is zero. This is a consequence of the law of conservation of charge.
\(\displaystyle \sum I_{\text{in}} = \sum I_{\text{out}}\)
Physical Basis:
- Charge cannot accumulate at a junction.
- All charge that flows into a junction must also flow out.
Applying the Rule:
- Assign current directions for each branch (arbitrary choices are okay).
- Define currents entering the junction as positive, leaving as negative (or vice versa, but stay consistent).
- Write the current balance equation and solve for unknowns.
Example Situations:
- At a node where one wire splits into two, the current in the single branch equals the sum of the currents in the two branches.
- In parallel circuits, junctions distribute current according to branch resistance.
Example:
A current of \(I = 6 \, \text{A}\) flows into a junction and splits into two branches. One branch carries \(I_1 = 4 \, \text{A}\). Find the current in the second branch \(I_2\).
▶️ Answer/Explanation
Step 1: Apply junction rule: \(I = I_1 + I_2\).
Step 2: Substitute values: \(6 = 4 + I_2\).
Step 3: Solve: \(I_2 = 2 \, \text{A}\).
Answer: The second branch current = \(2 \, \text{A}\).
Example:
At a junction, three currents meet. \(I_1 = 5 \, \text{A}\) enters, while \(I_2 = 3 \, \text{A}\) and \(I_3\) leave. Find \(I_3\).
▶️ Answer/Explanation
Step 1: Junction rule: \(I_1 = I_2 + I_3\).
Step 2: Substitute: \(5 = 3 + I_3\).
Step 3: Solve: \(I_3 = 2 \, \text{A}\).
Answer: The current in the third branch = \(2 \, \text{A}\).