Kirchhoff’s current law states that the algebraic sum of currents at any node of a lumped circuit is zero: the total current entering a node equals the total current leaving it, because charge cannot accumulate at an ideal node.
Learning objectives
- State Kirchhoff’s current law and explain its link to charge conservation.
- Write a node equation with a consistent sign convention.
- Calculate an unknown branch current from known currents.
- Apply KCL to parallel circuits and to current measurement.
Progressive explanation
Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.
Beginner The core idea
Think of water pipes meeting at a tee. The water arriving at the tee must equal the water leaving it; a pipe that carries water in must have somewhere for it to go. In circuits, the junction is called a node, and the rule applies to every node of the circuit at every instant.
Intermediate How it behaves in a circuit
When applying KCL, label each branch current with a reference arrow before writing equations. A current chosen in the opposite direction to the real flow simply comes out negative, which is not an error. For two resistors in parallel fed by a total current I, KCL gives I = I₁ + I₂, and the current sharing follows the resistance values.
Advanced Limits, assumptions, engineering detail
In nodal analysis, KCL is written for every non-reference node, giving a system of equations in node voltages. The reference node is chosen arbitrarily and its voltage is set to zero. KCL also explains why a current measurement around a whole multi-conductor cable reads zero: the algebraic sum of the conductor currents inside the clamp is zero when no current leaves through a path outside the clamp.
Formulas and variables
Units of measurement
| Quantity | Symbol | Unit |
|---|---|---|
| Electric current | I | ampere (A) |
| Charge | Q | coulomb (C) |
| Time | t | second (s) |
Worked examples
Unknown branch current at a node
Problem. At a junction, 5.0 A enters. Two branches leave with 2.0 A and 1.2 A. Find the current in the third outgoing branch.
Solution. KCL: ΣI_in = ΣI_out, so 5.0 = 2.0 + 1.2 + I₃. Therefore I₃ = 5.0 − 3.2 = 1.8 A.
Parallel branch currents
Problem. A 12 V source feeds a 6 Ω and a 3 Ω resistor in parallel. Use KCL to find the total current.
Solution. Each branch has 12 V: I₁ = 12/6 = 2.0 A and I₂ = 12/3 = 4.0 A. KCL gives I_total = 2.0 + 4.0 = 6.0 A.
Practical applications
- Nodal analysis of circuits
- Current distribution in parallel loads
- Fuse and feeder sizing
- Return-current and ground-path analysis
Automotive applications
KCL explains why the current flowing out of the battery positive terminal must return through the ground path. A missing ground strap forces return current through unintended paths, such as the chassis, bearings or a small signal wire, which can overheat or cause electrical noise. Measuring current in each return path is a systematic way to find an abnormal ground current.
Industrial applications
In industrial distribution, KCL governs the current balance in a panel: the current entering a busbar must leave through the outgoing feeders and the neutral or protective conductor according to the system configuration. In three-phase systems with unbalanced loads, the neutral current is the phasor sum of the phase currents, so an open neutral can cause dangerous phase voltage shifts.
Measurement procedure
- Measure each branch current individually and add them with their assigned directions.
- A clamp around several conductors reads the algebraic sum, which is useful for checking residual current.
- Use a current probe and oscilloscope for transient or non-sinusoidal currents where a single RMS reading is not enough.
- Confirm the measurement by checking that the entering and leaving currents balance within the meter accuracy.
Common mistakes
- Forgetting the reference direction and adding currents with the wrong sign.
- Applying KCL at a node that is not a defined lumped node, where significant capacitance or displacement current exists.
- Assuming current splits equally between unequal parallel resistors.
- Ignoring current that leaves through an unmeasured return or ground path.
Safety considerations
- Current measurements require breaking the circuit or using a properly rated clamp; never create a parallel path through the meter.
- Use equipment rated for the circuit voltage and for the prospective fault current.
- Treat unexpected current in a ground or chassis path as a serious fault until it is explained.