Kirchhoff's Current Law (KCL)

Definition

Kirchhoff's Current Law (KCL) states that the algebraic sum of currents entering any junction (or node) in a circuit is equal to zero.

Or equivalently:

The sum of currents entering a junction equals the sum of currents leaving that junction.

Formula

$$\sum I_{\text{in}} = \sum I_{\text{out}}$$

Or using algebraic form:

$$\sum I = 0$$

Where currents entering are positive and currents leaving are negative (or vice versa — just be consistent).


Physical Basis

KCL is based on the conservation of charge:

  • Charge cannot accumulate at a junction
  • Whatever charge flows in must flow out
  • No charge is created or destroyed at the junction

What is a Junction/Node?

A junction (or node) is a point in a circuit where:

  • Three or more circuit elements meet
  • Currents from different branches combine or split
      I₁
       ↓
I₂ → [Node] → I₃
       ↓
       I₄

For this junction: $I_1 + I_2 = I_3 + I_4$


Sign Conventions

Convention Current Entering Current Leaving
Option 1 Positive (+) Negative (−)
Option 2 Negative (−) Positive (+)

Important: Choose one convention and stick to it throughout the problem.


Worked Examples

Example 1: Simple Junction

Three currents meet at a junction:

  • $I_1 = 2$ A entering
  • $I_2 = 5$ A entering
  • $I_3 = ?$ leaving

Solution: $$I_1 + I_2 = I_3$$ $$2 + 5 = I_3$$ $$I_3 = 7 \text{ A}$$

Example 2: Circuit with Multiple Junctions

For a circuit with unknown currents $I_1$, $I_2$, $I_3$:

At junction A: $I_1 = I_2 + I_3$

This gives one equation. More equations come from KVL applied to loops.


Application Tips

  1. Label currents first: Assign variables to all unknown currents
  2. Guess directions: If you guess wrong, you'll get a negative value — that's fine!
  3. Count junctions: For $n$ junctions, you get $(n-1)$ independent KCL equations
  4. Combine with KVL: Need both KCL and KVL to solve multi-loop circuits

Related Concepts


Mermaid Diagram: KCL at a Junction

graph LR
    A[I₁ = 3A] -->|Entering| J((Junction))
    B[I₂ = 2A] -->|Entering| J
    J -->|Leaving| C[I₃ = ?]
    J -->|Leaving| D[I₄ = 1A]
    
    style J fill:#90EE90

Calculation: $I_3 = I_1 + I_2 - I_4 = 3 + 2 - 1 = 4$ A