Kirchhoff's Laws

🔀 KCL — Kirchhoff's Current Law

Current at a Node — Conservation of Charge

KCL Statement

The sum of all currents entering a node equals the sum of all currents leaving that node.

Σ I_in = Σ I_out

or equivalently:
Σ I = 0 (define + for entering)

Based on conservation of charge.

Identifying Nodes & Branches

Node: Point where 2+ elements connect.
Branch: Path carrying a single current.
Loop: Closed path returning to start.

💡 Elements in series share exactly the same current — they are in the same branch.

🔁 KVL — Kirchhoff's Voltage Law

KVL Loop — Conservation of Energy

KVL Statement

The algebraic sum of all voltage drops around any closed loop equals zero.

Σ ΔV = 0 (closed loop)

Based on conservation of energy.

Sign Convention for KVL

Traveling in the direction of current through a resistor: −V drop (voltage falls).
Traveling from − to + through a source: +V gain (voltage rises).

⚠️ Keep your chosen loop direction consistent (CW or CCW). Don't switch mid-loop.

🔧 How to Solve Multi-Loop Circuits

1

Draw the circuit and label all known/unknown quantities.

2

Assign a current direction to each branch (guess if needed — the math will correct it).

3

Apply KCL at each node: Σ I_in = Σ I_out.

4

Apply KVL to independent loops: Σ ΔV = 0.

5

Solve the simultaneous equations for unknown currents/voltages.

6

If a current is negative, its actual direction is opposite to your assumed direction.

📝 Solved Example — Two Loops

Circuit: 30V source, R₁=8Ω (left branch), R₂=3Ω (middle), R₃=6Ω (right). Find i₁, i₂, i₃.

KCL at node: i₁ = i₂ + i₃ … (1)
KVL loop 1: −30 + 8i₁ + 3i₂ = 0 … (2)
KVL loop 2: −3i₂ + 6i₃ = 0 … (3)
─────────────────────────────────
Solving → i₁ = 3A, i₂ = 2A, i₃ = 1A
Verify: v₁=24V, v₂=6V, v₃=6V (24+6=30 ✓)

÷ Voltage & Current Divider Rules

Voltage Divider

Two resistors in series, same current I. Voltage splits proportionally.

v₁ = V_S × R₁/(R₁+R₂)
v₂ = V_S × R₂/(R₁+R₂)

💡 Each resistor gets a fraction = its own R / total R.

Current Divider

Two resistors in parallel, same voltage V. Current splits inversely to resistance.

i₁ = I_S × R₂/(R₁+R₂)
i₂ = I_S × R₁/(R₁+R₂)

💡 Each branch gets a fraction = OTHER resistor / total R. (Opposite of voltage divider!)

Power Sign Convention

P = v·i
P > 0 → absorbing power
P < 0 → supplying power

Σ P_absorbed = Σ P_supplied (conservation of energy checks out). Use this to verify your solution.