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🔗Kirchhoff’s Laws

Master circuit analysis with KCL and KVL — solve any linear circuit systematically.

KCL — Current Law

The algebraic sum of currents at any node is zero (what goes in = what comes out). Charge cannot accumulate at a node. In water analogy: water flowing into a pipe junction equals water flowing out. KCL is based on conservation of charge. Mathematically: Σ I_in = Σ I_out or Σ I = 0 (with sign convention).
Total current entering a node = Total current leaving

KVL — Voltage Law

The algebraic sum of voltages around any closed loop is zero. As you travel around a loop, voltage rises (batteries) must equal voltage drops (resistors). KVL is based on conservation of energy — the electric field is conservative. Pick a direction, assign polarities, sum to zero.
Sum of all voltages around a closed loop = 0

Sign Conventions

For KVL: Going through a resistor in the direction of current: -IR (voltage drop). Going against: +IR (voltage rise). Going through a battery from - to +: +V (rise). Going from + to -: -V (drop). For KCL: Current entering node: positive. Leaving: negative. Or vice versa — just be consistent.

🎮 Interactive: Voltage Divider

🎮 Interactive: Voltage Divider
Adjust R1 and R2 — verify KVL: V1 + V2 always equals supply voltage.
Voltage Divider: 12V → R1 → R2 → GND
Current
40.0mA
V across R1
4.00V
V across R2
8.00V
KVL Check: 12V = 4.0V + 8.0V = 12.0V ✓

🔗 Visual: KCL at a Node

🔗 Visual: KCL at a Node
Current entering a node equals current leaving. Arrows show flow direction.
Kirchhoff's Current Law (KCL)Σ I_in = Σ I_out at any node3AI₁=3A2AI₂=2AI₃=1A1AI₄=4A4AI_in: 3+2=5A | I_out: 1+4=5A ✓

💡 Mesh Analysis Example

💡 Mesh Analysis Example
Circuit: 10V source, R1=2Ω in series with parallel (R2=4Ω, R3=12Ω).
1. Find R_parallel: 1/(1/4+1/12) = 3Ω
2. Total R: 2+3 = 5Ω
3. Total I: 10/5 = 2A
4. V across parallel: 2×3 = 6V
5. I_R2: 6/4 = 1.5A | I_R3: 6/12 = 0.5A
6. Check KCL: 2A = 1.5A + 0.5A ✓ | KVL: 10V = 4V(R1) + 6V(parallel) ✓