MakersDeckfor Electrical and Electronics Engineering

Kirchhoff's Current Law

DC Circuits · 10 min read

KVL governs voltage around a loop; KCL governs current at a junction. Both laws flow from the same source — conservation — voltage from energy, current from charge.

I₁3 AI₂2 AI₃5 AI₁ + I₂ = I₃3 + 2 = 5 A
All entering currents sum to all leaving currents — charge is never stored in a wire junction.

The statement

At any node, the sum of all currents entering equals the sum of all currents leaving.

∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}}

Assign a sign (+ in, − out) and every current sums to zero:

∑nodeIk=0\sum_{\text{node}} I_k = 0
KCL is conservation of charge. A wire junction stores no charge — whatever flows in must flow straight out.

Nodes and branches

A node is any point where three or more wires meet. A branch is the component path between two nodes.

  • A series loop has two nodes — KCL gives no new information (same current everywhere).
  • A parallel bank has a fan-out node (current splits) and a fan-in node (currents merge).
  • Multi-loop circuits contribute one independent KCL equation per node to the system you solve.

Worked example — 3-branch node

Two currents enter a node (4 A and 3 A) and one unknown current leaves. Find I₃.

  1. Write KCL. I1+I2=I3I_1 + I_2 = I_3
  2. Substitute. I3=4+3=7 AI_3 = 4 + 3 = 7\,\text{A}
  3. Check direction. Positive result confirms I₃ leaves the node. A negative answer would mean the assumed outward direction was wrong — flip the arrow and use the magnitude.

Interactive playground

Adjust I₁ and I₂ — I₃ updates automatically to keep KCL satisfied at every step.

KCL at this node
I₁ + I₂ = I₃
3.0 + 1.0 = 4.0
I₃ (out) = 4.0 A
✓ KCL always satisfied

KCL in parallel circuits

In a parallel bank, the source current enters the fan-out node and splits into branch currents.

Isource=I1+I2+I3+…I_{\text{source}} = I_1 + I_2 + I_3 + \ldots

Each branch sees the same source voltage, so Ohm's Law gives Ik=V/RkI_k = V/R_k for every branch. Substituting that into the KCL sum above and dividing through by V produces the familiar parallel-resistance formula:

1Rtotal=1R1+1R2+1R3+…\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots

That formula isn't a separate rule to memorize — it's KCL, with Ohm's Law substituted in, at the fan-out node.

Common mistakes

Predict the outcome

Scenario 1: A node has I₁ = 4 A in, I₂ = 2 A in, and I₃ outgoing (unknown). What is I₃?

Scenario 2: A 4-branch node has 8 A in, 3 A out, 2 A out, and I₄ unknown. What is I₄ and which direction does it flow?

Scenario 3: If you double every branch current at a node, does KCL still hold?