Worked examples: KCL at a branching node

Two short problems — one parallel bank, one three-way junction with a missing branch — are enough to turn KCL from a formula into a reflex.

Example 1 — parallel bank, source current from branches

A 12 V source drives three parallel resistors: 4 Ω, 6 Ω, and 12 Ω. Find the source current by computing each branch current with Ohm's Law and applying KCL at the fan-out node.

1. Branch currents. Every branch sees the full 12 V:
   $$I_1 = 12/4 = 3\,\text{A},\ \ I_2 = 12/6 = 2\,\text{A},\ \ I_3 = 12/12 = 1\,\text{A}$$
2. KCL at the fan-out node. $I_{\text{source}} = I_1 + I_2 + I_3$
   $$I_{\text{source}} = 3 + 2 + 1 = 6\,\text{A}$$
3. Cross-check. $1/R_{\text{total}} = 1/4 + 1/6 + 1/12 = 1/2$, so $R_{\text{total}} = 2\,\Omega$ and $I = 12/2 = 6\,\text{A}$. KCL and the shortcut agree.

Example 2 — find a missing branch current

At a single node, four branches meet. Three currents are known: 5 A into the node (branch A), 2 A out (branch B), 1 A out (branch C). Branch D's current is unknown. Find it.

1. Sign convention. Call inbound positive. A contributes +5, B contributes −2, C contributes −1, D contributes $I_D$ (sign unknown).
2. Write KCL.
   $$+5 - 2 - 1 + I_D = 0$$
3. Solve. $I_D = -2\,\text{A}$. Negative means branch D carries 2 A flowing out of the node — opposite to the assumed sign.

Check: 5 A in, 2 + 1 + 2 = 5 A out. Charge conserved. ✓