Write a KCL equation at each non-reference node using Ohm's law.
Solve the resulting system for all node voltages.
Recover any branch current or power from node voltages.
KCL and KVL plus Ohm's law can crack any resistive circuit, but applying them ad hoc drowns you in unknowns. Nodal analysis is the systematic recipe — it is, almost verbatim, the algorithm inside every circuit simulator ever written. The recipe: (1) pick one node as the reference (ground, 0 V) — choose the one with the most connections; (2) label the remaining node voltages v_1, v_2, …; (3) write KCL at each labelled node, expressing every resistor current through Ohm's law; (4) solve the linear system.
The workhorse move is expressing a resistor's current from node voltages: the current flowing from node n to node m through resistance R is their voltage difference over R.
in→m=Rvn−vmbranch current from node voltages
The worked example: two node voltages, two KCL equations. Reference (ground) at the bottom rail.WORKED EXAMPLE
Two nodes, start to finish
In the figure: 1 mA enters node 1, 4 mA enters node 2, R_1 = 2 kΩ (node 1 to ground), R_2 = 8 kΩ (between nodes), R_3 = 4 kΩ (node 2 to ground).
KCL at node 1 (out = in): v_1/2 + (v_1 − v_2)/8 = 1. Multiply by 8: 4v_1 + v_1 − v_2 = 8, so 5v_1 − v_2 = 8.
KCL at node 2: v_2/4 + (v_2 − v_1)/8 = 4. Multiply by 8: 2v_2 + v_2 − v_1 = 32, so 3v_2 − v_1 = 32.
From the first, v_2 = 5v_1 − 8. Substitute: 3(5v_1 − 8) − v_1 = 32 → 14v_1 = 56 → v_1 = 4 V, v_2 = 12 V.
Verify with KCL at node 1: R_1 carries 4/2 = 2 mA down; R_2 carries (4 − 12)/8 = −1 mA, i.e. 1 mA flowing INTO node 1 from node 2. In: 1 + 1 = 2 mA. Out: 2 mA. ✓
v_1 = 4 V, v_2 = 12 V
Notice what the negative R_2 current did in the verification: the algebra quietly told us current actually flows from node 2 to node 1, and no re-drawing was needed. That is the entire case for the systematic method — directions are assumptions, and the signs sort out reality.
WORKED EXAMPLE
A voltage source to ground is a free answer
If an ideal voltage source connects a node directly to ground, that node's voltage is not an unknown — it equals the source value by definition.
A 3.3 V source from node 3 to ground means v_3 = 3.3 V. Write no KCL there; substitute the value everywhere else.
(A voltage source between two NON-reference nodes is trickier — that's Lesson 4.)
Known source-to-ground nodes reduce the system by one equation each.
PRACTICE
Q1
Set up only (no solving): node 1 connects through R_a to ground, through R_b to node 2, and receives a source current i_in. Write the KCL equation at node 1.
Q2
A single node receives 4 mA and connects to ground through 3 kΩ and 6 kΩ (both to ground). Find the node voltage.
Q3
Repeat the worked example's circuit shape with new values: 1 mA into node 1, 3 mA into node 2, R_1 = 2 kΩ, R_2 = 4 kΩ (between nodes), R_3 = 4 kΩ. Find both node voltages.
Q4
In practice problem 3, find the power delivered by each current source and absorbed by each resistor, and confirm Σp = 0.