Node-Voltage Method

Cut your unknowns in half by solving for node voltages instead of branch currents.

Circuit AnalysisElectrical Engineering Year 1
⏱️ About 18 min

You're debugging a power distribution board with three interconnected nodes and need every voltage fast — without tracing every branch current by hand.

💡
The big idea: Pick one node as ground, express every branch current as a voltage difference over a resistance, and let KCL produce one equation per unknown.
🎯 By the end, you'll be able to
  • Select a reference node and label unknown node voltages
  • Write KCL equations at each non-reference node using Ohm's law in terms of node voltages
  • Solve the resulting system of equations for all node voltages
  • Compute any branch current from the solved node voltages
📎 Helpful to know first
  • Kirchhoff's Laws & Voltage/Current Dividers

From Branch Currents to Node Voltages

In Module 1 you solved circuits by labeling every branch current and applying KCL and KVL directly. As circuits grow, the number of branch currents multiplies fast. The node-voltage method cuts the unknowns down: instead of tracking every current, you pick one node as ground and solve for the voltage at every other node. Every branch current then follows from a simple Ohm's-law subtraction.

🔑 The core idea

A circuit with N nodes needs only N − 1 unknown voltages. Each unknown gets one KCL equation. Branch currents are computed afterward — never solved for directly.

The Three-Step Procedure

Step 1 — Pick a reference node. Choose one node as ground (0 V). A good choice is the node with the most connections, which minimizes the number of terms in your equations.

Step 2 — Label unknowns. Assign a voltage variable (V1, V2, …) to every non-reference node.

Step 3 — Write KCL at each non-reference node. Sum the currents leaving the node through each connected resistor. For a resistor R between node voltages Va and Vb, the current leaving node A toward node B is (Va − Vb)/R. Set the sum equal to any current source injecting into the node.

\[ I_{A \to B} = \frac{V_A - V_B}{R} \]
Current leaving node A through a resistor R connected to node B. This one expression replaces an entire branch-current variable.
📝 Worked example: A 4 A current source feeds node 1. R1 = 2 Ω connects node 1 to ground, R2 = 2 Ω connects node 1 to node 2, and R3 = 4 Ω connects node 2 to ground. Find V1 and V2.
  1. Step 1 — Choose the bottom node as ground (0 V). Unknowns: V1 and V2.
  2. Step 2 — KCL at node 1 (currents leaving): V1/2 + (V1 − V2)/2 = 4
  3. Step 3 — KCL at node 2 (currents leaving): (V2 − V1)/2 + V2/4 = 0
  4. Step 4 — Simplify eq. 1 (multiply by 2): 2V1 − V2 = 8
  5. Step 5 — Simplify eq. 2 (multiply by 4): −2V1 + 3V2 = 0, so V1 = 3V2/2
  6. Step 6 — Substitute: 2(3V2/2) − V2 = 8 → 2V2 = 8 → V2 = 4 V
  7. Step 7 — Back-substitute: V1 = 3(4)/2 = 6 V
✓ V1 = 6 V, V2 = 4 V
✏️ Practice: A 3 A current source feeds node 1. R1 = 1 Ω connects node 1 to ground, R2 = 1 Ω connects node 1 to node 2, and R3 = 1 Ω connects node 2 to ground. Find V1.
V
Solution
  1. KCL at node 1: V1 + (V1 − V2) = 3 → 2V1 − V2 = 3
  2. KCL at node 2: (V2 − V1) + V2 = 0 → V1 = 2V2
  3. Substitute: 2(2V2) − V2 = 3 → 3V2 = 3 → V2 = 1 V
  4. V1 = 2(1) = 2 V

Check your understanding

1. In the node-voltage method, how many independent KCL equations are needed for a circuit with N total nodes (including the reference)?
One node is chosen as the reference (ground), leaving N − 1 non-reference nodes. Each gets one KCL equation.
2. Node A is at 10 V and node B is at 4 V, connected by a 3 Ω resistor. What is the current from A to B through the resistor?
Current from A to B = (Va − Vb)/R = (10 − 4)/3 = 2 A.
🔎 Related lessons
Mesh-Current Method
✅ Key takeaways
  • Choose one node as the reference (0 V); each remaining node gets one unknown voltage.
  • Express every resistor current as (Va − Vb)/R and apply KCL at each non-reference node.
  • The result is a linear system — solve by substitution or matrix methods.
  • Once node voltages are known, any branch current follows directly from Ohm's law.
➡️ Next, we'll flip the approach — instead of node voltages, we'll solve for loop currents using the mesh-current method.
Want to test yourself on this? Try the Electrical Aptitude test →