Nodal analysis (node-voltage method) is the most systematic way to solve an electric circuit. It turns any linear circuit into a small set of simultaneous equations based on Kirchhoff’s Current Law (KCL). It is a regular topic in university exams, polytechnic finals and EEE job tests (BCS, PGCB, BPDB, DESCO, BREB): “how many node equations?”, “what is a supernode?”, and short numerical problems. This post explains the method step by step, with an animated worked example, a supernode example, a comparison with mesh analysis and practice MCQs.
- Nodal analysis uses KCL at each node, with the node voltages as unknowns.
- A circuit with N nodes needs N − 1 independent node equations (one node is the reference, 0 V).
- Current leaving a node through a resistor R = (Vnode − Vother)/R.
- A voltage source between two non-reference nodes forms a supernode: write one KCL for the combined region plus the source’s constraint equation.
- Matrix form: [G][V] = [I], where G is the conductance matrix.
1. What Is Nodal Analysis?
A node is a point where two or more circuit elements meet. A principal (essential) node joins three or more elements. In nodal analysis we:
- choose one node as the reference (ground), at 0 V;
- assign an unknown voltage (V1, V2, …) to every other node, measured from the reference;
- apply KCL at each of these nodes, expressing every branch current through Ohm’s law in terms of the node voltages;
- solve the resulting simultaneous equations.
Once the node voltages are known, every branch current and power follows from Ohm’s law. The method works with resistors, independent and dependent sources, and with impedances in AC circuits.
KCL: Σ currents leaving a node = 0 Ibranch = (Va − Vb) / R = (Va − Vb) G2. Step-by-Step Procedure
- Identify all nodes and select the reference node, usually the node with the most connections or the bottom rail. Mark it with the ground symbol.
- Label node voltages V1, V2, … for the remaining N − 1 nodes.
- Write KCL at each non-reference node. Assume all unknown currents leave the node: for each resistor write (Vthis − Vneighbour)/R. A current source pointing into the node enters the equation with a minus sign when it is moved to the “leaving” side.
- Simplify into standard form (collect V1, V2 terms).
- Solve by substitution, elimination, Cramer’s rule or matrix inversion.
- Back-substitute to find branch currents, voltages and power, and check KCL at each node.
3. Solved Example (Animated)
Problem: In the circuit below, a 20 V source feeds node 1 through R1 = 2 Ω. R2 = 4 Ω connects node 1 to ground, R3 = 4 Ω connects node 1 to node 2, R4 = 4 Ω connects node 2 to ground, and a 1 A current source feeds node 2. Find V1, V2 and all branch currents.
KCL at node 1 (currents leaving):
(V1 − 20)/2 + V1/4 + (V1 − V2)/4 = 0 ⟹ 4V1 − V2 = 40 … (1)KCL at node 2 (the 1 A source enters the node, so it appears as −1 on the “leaving” side):
(V2 − V1)/4 + V2/4 − 1 = 0 ⟹ −V1 + 2V2 = 4 … (2)Solve: from (1), V2 = 4V1 − 40. Substituting in (2): −V1 + 8V1 − 80 = 4, so 7V1 = 84. Therefore
V1 = 12 V, V2 = 4(12) − 40 = 8 V| Branch | Current | Direction |
|---|---|---|
| R1 (2 Ω) | (20 − 12)/2 = 4 A | source → node 1 |
| R2 (4 Ω) | 12/4 = 3 A | node 1 → ground |
| R3 (4 Ω) | (12 − 8)/4 = 1 A | node 1 → node 2 |
| R4 (4 Ω) | 8/4 = 2 A | node 2 → ground |
Check: at node 1, 4 A in = 3 A + 1 A out ✔. At node 2, 1 A (R3) + 1 A (source) in = 2 A out ✔. Power delivered by the 20 V source = 20 × 4 = 80 W.
4. Matrix (Inspection) Method
For circuits containing only resistors and independent current sources, the node equations can be written directly by inspection:
[G][V] = [I]- Gkk (diagonal) = sum of all conductances connected to node k.
- Gkj (off-diagonal) = − (sum of conductances directly between nodes k and j). G is therefore symmetric.
- Ik = sum of source currents entering node k. A voltage source in series with a resistor can first be converted to a current source (source transformation).
For the example, the 20 V source with 2 Ω becomes a 10 A source in parallel with 2 Ω:
| 1/2 + 1/4 + 1/4 | −1/4 |
| −1/4 | 1/4 + 1/4 |
| V1 |
| V2 |
| 10 |
| 1 |
5. Supernode
If a voltage source sits between two non-reference nodes, its current is unknown and cannot be written with Ohm’s law. The two nodes and the source together are treated as one supernode:
- Apply KCL to the closed boundary around both nodes (ignore the source’s internal current).
- Add the constraint equation from the source: Va − Vb = Vs.
Example: A 6 A current source feeds node 1. A 6 V source is connected between node 1 (+) and node 2 (−). Node 1 has 2 Ω to ground, and node 2 has 2 Ω to ground.
Supernode KCL: V1/2 + V2/2 = 6 Constraint: V1 − V2 = 6Adding: V1 + V2 = 12 and V1 − V2 = 6, so V1 = 9 V, V2 = 3 V. Check: 4.5 A + 1.5 A = 6 A ✔.
6. Nodal vs Mesh Analysis
| Point | Nodal analysis | Mesh analysis |
|---|---|---|
| Law used | KCL | KVL |
| Unknowns | Node voltages | Mesh (loop) currents |
| No. of equations | N − 1 | B − N + 1 (planar circuits) |
| Best when | Many parallel branches, current sources | Many series elements, voltage sources |
| Special case | Supernode (voltage source between nodes) | Supermesh (current source shared by meshes) |
| Circuit type | Planar and non-planar | Planar only |
N = number of nodes, B = number of branches. Choose the method that gives fewer equations. For the example above, N = 3 gives 2 node equations, and B = 5 gives 5 − 3 + 1 = 3 mesh equations, so nodal is faster.
7. Exam Tips and Common Mistakes
- Use one sign convention throughout: “current leaving = (own voltage − other voltage)/R”.
- A resistor in series with a current source does not affect the node equation; the source fixes the current.
- For dependent sources, write the source in terms of node voltages before solving.
- Check your answer with KCL at every node; it takes seconds and catches sign errors.
- In AC (phasor) problems, replace R with Z and G with Y = 1/Z. The method is identical.
8. Practice MCQs
1. Nodal analysis is based on: (a) KVL (b) KCL (c) Ohm’s law only (d) Faraday’s law
2. A circuit has 5 nodes. The number of node equations required is: (a) 5 (b) 4 (c) 6 (d) 3
3. A circuit has 7 branches and 4 nodes. Number of independent mesh equations: (a) 3 (b) 4 (c) 5 (d) 7
4. A supernode is formed when: (a) A current source is between two meshes (b) A voltage source is between two non-reference nodes (c) Two resistors are in parallel (d) A node has no current
5. In the conductance matrix, an off-diagonal term G12 equals: (a) +sum of conductances at node 1 (b) −conductance between nodes 1 and 2 (c) 0 always (d) resistance between nodes
6. The conductance matrix of a passive resistive network (no dependent sources) is: (a) Symmetric (b) Skew-symmetric (c) Diagonal (d) Singular
7. A node is connected to 10 V through 5 Ω and to ground through 5 Ω. Its voltage is: (a) 10 V (b) 5 V (c) 2.5 V (d) 0 V
8. Nodal analysis can be applied to: (a) Planar circuits only (b) Non-planar circuits only (c) Both planar and non-planar circuits (d) DC circuits only
9. The reference node in nodal analysis is assumed to be at: (a) Highest potential (b) 0 V (c) Supply voltage (d) Average voltage
10. The number of unknowns in a supernode problem is reduced because: (a) KVL is used (b) The source gives a constraint equation (c) Current is zero (d) Resistors are ignored
9. বাংলায় দ্রুত রিভিশন
- নোডাল অ্যানালাইসিস KCL-এর ওপর ভিত্তি করে; অজানা রাশি হলো নোড ভোল্টেজ।
- N টি নোড থাকলে সমীকরণ লাগে N − 1 টি; একটি নোড রেফারেন্স (০ ভোল্ট)।
- রেজিস্টরের মধ্য দিয়ে নোড থেকে বের হওয়া কারেন্ট = (নিজের ভোল্টেজ − পাশের নোডের ভোল্টেজ)/R।
- দুই নন-রেফারেন্স নোডের মাঝে ভোল্টেজ সোর্স থাকলে তা সুপারনোড; একটি KCL + একটি সংযোজক সমীকরণ।
- মেশ অ্যানালাইসিস KVL-ভিত্তিক; সমীকরণ সংখ্যা B − N + 1, শুধু প্ল্যানার সার্কিটে।
- উদাহরণের উত্তর: V₁ = ১২ V, V₂ = ৮ V; প্রতিটি নোডে KCL মিলিয়ে দেখুন।
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References: Charles K. Alexander & Matthew N. O. Sadiku, Fundamentals of Electric Circuits; William H. Hayt, Engineering Circuit Analysis. Images and animations: original work by Totalecer.
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