Mesh and nodal analysis

Why it matters

Real instrumentation circuits such as bridge networks, ladder attenuators, filter sections and transistor small-signal models are too tangled for series–parallel reduction. Mesh and nodal analysis turn any linear circuit into a set of simultaneous equations that you can write mechanically and solve by hand or by matrix. Circuit simulators like SPICE are built on (modified) nodal analysis.

Key ideas

Nodal analysis (KCL-based).

1. Pick a reference node (ground), usually the node with the most connections.
2. Assign voltages V1, V2, … to the remaining n − 1 nodes. Each is measured with respect to ground.
3. At each non-reference node, write KCL with every branch current expressed through the node voltages. For example, the current leaving node 1 through R to node 2 is (V1 − V2)/R.
4. Solve the n − 1 equations.

Nodal analysis works for planar and non-planar circuits. It is the natural choice when the circuit has current sources and few nodes.

Supernode. When an ideal voltage source sits between two non-reference nodes, you cannot write its current through Ohm's law. Enclose the source and its two nodes in a closed surface (the supernode). Write one KCL for the whole surface, plus the constraint Va − Vb = Vs. If the source connects a node to ground, that node voltage is simply known.

Mesh analysis (KVL-based).

1. The circuit must be planar, that is, drawable with no crossing wires.
2. Assign a clockwise mesh current to each of the b − n + 1 meshes.
3. Around each mesh, write KVL. A resistor shared by meshes j and k carries Ij − Ik (as seen from mesh j).
4. Solve.

Mesh analysis is the natural choice when the circuit has voltage sources and few meshes.

Supermesh. A current source shared by two meshes has an unknown voltage across it. Remove it mentally, write KVL around the larger loop formed by the two meshes, and add the constraint Ij − Ik = Is with the correct sign. A current source on the outer boundary of one mesh fixes that mesh current directly.

Dependent sources. Write the equations treating the dependent source like an independent one. Then express its controlling variable in terms of the node voltages or mesh currents. With dependent sources the coefficient matrix is no longer symmetric.

Inspection (matrix) form. For circuits with only resistors and independent sources:

• Nodal form is [G][V] = [I]. The diagonal entry G_kk is the sum of conductances at node k. The off-diagonal entry G_jk is minus the conductance between nodes j and k. I_k is the source current injected into node k.
• Mesh form is [R][I] = [V]. R_kk is the total resistance around mesh k. R_jk is minus the shared resistance (with all meshes clockwise). V_k is the net source rise in mesh k along the mesh direction.

Choosing the method. Count the equations. Nodal needs n − 1 minus the number of voltage sources. Mesh needs b − n + 1 minus the number of current sources. Choose the smaller set.

Formulas

• Branch current from node voltages: I_jk = (Vj − Vk)/R, in A, with V in V and R in Ω.
• Nodal equations: [G][V] = [I], with G in S (siemens).
• Mesh equations: [R][I] = [V], with R in Ω.
• Number of nodal equations: n − 1. Number of mesh equations: b − n + 1 (planar circuits only).
• Supernode constraint: Va − Vb = Vs. Supermesh constraint: Ij − Ik = Is.
• Cramer's rule for two unknowns: x1 = (b1·a22 − b2·a12)/(a11·a22 − a12·a21)

Worked examples

Example 1 (standard): two-mesh circuit. Given: mesh 1 has a 12 V source (driving the clockwise mesh current I1) and 2 Ω. Mesh 2 has 6 Ω and a 6 V source whose polarity opposes the clockwise current I2. A 4 Ω resistor is shared by both meshes. Find the current in the 4 Ω resistor.

1. Mesh 1, KVL: 12 = 2·I1 + 4·(I1 − I2), so 6·I1 − 4·I2 = 12.
2. Mesh 2, KVL: −6 = 6·I2 + 4·(I2 − I1), so −4·I1 + 10·I2 = −6.
3. The determinant is Δ = 6·10 − (−4)(−4) = 44.
4. I1 = (12·10 − (−6)(−4))/44 = 96/44 = 2.182 A.
5. I2 = (6·(−6) − (−4)·12)/44 = 12/44 = 0.273 A.
6. The current in the 4 Ω resistor is I1 − I2 = 84/44 = 1.909 A, flowing in the direction of I1.

Answer: 1.91 A through the 4 Ω resistor.

Example 2 (GATE level): nodal analysis with a dependent source. Given:

• Node 1 connects to a 10 V source through 2 Ω, to ground through 4 Ω, and to node 2 through 2 Ω.
• Node 2 connects to ground through 4 Ω.
• A current-controlled current source injects 2·ix into node 2, where ix is the downward current in the 4 Ω resistor at node 1.

Find V1 and V2.

1. Controlling variable: ix = V1/4.
2. KCL at node 1 (currents leaving): (V1 − 10)/2 + V1/4 + (V1 − V2)/2 = 0, so 1.25·V1 − 0.5·V2 = 5.
3. KCL at node 2: (V2 − V1)/2 + V2/4 = 2·(V1/4), so −1.0·V1 + 0.75·V2 = 0.
4. From step 3, V2 = V1/0.75 = 1.333·V1.
5. Substitute into step 2: 1.25·V1 − 0.667·V1 = 5, so V1 = 5/0.5833 = 8.571 V.
6. Then V2 = 11.43 V.
7. Check at node 2: the left side is (11.43 − 8.571)/2 + 11.43/4 = 1.429 + 2.857 = 4.286 A. The right side is 2·(8.571/4) = 4.286 A. ✓

Answer: V1 = 8.57 V, V2 = 11.43 V. Node 2 is above the source voltage because the dependent source pumps energy in.

Common mistakes

• Writing the shared-resistor current as I1 + I2 when both mesh currents are clockwise. It is I1 − I2 from mesh 1's view.
• Writing KCL at a node that has an ideal voltage source attached instead of using a supernode, which forces an extra unknown current.
• Forgetting the constraint equation for a supernode or supermesh, which leaves one equation short.
• Using mesh analysis on a non-planar circuit.
• Assuming the inspection (symmetric matrix) method works with dependent sources. It does not; write the equations out in full.
• Answering with a mesh current when the question asks for a branch current.

For GATE IN

Expect NAT questions asking for one node voltage, branch current or power in a 3–5 node network, often with a dependent source or a floating voltage source. Practise deciding quickly between mesh and nodal, writing supernode/supermesh constraints, and using Cramer's rule for 2×2 and 3×3 systems. A quick power balance or KCL check at the end catches most sign errors.

Quick check

1. A circuit has 7 branches and 5 nodes. How many mesh equations are needed?
2. A 5 V source connects node 2 (+) to node 3. What constraint goes with the supernode?
3. Can mesh analysis be used for a non-planar circuit?
4. With dependent sources, is the nodal G matrix necessarily symmetric?

Answers: 1. 7 − 5 + 1 = 3. 2. V2 − V3 = 5 V. 3. No; use nodal analysis. 4. No.