← CA-3 · KIRCHHOFF & SYSTEMATIC ANALYSIS

Supernodes and supermeshes

YOU WILL BE ABLE TO
  • Handle a voltage source between two non-reference nodes (supernode).
  • Handle a current source shared between two meshes (supermesh).
  • Write the constraint equation each construct requires.
  • Recognise both as the same idea: trade a lost equation for a free constraint.

Each systematic method has one awkward customer. Nodal analysis stumbles on a voltage source floating between two non-reference nodes — you can't write the resistor-style current through it, because an ideal voltage source permits ANY current. Mesh analysis stumbles on a current source shared between two meshes, for the mirror-image reason: it permits ANY voltage. Both problems dissolve with the same move.

The supernode: draw a boundary around the floating voltage source and its two nodes, and write one KCL equation for everything crossing that boundary — the source's own unknown current stays inside and never appears. You've lost one KCL equation; the source pays you back with a constraint you get for free.

v1−v2=vsourcev_1 - v_2 = v_{\text{source}}supernode constraint
6 mAv1R12 kΩ+−4 Vv2R22 kΩ
A 4 V source floats between v_1 and v_2: KCL around both nodes at once, plus the free constraint v_1 − v_2 = 4.
WORKED EXAMPLE

Supernode, start to finish

  1. In the figure: 6 mA enters node 1; R_1 = 2 kΩ and R_2 = 2 kΩ run to ground; a 4 V source (+ at node 1) floats between the nodes.
  2. Supernode KCL (all outflows to ground = inflow): v_1/2 + v_2/2 = 6 → v_1 + v_2 = 12.
  3. Constraint: v_1 − v_2 = 4.
  4. Add: 2v_1 = 16 → v_1 = 8 V, v_2 = 4 V.
  5. The source's own current, if you want it: KCL at node 2 alone gives i_source = v_2/2 = 2 mA flowing from node 1 to node 2 through the source. It was never needed for the solve.
v_1 = 8 V, v_2 = 4 V

The supermesh is the perfect dual. A current source shared between meshes 1 and 2 has an unknowable voltage, so don't walk through it: write one KVL around the OUTER boundary of both meshes combined, skipping the shared branch entirely. The source repays the lost equation with its own constraint on the mesh-current difference.

i2−i1=isourcei_2 - i_1 = i_{\text{source}}supermesh constraint (source arrow agreeing with i_2's direction in the shared branch)
+−9 VR11 kΩ3 mAR22 kΩi1i2
A 3 mA source on the shared boundary: KVL around the combined outer loop, plus i_2 − i_1 = 3 mA.
WORKED EXAMPLE

Supermesh, start to finish

  1. In the figure: 9 V source, R_1 = 1 kΩ (mesh 1's top), R_2 = 2 kΩ (mesh 2's top), and a 3 mA source pointing up on the shared branch.
  2. Clockwise i_1 goes DOWN the shared branch and clockwise i_2 goes UP it, so the branch current upward is i_2 − i_1, and the constraint is i_2 − i_1 = 3.
  3. Outer-loop KVL (skipping the shared branch): −9 + 1·i_1 + 2·i_2 = 0 → i_1 + 2i_2 = 9.
  4. Substitute i_2 = i_1 + 3: i_1 + 2i_1 + 6 = 9 → i_1 = 1 mA, i_2 = 4 mA.
  5. Verify the constraint: 4 − 1 = 3 ✓. Outer loop: −9 + 1(1) + 2(4) = 0 ✓.
i_1 = 1 mA, i_2 = 4 mA

PRACTICE

Q1

Set up only: a source v_x floats between nodes 1 and 2 (+ at node 1); node 1 also connects to ground through R_1, node 2 through R_2, and a current i_in enters node 1. Write the supernode equation and the constraint.

Q2

Solve: 4 mA enters node 1; R_1 = 5 kΩ and R_2 = 5 kΩ to ground; a 10 V source floats between the nodes (+ at node 1). Find both node voltages.

Q3

Solve: an 8 V source drives mesh 1 through R_1 = 2 kΩ; mesh 2's outer branch is R_2 = 1 kΩ; a 2 mA source (arrow agreeing with i_2) sits on the shared branch. Find both mesh currents.

Q4

Conceptual: why is it illegal to write a KVL term for the voltage across an ideal current source (and the KCL dual for the current through an ideal voltage source)?

Supernodes and supermeshes · CA-3 · iLearnSTEM Electronics