← CA-3 · KIRCHHOFF & SYSTEMATIC ANALYSIS

Mesh analysis

YOU WILL BE ABLE TO
  • Define mesh currents and state when mesh analysis applies.
  • Write a KVL equation around each mesh.
  • Solve for mesh currents and recover branch currents.
  • Choose sensibly between nodal and mesh analysis.

Nodal analysis is KCL made systematic; mesh analysis is the same trick built on KVL. It applies to planar circuits — ones you can draw with no crossing wires — which is nearly everything at this level (and literally everything on a 2-layer PCB's single side). A mesh is a loop enclosing no other loop: a window pane in the circuit's window frame. Assign each mesh a circulating mesh current, all clockwise by convention, then write KVL around each mesh.

The one subtlety: an element on the boundary between two meshes carries BOTH mesh currents, in opposite directions, so its actual current is their difference.

ishared=i1−i2i_{\text{shared}} = i_1 - i_2current in a branch shared by meshes 1 and 2
+−8 VR11 kΩR32 kΩR21 kΩ+−2 Vi1i2
The worked example: two clockwise mesh currents; R_3 lies on the shared boundary and carries i_1 − i_2.
WORKED EXAMPLE

Two meshes, start to finish

  1. In the figure: an 8 V source drives mesh 1 through R_1 = 1 kΩ; R_3 = 2 kΩ is shared; mesh 2 contains R_2 = 1 kΩ and a 2 V source whose + terminal faces the top (so clockwise i_2 enters its + terminal — a drop).
  2. Mesh 1 KVL (clockwise, V/mA/kΩ): −8 + 1·i_1 + 2(i_1 − i_2) = 0 → 3i_1 − 2i_2 = 8.
  3. Mesh 2 KVL: 2(i_2 − i_1) + 1·i_2 + 2 = 0 → −2i_1 + 3i_2 = −2.
  4. Multiply the first by 3 and the second by 2, then add: 9i_1 − 6i_2 = 24 and −4i_1 + 6i_2 = −4 give 5i_1 = 20 → i_1 = 4 mA, then i_2 = (−2 + 2·4)/3 = 2 mA.
  5. The shared R_3 carries i_1 − i_2 = 2 mA downward. Verify mesh 1: −8 + 1(4) + 2(2) = 0 ✓. Mesh 2: 2(−2) + 1(2) + 2 = 0 ✓.
i_1 = 4 mA, i_2 = 2 mA; R_3 carries 2 mA downward
WORKED EXAMPLE

Recovering everything else

  1. From i_1 = 4 mA and i_2 = 2 mA: R_1 carries 4 mA, R_2 carries 2 mA, R_3 carries 2 mA.
  2. The 8 V source delivers p = 8 × 4 = 32 mW. The 2 V source: clockwise i_2 = 2 mA enters its + terminal, so it ABSORBS 2 × 2 = 4 mW (it's being "charged", like CA-1's battery example).
  3. Resistors: 1(4²) = 16 mW + 2(2²) = 8 mW + 1(2²) = 4 mW = 28 mW absorbed.
  4. Balance: 32 delivered = 28 + 4 absorbed. Σp = 0 ✓.
Every branch quantity and a passing power audit, from two mesh currents.

PRACTICE

Q1

Set up only: mesh 1 contains a source v_s and resistor R_a, sharing R_c with mesh 2, which also contains R_b. Write both mesh equations (clockwise currents).

Q2

Solve: mesh 1 has an 8 V source (aiding i_1) and R_1 = 2 kΩ, shared resistor 1 kΩ; mesh 2 has R_2 = 4 kΩ and a 2 V source aiding i_2. Find both mesh currents.

Q3

In practice problem 2, how much power does the 8 V source deliver, and is the 2 V source delivering or absorbing?

Q4

Two conceptual checks: (a) why does mesh analysis require a planar circuit? (b) a shared branch computes to i_1 − i_2 = 0 — what does that mean physically?

Mesh analysis · CA-3 · iLearnSTEM Electronics