Interpreting Mesh Currents - Circuit Analysis - Lecture Notes | ECE 2040, Study notes of Electrical Circuit Analysis

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Circuit Analysis Lecture

Outline:

• Mesh analysis
• Network theorems
• Source transformation

Mesh Analysis

• Alternative to node analysis...but still guided KVL & KCL
• Starting point: Mesh = “window pane”, i.e., loop that does not enclose other loops.
• Restriction: Planar circuits (or else, “mesh” is ill defined)

Interpreting Mesh Currents

• Readily interpreted in some cases
• Algebraic artifact in other cases

i 1 i 2

i 3

i 4

ix iy

• i 1 , i 2 , i 4 are readily interpreted...what is i 3?
• KCL: ix + iy + i 2 = i 4
• Mesh analysis: ix = i 3 − i 2 ⇒ i 3 = ix + i 2 iy = i 4 − i 3 ⇒ i 3 = i 4 − iy equating the 2 produces KCL

Mesh Current Relationships

• Current passing through shared branch is difference of mesh currents (with appropriate sign)
• Presence of current sources imposes constraints (compare to voltage sources in node analysis)
• Current through branch determines voltage differential across resistor (with appropriate sign)
• Mesh analysis steps:
  • Label mesh currents (#unknowns = #meshes)
  • Apply KVL around each mesh (#equations = #meshes)
  • In case of current sources: ∗ Case 1: Current source specifies mesh current ∗ Case 2: Current source specifies difference of mesh currents & apply KVL to “supermesh”

Examples, cont

• • Mesh1: −i 12 − 16 − 4 ix − (i 1 − i 3 )3 =
• • Mesh2: 4 ix − i 2 2 + 8 =
• • Mesh34: i 4 − i 3 =
• • Mesh34: −i 36 − (i 3 − i 1 )3 − −i 4 8 =
• • ix = −i
• • Mesh 13: 25 − i 330 − 5 va − (i 3 − i 2 )20 − (i 1 − i 2 )10 = • Mesh13: i 1 − i 3 = 3ib
• • Mesh2: i 2 = −
• • ib = i 3 − i
• • va = (i 1 − i 2 )

Network Theorems

• Conversation shift:
  • From: Given a circuit, find x, y, z...
  • To: Based on circuit understanding, what are general circuit properties?
• Circuit theorems:
  • Source transformation
  • Linearity
  • Thevenin equivalence
  • Norton equivalence
  • Maximum power transfer

Source Transformation

• Under what conditions are the following equivalent:
  • Voltage source in series with resistor
  • Current source in parallel with resistor
• Compute current/voltage relationship for each:
  • Voltage source w/ resistor:

i =

v − vs Rs

⇒ i = (1/Rs)v − vs/Rs

• Current source with resistor:

v = (i + is)Rp ⇒ i = (1/Rp)v − is

• Equivalence conditions:
  1. Rs = Rp
  2. vs = isRs = isRp

Illustration

• Compute voltage across 20Ω resistor via node analysis:

5 − va 5

3 − va 30

va 20 Rewrite 1 =

va 5

va −^3 30

va 20

• Apply source transformation: is = 1 & Rp = 5
• Compute voltage across 20Ω resistor via node analysis:

1 =

va 5

va 20

va − 3 30 Same!