Node Analysis, With and Without Voltage Sources - Lecture Slides | ECE 2040, Study notes of Electrical Circuit Analysis

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Circuit Analysis Lecture

Outline:

• Node analysis
• Without voltage sources
• With voltage sources

Circuit Equations

• Method #1:
• Apply KVL & KCL as needed
• Result: Set of simultaneous equations

Ax = b where A is a matrix, b is a column vector, and x is a column vector of unknown volt- ages/currents.

• Pros: Straightforward & guaranteed
• Cons: Cumbersome (on paper) & results in additional variables & equations
• Alternatives:
  • Node analysis
  • Mesh analysis
• Can view these as “guided” KVL & KCL

Illustration

• From previous figure...two nodes plus reference
• What is voltage across resistor R 1?
  • Let vab denote differential with a positive and b negative va − vab − vb = 0 ⇒ va − vb = vab
  • Let vba denote differential with a negative and b positive va + vba − vb = 0 ⇒ vba = vb − va
• What is current through resistor R 1?
  • From a to b: vab/R 1
  • From b to a: vba/R 2
• To determine va & vb:
  • KCL at a: is = va/R 2 + (va − vb)/R 1
  • KCL at b: (va − vb)/R 1 = vb/R 3
• 2 equations for 2 unknowns...

Node Voltage Relationships

• KVL: v 12 = v 1 − v 2
• In case of voltage source between nodes: v 12 = Vs (subject to consistent signs)
• In case of resistor between nodes: i = (v 1 − v 2 )/R (from 1 to 2 )

Example with Voltage Source

• Node a: 3 branches va/R 1 + (va − vb)/R 2 +? = 0
• On one hand: Cannot immediately compute current through vs
• On other hand: Know va = vs, so only unknown node voltage is vb
• Node b: 3 branches (vb − va)/R 2 + vb/R 3 = is
• Substitute va = vs and solve for vb
• KCL at node a provides current through voltage source

Example with Voltage Source, cont

• Node a: va/R 1 +? = 0
• Node b: vb/R 2 +? = is
• Unlike previous example, we only know relative voltage

va − vb = vs but not absolute value of va or vb

• Solution: Apply KCL to union of nodes, i.e., “supernode”

va/R 1 + vb/R 2 = is

• This is shortcut to labeling & eliminating current through voltage source

More Examples

• ab : 1.5 = va/6 + 3.5 + vb/ 3
• ab : vb − va = 12
• a : va = 8
• b : (vb −^ va)/6 + (vb −^ vc)/3 = 2
• c : vc = 3ix
• ix = (va − vb)/ 6

More Examples, cont

• ab : 3 = va/4 + vb/ 10
• va − vb = 4vx
• vx = −va
• a : (va − 6)/10 + ia + (va − vb)/20 = 0
• b : (vb − va)/20 + 5ia = 0
• ia = (6 − va)/10 + (vb − va)/ 20