EEGR2111 L8

Nodal analysis with voltage sources
  • What would the nodal equations be for the circuit below?

    Above: You do not need to solve for V1,
    it is established as 3V by the source

  • What would the nodal equations be for the circuit below?

    Applying nodal analysis to the above:

  • Node 1:
    -is + (V1 - V2) / R1 =0
    -3 - V2 + V1 = 0
  • Node 2:
    (V2 - V1) / R1 + (V2 - 0) / R2 + i2 =0
    -V1 (1/R1) V2 ( 1/R1 + 1/R2) + i2 =0
  • Node 3:
    -i2 - (V3 - 0) / R4
  • So, 3 equations are:
    (1) V1 + (-1) V2 + (0) V3 = 3
    (-1) V1 + (5/4) V2 + (0) V3 = -i2
    (0) V1 + (0) V2 + (1/8) V3 = i2

  • But the last two equations can be combined to eliminate i2:
    (-1) V1 + (5/4) V2 + (1/8) V3 = 0
  • But: we need 3 equations since we have 3 unknowns,
    So, final equation is:
    V2 = V3 + 2
  • "Supernodes"
  • The above example is redrawn with the supernode:

    KCL at the supernode yields:
    (V2 - V1) / R1 + ( V2 - 0) / R2 + (V3 - 0) / R4 =0
    (-1) V1 + (5/4) V2 + (1/8) V3 = 0
    The SAME result we had above when we combined the two equations to eliminate i2.

  • So: supernode effectively accomplishes the elimination of the dummy variable i2.
  • Dependent sources:
  • For a dependent current source:

    KCL at V1 yields:
    is = (1/R1 + 1/R2 + 1/R\pi + 1/RF) V1 + (-1/RF) V1
    KCL at V2 yields:
    (gm - 1/RF) V1 + (1/RF + 1/RL ) V2 = 0

  • For a dependent voltage source (type 1):

    Same as previous example except second node voltage is known:
    V2 = g V1

  • For a dependent voltage source (type 2):

    Need supernode as shown above
    KCL at the supernode yields:
    (V2 - V1) / R1 + ( V2 - 0) / R2 + (V3 - 0) / R4 =0
    (-1) V1 + (5/4) V2 + (1/8) V3 = 0
    KCL at the V1 yields:
    is = (V1 - V2) / R1
    And our final equation (from the supernode) is:
    V2 = V3 + g V1

  • How would this change if the dependent current was k i2 ?
  • Mesh analysis
  • Mesh - a loop not containing any other loops
  • Planar circuits!
  • Mesh current - current that flows around the mesh
  • Physical current through an element is the ALGEBRAIC sum of the mesh currents through it
  • How many meshes in the circuit below?

  • Given circuit with N meshes, N KVL equations are required
    Solve for N mesh currents

    PROCEDURE:

  • Label CLOCKWISE mesh currents a, b, c, ... or 1, 2, 3 ...
  • Apply KVL to all N meshes
  • Solve the resulting set of N simultaneous equations

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