EEGR2111 L11

Thevenin's theorem

  • Thevinin's theorem - reduce portion of circuit to a simple equivalent
  • A general form of source transformations
  • Reduce a circuit to a single source and single element
  • For example, below we may wish to reduce the portion of the circuit to the left of terminals A-B to a single voltage source and resistor
  • We can already reduce this using series and parallel resistances and source transformations to:
  • This can also be done more generally (and efficiently) using Thevinin's theorem

  • Thevinin's theorem PROCEDURE
  • 1) Divide circuit into 2 parts, A and B, connected at a pair of terminals
    Note: the control variables of dependent sources must be in the same part of the circuit as the dependent source
  • 2) With part B removed:
    2a) Find the open circuit voltage Voc at the two terminals of part A
    2b) Find the short circuit current Isc at the two terminals of part A
  • 3) The equivalent circuit for part A is then a voltage of magnitude Vt = Voc in series with a resistor of value Rt = Voc / Isc
  • 4) Part A of the circuit may then be replaced with its Thevenin equivalent of Vt in series with Rt
  • NOTE: Rt can also be determined by finding the resistance looking into the terminals of part A of the circuit when all independent sources are set equal to zero
    • Example L11-1 Find the Thevenin equivalent of the two circuits above
  • First circuit
  • 1) Divide at A-B
  • 2) Then:
    Voc = 3 { R2 / (R1 + R2) } = 3/4 V
    Isc = { 3 / (R1 + [R2 || R3] ) } { G3 / (G3 + G2) } = { 3/15 } { 1/4 } = 1/20 A
  • 3) Rt = Voc / Isc = (3/4) / (1/20) = 15 Ohm
  • 4) So:
    Vt = 3/4 V
    Rt = 15 Ohm
  • Second circuit
  • 1) Divide at A-B
  • 2) Then:
    Voc = 3/4 V
    Isc = (3/4) / 15 = 1/20 A
  • 3) Rt = Voc / Isc = (3/4) / (1/20) = 15 Ohm
  • 4) So:
    Vt = 3/4 V
    Rt = 15 Ohm

    • Norton's theorem

  • Norton's theorem - reduce portion of circuit to a simple equivalent
  • Similar to Thevenin except equivalent id current source in parallel with resistor

  • Norton's theorem PROCEDURE
  • 1) Divide circuit into 2 parts, A and B, connected at a pair of terminals
    Note: the control variables of dependent sources must be in the same part of the circuit as the dependent source
  • 2) With part B removed:
    2a) Find the open circuit voltage Voc at the two terminals of part A
    2b) Find the short circuit current Isc at the two terminals of part A
  • 3) The equivalent circuit for part A is then a current of magnitude In = Isc in parallel with a resistor of value Rn = Voc / Isc
  • 4) Part A of the circuit may then be replaced with its Norton equivalent of In in parallel with Rn
  • NOTE: Rn can also be determined by finding the resistance looking into the terminals of part A of the circuit when all independent sources are set equal to zero
    • NOTES:
  • The norton and Thevenin correspond to equivalent non-ideal sources
  • Rn = Rt
    • Example L11-2 Find the Thevenin equivalent of the two circuits above
  • First circuit
  • 1) Divide at A-B
  • 2) Then:
    Voc = 3/4 V as before
    Isc = 1/20 A as before
  • 3) Rn = Voc / Isc = (3/4) / (1/20) = 15 Ohm
  • 4) So:
    In = 1/20 A
    Rn = 15 Ohm
  • Second circuit
  • 1) Divide at A-B
  • 2) Then:
    Voc = 3/4 V
    Isc = (3/4) / 15 = 1/20 A
  • 3) Rn = Voc / Isc = (3/4) / (1/20) = 15 Ohm
  • 4) So:
    In = 1/20 A
    Rn = 15 Ohm
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