EE 204
Lecture 23
Thevenin’s and Norton’s Equivalents in Sinusoidal
Circuits
Thevenin’s and Norton’s equivalents circuits are powerful tools for circuit reduction that can
also be applied to frequency domain circuits. The same process is used as seen in lecture 9
except that we substitute impedance (Z) for resistance (R).
Z th
a
V th
b
The figure above illustrates the frequency-domain version of a Thevenin equivalent circuit.
The figure below shows a Norton equivalent circuit.
IN
ZN
Thevenin voltage and impedance are calculated in the same way as those used with resistive
circuit
except that it involves manipulation of complex quantities.
Example 1: Find the Thevenin equivalent circuit with respect to the terminals a, b for the circuit
shown
V S = 80 0o [V ]
Solution:
Thevenin voltage Vth is the open circuit voltage across R.
I
V oc
V S = 80 0o [V ]
Voc = 18 I
V
I=
Z total
80∠0D
80∠0D
80∠0D
I=
=
=
j 30 + 18 − j 24 18 + j 6 18.97∠18.43D
80∠0D
= 4.216∠ − 18.43D A
D
18.97∠18.43
⇒ VOC = 4.216∠ − 18.43D ×18 = 75.89∠ − 18.43DV
⇒I=
Therefore
⇒ Vth = 75.89∠ − 18.43DV
Thevenin impedance Zth is the equivalent impedance seen between terminals a,b when we short
the voltage source. Zth is the parallel combination of R=18 Ohm and (j30-j24)
Z th
Therefore:
18 × j 6
108∠90D
=
18 + j 6 18.97∠18.43
⇒ Z th = 5.69∠71.57D Ω
Z th = 18 //( j 6) =
Therefore Thevenin equivalent circuit is:
Z th = 5.69 71.57D Ω
V th = 75.89 −18.43D [V ]
a
b
16 0o [ A ]
Example 2:
Find the Norton equivalent circuit with respect to the terminals a,b for the circuit.
16 0o [ A ]
Solution:
IN is the short circuit current ISC = the current flowing into 15+j30 impedance.
I sc
16 0o [ A ]
Therefore
I sc
(16∠0D ) × 25
= 6.4 − j 4.8 A
25 + 15 + j 30
⇒ I N = 6.4 − j 4.8 A
I SC =
ZN is the capacitor in // with (40+j30) ie
ZN
(− j 50)(40 + j 30)
= 50 − j 25Ω
40 + j 30 − j 50
Z N = 55.9∠ − 26.57D Ω
⇒ ZN =
Z N = 55.9 −26.57D Ω
I N = 6.4 - j 4.8 A
Example 3:
Calculate the TEC to the left of “a-b” in:
a) the phasor representation
b) the instantaneous representation
1
F
100
Figure 1
Solution:
a) the circuit is first transformed to the phasor representation:
V S = 5 −90o
Figure 2
Calculate the open circuit voltage Voc = Vth
I = 0 (because of the open circuit) ⇒
VDR ⇒ V1 = Voc =
j 20Ω
& 16Ω are in series &
16
16
16 × (5 −90o )
× Vs =
× 5 −90o =
16 + j 20
16 + j 20
25.612 51.340o
V1 = Voc
∴ Voc = Vth = 3.124 −141.340o [V ]
a
I = 0 -j10
j20
+
+
16
V oc
V1
V S = 5 −90o [V]
b
Figure 3
Next find the Thevenin impedance Z th
Set all independent sources to zero ⇒ replace Vs by a short circuit
Z th
Figure 4
j 20Ω 16Ω
⇒
16 × j 20
16 × 20 90o
=
= 12.494 38.660o = 9.756 + j 7.805 Ω
16 + j 20 25.612 51.340o
9.756 + j 7.805 Ω
Z th
Figure 5
∴ Z th = (9.756 + j 7.805) + (− j10) = 9.756 − j 2.195 Ω
9.756 − j 2.195 Ω
Z th
Figure 6
∴ The TEC in the phasor is shown
Z th = 9.756 − j 2.195 Ω
V th = 3.124 −141.340o [V ]
a
b
Figure 7
b) to find the TEC in the instantaneous representation, transform the circuit to the instantaneous
representation:
Vth = 3.124 −141.340o [V ]
⇒
vth (t ) = 3.124 cos(10t − 141.340o ) [V ]
How do we transform Z th = 9.756 − j 2.195 Ω to the instantaneous representation?
Z th = Rth + jX th = 9.756 − j 2.195 Ω
⇒
Rth = 9.756 Ω & X th = −2.195 < 0
We have a capacitive impedance because X th < 0
∴
1
= 2.195
ωC
⇒
1
= 2.195
10C
⇒
C=
1
= 0.045558 F = 45.558 mF
10 × 2.195
The resulting TEC in the instantaneous representation is shown
v th (t ) = 3.124 cos(10t -141.340o ) [V ]
Figure 8