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CHAPTER 10 SINUSOIDAL STEADY-
STATE ANALYSIS
An expert problem solver must be endowed with two incompatible
quantities, a restless imagination and a patient pertinacity.
- Howard W. Eves (1911-2004)
Howard W. Eves is a well known author and
longtime professor at the University of Maine.
He enjoyed a long and distinguished career as a
teacher, geometer, writer, editor, and historian of
mathematics.
10.1 INTRODUCTION
In this chapter, we want to see how nodal analysis, mesh analysis,
Thevenin’s theorem, Norton’s theorem, superposition, and source
transformations are applied in analyzing ac circuits. Since these techniques
were already introduced for dc circuits, our major effort here will be to
illustrate with examples.
Analyzing ac circuits usually requires three steps.
- Transform the circuit to the phasor or frequency domain.
- Solve the problem using circuit techniques (nodal analysis, mesh
analysis, superposition, etc.).
- Transform the resulting phasor to the time domain.
10.2 NODAL ANALYSIS
The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid
for phasors, as demonstrated in Section 9.6, we can analyze ac circuits by
nodal analysis. The following examples illustrate this.
EXAMPLE 10.1 Find in the circuit of Fig.10.1 using nodal analysis.
The frequency-domain equivalent circuit is as shown in Fig.10.2.
Applying KCL at node 1,
At node 2,
1 1 1 2
1 2
, or (1 1.5) 2.5 20
j j
j j
V V V V
V V
1 2 2
x
j j
V V V
I
10.3 MESH ANALYSIS
EXAMPLE 10.3 Determine
current I
o
in the circuit
of Fig.10.7 using mesh
analysis.
Solution:
Applying KVL to mesh 1,
we obtain
1 2 3
(8 + j 10 − j 2) I − −( j 2) I − j 10 I = 0
For mesh 2,
For mesh 3, I
3
2 1 3
(4 − j 2 − j 2) I − −( j 2) I − −( j2) I + 20 ∠ 90 = 0
D
In matrix form
The desired current is
1
2
j j j
j j j
I
I
2 2
6.12 35.22 A , 6.12 144.78 A
o
I = ∠ − I = − I = ∠
D D
10.4 SUPERPOSITION THEOREM
Since ac circuits are linear, the superposition theorem applies to ac circuits
the same way it applies to dc circuits.
The theorem becomes important if the circuit has sources operating at
different frequencies. In this case, since the impedances depend on
frequency, we must have a different frequency-domain circuit for
each frequency.
The total response must be obtained by adding the individual responses in
the time domain.
It is incorrect to try to add the responses in the phasor or frequency
domain, because the exponential factor e
j ωt
is implicit in sinusoidal
analysis, and that factor would change for every angular frequency ω.
EXAMPLE 10.5 Use the superposition theorem to find I
o
in the circuit
in Fig.10.7.
Solution:
Let
Where and are due to
the voltage and current
sources, respectively.
o o o
I = I ′ +I′′
o
I ′
o
I ′′
10.5 SOURCE TRANSFORMATION
Source transformation in the frequency domain involves transforming a
voltage source in series with an impedance to a current source in
parallel with an impedance, or vice versa.
As we go from one source type to another, we must keep the following
relationship in mind:
s
s s s s
s
V
V Z I I
Z
10.6 THEVENIN AND NORTON EQUIVALENT CIRCUITS
Thevenin’s and Norton’s theorems are applied to ac circuits in the same
way as they are to dc circuits. The only additional effort arises from the need
to manipulate complex numbers.
Th Th
N N N
V = Z I Z =Z
V
Th
is the open-circuit
voltage while I
N
is the short-
circuit current.
EXAMPLE Find the Thevenin equivalent of the circuit in Fig.10.25 as seen
from terminals a-b.
Solution:
To find V
Th
, we apply KCL at
node 1 in Fig.(a).
15 = I
o
+ 0.5I
o
⇒ I
o
=10A
Applying KVL to the loop on
the right-hand side in Fig.(a),
we obtain
Thus, the Thevenin voltage is
Th
o o
− I − j + I + j + V =
Th
V = 55 ∠ −90 V
D
To obtain Z
Th
, we remove
the independent source and
connect a 3-A current source
to terminals a-b as shown in
Fig(b).
At the node, KCL gives
3 = I
o
+ 0.5I
o
⇒ I
o
=2A
Applying KVL to the outer loop in Fig.(b) gives
The Thevenin impedance is
s o
V = I + j + − j = −j
Th
s
s
j
j
V
Z
I
1 1 1 1
3 0 0
10 5 10 20
o
j
∠ − − −
= + +
−
V V V V V
D
Applying KCL at node 1,
At node 2, KCL gives
Hence,
1
0 0
10 10
o
j
− −
=
−
V V
1.029 59.04 , ( ) 1.029cos(1000 59.04 )V
o o
V = ∠ v t = t+
D D
10.8 AC ANALYSIS USING PSPICE
Wien-bridge oscillator
10.10 SUMMARY
¾ We apply nodal and mesh analysis to ac circuits by applying KCL
and KVL to the phasor form of the circuits.
¾ In solving for the steady-state response of a circuit that has
independent sources with different frequencies, each independent
source must be considered separately. The most natural approach to
analyzing such circuits is to apply the superposition theorem. A
separate phasor circuit for each frequency must be solved
independently, and the corresponding response should be obtained in
the time domain. The overall response is the sum of the time-domain
responses of all the individual phasor circuits.
¾ The concept of source transformation is also applicable in the
frequency domain.
¾ The Thevenin equivalent of an ac circuit consists of a voltage source
V
Th
in series with the Thevenin impedance Z
Th
¾ The Norton equivalent of an ac circuit consists of a current source I
N
in parallel with the Norton impedance Z
N
(=Z
Th
