Electric Circuits II
Sinusoidal Steady State Analysis
Dr. Firas Obeidat
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Table of Contents
1
2
3
4
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• Nodal Analysis
• Mesh Analysis
• Superposition Theorem
• Source Transformation
• Thevenin and Norton Equivalent Circuits
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Dr. Firas Obeidat – Philadelphia University
Sinusoidal steady state analysis
Steps to Analyze AC Circuits:
1. Transform the circuit to the phasor or frequency domain.
2. Solve the problem using circuit techniques (nodal analysis, mesh
analysis, superposition, etc.).
3. Transform the resulting phasor to the time domain.
Frequency domain analysis of an ac circuit via phasors is
much easier than analysis of the circuit in the time domain
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Nodal Analysis
The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid for
phasors, AC circuits can be analyzed by nodal analysis.
Example: Find the time-domain node voltages v1(t) and v2(t) in the circuit
shown in the figure
Apply KCL on node 1
Apply KCL on node 2
From the above equations, we can find that V1=1− j2 V and V2=− 2+j4 V
The
time
domain
solutions are obtained
by expressing V1 and V2
in polar form:
The time domain
expression is
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Nodal Analysis
Example: Compute V1 and V2 in the circuit
Nodes 1 and 2 form a supernode. Applying KCL
at the supernode gives
But a voltage source is connected between nodes 1
and 2, so that
Substitute the above equation in the first equation
gives
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Mesh Analysis
Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis.
Example: Determine IO current in the circuit using mesh analysis.
Applying KVL to mesh 1, we obtain
For mesh 2
For mesh 3, I3=5 A, Substituting this in the above
two equations, we get
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Mesh Analysis
Example: solve for Vo in the circuit using mesh analysis
meshes 3 and 4 form a supermesh due to the current
source between the meshes. For mesh 1, KVL gives
(1)
For mesh 2, KVL gives
(2)
For supermesh, KVL gives
(3)
Due to the current source between meshes 3 and 4, at
node A
(4)
Substitute equation (2) in equation (1) gives
(5)
Substitute equation (4) in equation (3) gives
(6)
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Mesh Analysis
From equation (5) and equation (6), we obtain the matrix equation
We obtain the following determinants
Current I1 is obtained as
The required voltage Vo is
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Superposition Theorem
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.
Example: Use the superposition theorem to find Io
in the circuit.
Let
(1)
Where Io′ and Io″ are due to the voltage and current
sources, respectively. To find Io′ consider the
circuit in fig(a). If we let Z be the parallel
combination of –j2 and 8+j10, then
The current Io′ is
(2)
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Superposition Theorem
To get Io″, consider the circuit in fig(b). For mesh 1
(3)
For mesh 2
(4)
For mesh 3
From (4) & (5)
(5)
Expressing I1 in terms of I2 gives
(6)
Substituting eq(5) and eq(6) into eq(3), we get
(7)
From eq(2) and eq(7), we get
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Superposition Theorem
Example: Find vo of the circuit using the superposition theorem.
Since the circuit operates at three different
frequencies for the dc voltage source), one
way to obtain a solution is to use
superposition,
(1)
Where v1 is due to the 5-V dc voltage source, v2 is due to the voltage source, and v3 is
due to the current source. To findv1 we set to zero all sources except the 5-V dc source.
We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a
short circuit to dc. There is an alternative way of looking at this. Since ω=0, jωL=0,
1/ωj=∞.
From fig(a)
(2)
To find v2 we set to zero both the 5-V source and the 2sin5t
current source and transform the circuit to the frequency
domain.
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Superposition Theorem
The equivalent circuit is now as shown in fig(b).
Let
By voltage division
In time domain
(3)
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Superposition Theorem
To obtain v3 we set the voltage sources to zero and transform what is left to the
frequency domain.
The equivalent circuit is now as shown in
fig(c). Let
By current division
In time domain
(4)
From eq (1), eq (2), eq (3) and eq (4), we get
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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.
Example: Calculate Vx in the circuit using the method
of source transformation
Transform the voltage source to a current source as in fig
(a)
The parallel combination of 5 Ω resistance and 3+j4
impedance gives
Converting the current source to a voltage source yields
the circuit in fig (b), where
By voltage division
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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.
 A linear circuit is replaced by a voltage source
in series with an impedance
 In Norton equivalent circuit, a linear circuit
is replaced by a current source in parallel
with an impedance.
 Thevenin’s and Norton’s equivalent circuits
are related as
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Thevenin and Norton Equivalent Circuits
Example: Obtain the Thevenin equivalent at terminals ab in the circuit.
To find ZTh, set the voltage source to zero. As shown in
fig(a), the 8Ω resistance is in parallel with the –j6
reactance, and the resistance 4Ω is in parallel with the j12
reactance. so that their combination gives
The Thevenin impedance is the series combination of Z1
and Z2 that is,
To find VTh consider the circuit in fig(b). Currents are
obtained as
Applying KVL around loop bcdeab in fig(b) gives
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Thevenin and Norton Equivalent Circuits
Example: Find the Thevenin equivalent circuit as seen
from terminals a-b.
To find VTh, apply KCL at node 1 in fig(a)
Applying KVL to the middle loop fig(a)
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Thevenin and Norton Equivalent Circuits
To obtain ZTh, remove the independent source.
Due to the presence of the dependent current
source, connect a 3-A current source to terminals
a-b as in fig(b). At the node, KCL gives
Applying KVL to the outer loop in fig(b) gives
The Thevenin impedance is
Example: Obtain Io current using Norton’s theorem.
To find ZTh, set the sources to zero as shown
in fig(a). the 8-j2 and 10+j4 impedances are short
circuited, so that
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Thevenin and Norton Equivalent Circuits
To get IN we short-circuit terminals a-b as in fig(b) and apply mesh
analysis. Meshes 2 and 3 form a supermesh because of the current source
linking them. For mesh 1
(1)
For the supermesh
(2)
At node a, due to the current source between meshes 2 and 3,
(3)
Adding eqs. (1) and (2) gives
from eqs. (1)
The Norton current is
Figure (c) shows the Norton equivalent circuit along with the impedance at terminals a-b. By
current division
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