Republic of the Philippines
North Eastern Mindanao State University
Bislig Campus
Maharlika, Bislig City
EE 221 – Electrical Circuits 2
Bachelor of Science in Electrical Engineering
Sinusoidal Steady State Analysis
June 5, 2023
SUBMITTED BY:
Peramide, Marc
Calipong, Julius Vincent
Cordova, Reneil
Cerbas, Tom Jim
Calape, Bencin
SUBMITTED TO:
Engr. Lorenzo L. Moricho, Jr.
10
CHAPTER
SINUSOIDAL STEADY STATE ANALYSIS
CONTENTS:
●
●
●
●
●
Nodal Analysis
Mesh Analysis
Superpositon Theorem
Thevenin’s Theorem
Norton’s Theorem
OBJECTIVES:
●
Learn how to Analyze AC Circuits
●
Learn how to transform the resulting phasor to the time domain
●
Solving using circuit techniques (nodal analysis, mesh analysis, superposition and
etc.)
INTRODUCTION:
Steady-state sinusoidal analysis using phasors is a mathematical technique used in electrical
engineering to analyze the behavior of a circuit operating at a constant frequency over time. This chapter
will concentrate on the steady-state response of circuits driven by sinusoidal sources. The response will
also be sinusoidal. For a linear circuit, the assumption of a sinusoidal source represents no real
restriction, since a source that can be described by a periodic function can be replaced by an equivalent
combination (Fourier series) of sinusoids.
NODAL ANALYSIS
OBJECTIVES:
At the end of the lesson, the learners should be able;
1. To determine the voltages at each node in an AC circuit using nodal analysis.
2. To apply Kirchhoff's current law (KCL) to the analysis of AC circuits using nodal analysis.
3. To simplify complex AC circuits using nodal analysis by reducing the number of unknown
variables.
Page 1
INTRODUCTION:
In AC circuit nodal analysis, the nodal voltage method is applied to determine the unknown
node voltage levels in a circuit with sinusoidal inputs. Sinusoidal sources can be converted to phasor
form and written as rectangular form of complex functions, allowing for circuit analysis with complex
numbers. Kirchhoff's current law is used to write nodal equations for each node in the circuit, which
can be solved simultaneously to find the voltage at each node. The resulting set of equations can be
solved using matrix techniques to find the unknown nodal voltages in the circuit.
Example 1.
Find 𝑉1 𝑎𝑛𝑑 𝑉2 in the circuit using nodal analysis.
Solution:
We first convert the circuit to the frequency domain:
10 𝑐𝑜𝑠2𝑡 → 10∠0°
𝜔 = 2 𝑟𝑎𝑑/𝑠
2𝐻 → 𝑗𝜔𝐿 = 𝑗4
0.2𝐹 →
1
= −𝑗2.5
𝑗𝜔𝐶
Draw the equivalent circuit diagram in frequency domain
Applying nodal analysis,
Page 2
Node 1.
𝑉₁
𝑉₁−𝑉₂
+ −𝑗2.5 −
2
1
10 = 0
1
−1
(−𝑗2.5 + 2) 𝑉₁ + (−𝑗2.5) 𝑉₂ = 10
(0.5 + 𝑗0.4)𝑉₁ + (−𝑗0.4)𝑉₂ = 10
(Eq.1)
Node 2.
𝑉₂−𝑉₁
𝑉₂
𝑉₂−𝑉₁
+ 𝑗4 + 4
−𝑗2.5
−1
3
=0
1
1
1
(−𝑗2.5 − 4) 𝑉₁ + (−𝑗2.5 + 𝑗4 + 4) 𝑉₂ = 0
(−0.75 − 𝑗0.4)𝑉₁ + (0.25 + 𝑗0.15)𝑉₂ = 0
(Eq.2)
Equation (1) and (2) can be cast into matrix form as,
𝛥 = [0.5 + 𝑗0.4 − 𝑗0.4 − 0.75 − 𝑗0.4 0.25 + 𝑗0.15 ] = (0.5 + 𝑗0.4)( 0.25 + 𝑗0.15) − ( −0.75 −
𝑗0.4)( −𝑗0.4)
𝛥 = 0.225 − 𝑗0.125
𝛥₁ = [10 − 𝑗0.4 0 0.25 + 𝑗0.15 ] = (10)( 0.25 + 𝑗0.15) −( 0)(−𝑗0.4)
𝛥₁ = 2.5 + 𝑗1.5
𝛥₂ = [0.5 + 𝑗0.4 10 − 0.75 − 𝑗0.4 0 ] = (0.5 + 𝑗0.4)(0) −(−0.75 − 𝑗0.4)(10)
𝛥₂ = 7.5 + 𝑗4
𝑉₁ =
𝛥₁
𝛥
𝑉1 = 11.327∠60.02° 𝑉
Transforming this to time domain
𝑉1 = 11.327𝑐𝑜𝑠(2𝑡 + 60.02°) 𝑉
𝑉₂ =
𝛥₂
𝛥
𝑉2 = 33.02∠57.13° 𝑉
Transforming this to time domain
𝑉2 = 33.02𝑐𝑜𝑠(2𝑡 + 57.13°) 𝑉
(Sadiku, Page#416)
Example 2.
Page 3
Calculate 𝑉1 𝑎𝑛𝑑 𝑉2 in the circuit using nodal analysis.
Solution:
Applying nodal analysis,
Node 1.
𝑉1 −75
𝑉₁
𝑉
𝑉
+ + 2 + 2
4
𝑗4
−𝑗1
2
1
1
1
=0
1
(𝑗4 + 4) 𝑉₁ + (−𝑗1 + 2) 𝑉₂ = 18.75
(0.25 − 𝑗0.25)𝑉₁ + (0.5 + 𝑗)𝑉₂ = 18.75
(Eq.1)
Supernode:
𝑉1 − 𝑉2 = 100∠60°
(Eq.2)
Equation (1) and (2) can be cast into matrix form as,
𝛥 = [0.25 − 𝑗0.25 0.5 + 𝑗 1 1 ] = (0.25 − 𝑗0.25)( 1) − ( 1)( 0.5 + 𝑗)
𝛥 = −0.75 − 𝑗0.75
𝛥₁ = [18.75 0.5 + 𝑗 100∠60° 1 ] = (18.75)(1) −( 100∠60°)(0.5 + 𝑗)
𝛥₁ = 42.85 − 𝑗93.30
𝛥₂ = [0.25 − 𝑗0.25 18.75 1 100∠60° ] = (0.25 − 𝑗0.25)(100∠60°) −(1)(18.75)
𝛥₂ = 15.40 + 𝑗9.15
𝑉₁ =
𝛥₁
𝛥
𝑉1 = 96.798∠69.668° 𝑉
𝑉₂ =
𝛥₂
𝛥
𝑉2 = 16.89∠165.72° 𝑉
(Sadiku, Page#417)
Page 4
Example 3.
Find 𝐼𝑜 in the circuit using nodal analysis.
Solution:
Find 𝐼𝑜,
𝑉
1
Where 𝐼𝑜 = 8−𝑗2
Applying nodal analysis,
Node 1.
𝑉₁
𝑉₁
𝑉₁−𝑉₂
+ +
−
8−𝑗2
𝑗4
6
1
1
1
10 = 0
−1
(8−𝑗2 + 𝑗4 + 6) 𝑉₁ + ( 6 ) 𝑉₂ = 10
(0.284 − 𝑗0.2206)𝑉₁ + (−0.1667)𝑉₂ = 10
(Eq.1)
Node 2.
𝑉2 = 50∠30°
(Eq.2)
Substituting the 𝑉₂
(0.284 − 𝑗0.2206)𝑉₁ + (−0.1667)(50∠30° ) = 10
(0.284 − 𝑗0.2206)𝑉₁ (−0.1667)(50∠30°)
=
0.284 − 𝑗0.2206
0.284 − 𝑗0.2206
𝑉1 = 49.26∠51.44° 𝑉
Substitute the 𝑉1 𝑡𝑜 𝐼𝑜
𝐼𝑜 =
49.26∠51.44°
8 − 𝑗2
𝐼𝑜 = 5.49∠65.48° 𝐴
(Sadiku, Page#418)
Page 5
Exercises:
1. Use nodal analysis to find V in the circuit.
2. Determine Vo in the circuit.
V
₁
3. Calculate the voltage at nodes 1 and 2 in
the circuit.
4. Determine Vx in the circuit using nodal
analysis.
V
₁
V
₂
5. Obtain Vo in the circuit using nodal analysis.
V
₁
V
₂
V
₃
Page 6
6. By nodal analysis, obtain current Io in the circuit.
₁
7. Use nodal analysis to find Vx in the circuit.
8. Solve for the current I in the circuit using nodal analysis.
V
₁
V
₂
9. Determine Vx using nodal analysis.
V
₁
V
₂
10. Obtain Vo using nodal analysis.
V
₁
Page 7
11. Find Io in the circuit using nodal analysis
V
₁
12. Solve Io using nodal analysis.
V
₁
13. Using nodal analysis to find current Io in the circuit.
V
₁
14. Determine i in the circuit.
V
₁
15. Use nodal analysis to find Ix in the circuit.
V
₁
V
₂
MESH ANALYSIS
Page 8
OBJECTIVES:
At the end of the lesson, the learners should be able;
1. Assign mesh currents to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages
terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh current
INTRODUCTION:
Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as
the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient
and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed
path with no node passed more than once. A mesh is a loop that does not contain any other loop within
it.
Q 1) In the circuit shown in Figure 53 determine all branch currents and the voltage across the
5 Ω resistor by loop current analysis.
Solution:
are two meshes and named as
Figure 2.
In the given circuit there
i1 and i2 as shown in
Applying the KVL for the loop abcda. The loop is
passing through 3, 5 and 6 Ω resistors and it touches
negative terminal of the battery of 50 volts. The 5 Ω
resistor is common to loop currents i1 and i2. For loop1 the current i2 in 5 Ω resistor is opposite
to
the
3i1
+
i1
5(i1
current.
−
i2)
Hence
+
loop1
6i1
−
equation
50
is.
=
0
Page 9
14i1
−
5i2
=
50
Similarly for the loop befcb , The loop2 is passing through 2, 8 and 5 Ω resistors and it touches
positive terminal of the battery of 25 volts. The 5 Ω resistor is common to loop currents i1 and i2.
For loop2 the current i1 in 5 Ω resistor is opposite to the i2 current. Hence loop2 equation is.
2i2 + 8i2 + 5(i2 − i1) + 25 = 0
−5i1 + 15i2 = −25
The two simultaneous equations are:
14i1 − 5i2 = 50
(1)
−5i1 + 15i2 = −25
(2)
Multiply eqn 1 by 3 and adding with equation 2
42i1 − 15i2 = 150
−5i1 + 15i2 = −25
−−−−−=−−
37i1 = 125
i1 = 3.3784A
i2 = −0.541A
iab = 3.3784A
ieb = −i2 = 0.541A
ibc = i1 − i2 = 3.3784 − (−0.541) = 3.9194A
voltage across the 5 Ω resistor is 5ibc = 19.597V
Q 2) In the circuit shown in Figure 51 determine all branch currents by mesh current analysis.
Solution:
Applying the KVL for the loop abcda
8i1 + 4(i1 − i2) − 42 = 0
12i1 − 4i2 = 42
Similarly for the loop befcb
6i2 + 4(i2 − i1) − 10 = 0
−4i1 + 10i2 = 10
Simultaneous equations are
12i1 − 4i2 = 42
−4i1 + 10i2 = 10
Page 10
Q3) Find the loop currents i1, i2, i3 in the circuit shown in Figure 17
Solution:
Vx = 3(i3 − i2)
i1 = 10A
i3 − i1 =Vx / 8 = 0.125Vx
i3 − 10 = 0.125 × 3(i3 − i2)
i3 − 10 = 0.375i3 − 0.375i2
0.375i2 + 0.625i3 = 10
Applying KVL to mesh i2
4i2 + 3(i2 − i3) + 5(i2 − i1) = 0
−5i1 + 12i2 − 3i3 = 0
−5 × 10 + 9i2 − 4i3 = 0
12i2 − 3i3 = 50
The simultaneous equations are
0.375i2 + 0.625i3 = 10
12i2 − 3i3 = 50
Exercise 1
In the circuit shown in Figure 18 determine thecurrent through 2 Ω resistor.
Page 11
Exercise 2
In the circuit shown in Figure 21 determine the current Ix
Exercise 3
In
the
circuit
shown
in
Figure
23
determine
the
current
Ix
Exercise 4
In the circuit shown in Figure 25 determine the current Ix
Exercise 5
Page 12
In the circuit shown in Figure 27 determine all the loop currents
Exercise 6
Use loop analysis to find Vx in the circuit shownin Figure 32 such that the current through 2 + J3 Ω is
zero.
Exercise 7
Use mesh analysis to determine the three mesh currents i1, i2, i3 as shown in Figure 42
Page 13
Exercise 8
Find the voltage across 20 Ω resistor in the network shown in Figure ?? by mesh analysis.
Exercise 9
Use
mesh analysis to
determine the three mesh currents i1, i2, i3 also determine the voltage drop across 2Ω for the circuit
shown in Figure 49
Exercise 10
Use mesh analysis to determine the mesh currents i1, i2, i3, i4 for the circuit shown in Figure 50
Exercise 11
Determine the loop currents I1, I2, I3 and I4 for the network shown in Figure 38.
Page 14
Exercise 12
For the circuit shown in
supplied 10 V source
Figure 60 find the power
using mesh analysis.
Exercise 13
Use mesh analysis to find the current in the circuit of Fig.
3.20.
Exercise 14
Page 15
For
the
circuit
in
Fig.
3.24,
find
to
using
mesh
analysis.
Exercise 15
For the circuit in Fig. 3.18, find the branch currents and using mesh analysis.
Page 16
SUPERPOSITION THEOREM
OBJECTIVES:
At the end of the lesson, the learners should be able;
1. To determine the voltages at each node and the current each node in an AC circuit using
superposition theorem.
2. To understand the theorem if the circuit has source operating at different frequencies.
3. To determine if the all source have the same frequency, superposition is applied in the phasor
domain.
INTRODUCTION:
The Superposition Theorem states that a circuit can be analyzed with only one source of power
at a time, the corresponding component voltages and currents algebraically added to find out what they'll
do with all power sources in effect. Since ac circuits are linear, the theorem applies to ac circuits the
same way it applies to de 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.
Example.
Use the Superposition theorem to 𝐼0
𝐼0 = 𝐼 ′ 0 + 𝐼 ′ ′0
−𝑗2(8+𝑗10)
Z = −2𝑗+8+10𝑗 = 0.25 − 𝑗2.25
𝐼′ 0 =
𝑗20
4−𝑗2+𝑍
=
𝑗20
4−𝑗2+0.25−𝑗2.25
𝐼 ′ 0 = −2.353 + 𝑗2.353
Use mesh analysis to find 𝐼 ′ ′0
Mesh 1
(8+j8)𝐼1 − 𝑗10𝐼3 + 𝑗2𝐼2 = 0
Mesh 2
(4 − 𝑗4)𝐼2 + 𝑗2𝐼1 + 𝑗2𝐼3 = 0
Mesh 3
𝐼3 = 5
Substituting 𝐼3 𝑡𝑜 𝑚𝑒𝑠ℎ 1 𝑎𝑛𝑑 2
Page 17
(4 − 𝑗4)𝐼2 + 𝑗2𝐼1 + 𝑗2(5) = 0
𝐼1 = (2 + 𝑗2)𝐼2 − 5
For mesh 1
(8+j8)(2 + 𝑗2)𝐼2 − 5 − 𝑗10(5) + 𝑗2𝐼2 = 0
𝐼2 =
90−𝑗40
34
= 2.647 − 𝑗1.176
For 𝐼 ′ ′0
𝐼 ′ ′0 = −𝐼2
= −(2.647 − 𝑗1.176)
𝐼 ′ ′0 = −2.647 + 𝑗1.176
Calculate 𝐼0
𝐼0 = 𝐼 ′ 0 + 𝐼 ′ ′0
= −2.353 + 𝑗2.353 − 2.647 + 𝑗1.176
𝐼0 = −5 + 𝑗3.529 = 6.12 ∠ 144.78º A
Example :
Find current 𝐼0 𝑢𝑠𝑖𝑛𝑔 𝑠𝑢𝑝𝑒𝑟𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚.
KCL at node A
43.30+𝑗25−𝑉𝑎
6
=
𝑉𝑎
𝑗4
43.30 +j25 -𝑉𝑎 =
𝑉𝑎 =
+
𝑉𝑎
8−𝑗2
6
𝑗4
+ 8−𝑗2
6
43.30+𝑗25
6
6
+
+1
𝑗4 8−𝑗2
𝑉𝑎 ′ = 8.7470 + 𝑗21.4416 eq1.
10 +
0−𝑉𝑎
6
=
𝑉𝑎
𝑗4
𝑉
𝑎
+ 8−𝑗2
10 −
𝑉
𝑉
𝑎
10 = 𝑗4𝑎 + 8−𝑗2
+
𝑉𝑎
𝑉𝑎
𝑉𝑎
=
+
6
𝑗4 8 − 𝑗2
𝑉𝑎
6
𝑉𝑎 ′′ = 21.9558 + 17.0347𝑗 𝑉
Page 18
𝑉𝑎 = 8.7470 + 𝑗21.4416 + 21.9558 + 17.0347𝑗
𝑉𝑎 = 30.7028 + 38.4763𝑗 V
𝐼0 =
𝑉𝑎
30.7028 + 38.4763𝑗
=
8 − 𝑗2
8 − 𝑗2
𝐼0 = 2.4804 + 5.4296𝑗 A
Example:
Calculate 𝑣0 in the circuit using superposition theorem.
75∠-90
j5
75 sin5t = 75∠-90°
𝜔= 5
1H = j 𝜔𝐿 = j (5)(1) = j5
0.2𝐹 =
1
𝑗𝜔𝐶
−𝑗
= 5(0.2) = −1𝑗
Solution:
𝑉−75∠−90°
𝑉
+ −𝑗
8
1
1
+
𝑉
𝑗5
=0
1
(8 + −𝑗 + 𝑗5) 𝑉 = 9.375∠-90°
V = 11.577 ∠-171.12° + 90°
V = 11.57 ∠-81.12°
Page 19
6 cos10t = 6∠0°
𝜔 = 10
1H = j 𝜔𝐿 = j (10)(1) = j10
0.2𝐹 =
1
𝑗𝜔𝐶
= 10(0.2) = −𝑗0.5
𝑉
8
𝑉
−0.5𝑗
+
1
−𝑗
1
+
𝑉
𝑗10
− 6∠0° = 0
1
(8 + −0.5𝑗 + 𝑗10) 𝑉 = 6∠0°
V = 3.15 ∠ − 86.23°
𝑉0 = 11.577 𝑠𝑖𝑛 𝑠𝑖𝑛 (5𝑡 − 81.12°) + 3.15 𝑐𝑜𝑠(10𝑡 − 86.23°) V
Exercises ;
1. Find I in the circuit using Superposition
I
2. Find I in the circuit using Superposition.
Page 20
3. Find i in the circuit using Superposition.
4. Calculate 𝑣0 in the circuit using superposition theorem.
5.
Use
Superposition principle to obtain 𝑣𝑥 in the circuit.
Let 𝑣𝑠 = 50 𝑠𝑖𝑛 𝑠𝑖𝑛 2𝑡 𝑉 and 𝑖𝑠 = 12 𝑐𝑜𝑠(6t+10°)
6. Find 𝑣0 in the circuit, assuming that 𝑣𝑠 = 6 𝑐𝑜𝑠 𝑐𝑜𝑠 2𝑡 + 4 𝑠𝑖𝑛 𝑠𝑖𝑛 4𝑡 𝑉.
7. Determine 𝑣0 in the Circuit using Superposition principle
+
-
Page 21
0.1
8. Determine 𝑖0 in the circuit using superposition principle
9. Solve for 𝑣0(t) in the circuit using superposition principle.
10. Determine 𝐼0 in the circuit using superposition principle.
11. Determine
𝐼0 in the circuit using
superposition principle.
12. Find the current through the capacitor of (-j5) using superposition principle.
Page 22
13. Use
obtain 𝑣𝑥 in the
Let 𝑖𝑠 = 50 𝑠𝑖𝑛 𝑠𝑖𝑛 2𝑡 𝐴 and𝑣𝑠 = 12 𝑐𝑜𝑠(6t+10°) V.
Superposition principle to
circuit.
14. Use superposition to find V in the network shown in the circuit.
15. Find 𝐼𝑥 Using the Superposition Principle.
THEVENIN’S THEOREM AC CIRCUIT
OBJECTIVES:
At the end of the lesson, the learners should be able;
1. To determine the concept of Thevenin's theorem,
Page 23
2. To apply Thevenin's theorem to simplify complex circuits.
INTRODUCTION:
Thevenin's theorem is a powerful tool in electrical circuit analysis that allows us to simplify
complex networks into a single equivalent circuit. It states that any linear electrical network with voltage
and current sources can be replaced by an equivalent voltage source in series with an equivalent
resistance. This theorem greatly simplifies circuit analysis by reducing the complexity of the circuit
while preserving its behavior at a specific load. By using Thevenin's theorem, engineers and technicians
can easily analyze and predict the behavior of the circuits making an invaluable concept in electrical
engineering.
Example 1: ( Sadiku, page 449)
Find the Thevenin equivalent circuit at terminals a-b.
Solution:
To find the 𝑍𝑍ℎ open circuit the current source, consider the circuit below;
(𝑍4)(−𝑍2)
𝑍𝑍ℎ = 6 + (
𝑍4−𝑍2
)
𝑍𝑍ℎ = 6 − 𝑍4 = 7.211𝑍 − 33.69°𝑍
To obtain the value of 𝑍𝑍ℎ use ohm’s law:
𝑍𝑍ℎ = 𝑍𝑍ℎ 𝑍 where as 𝑍 = 2
𝑍𝑍ℎ = (7.211𝑍 − 33.69°)(2)
𝑍𝑍ℎ = 14.422𝑍 − 33.69°𝑍
Page 24
Thevenin Equivalent Circuit:
Example 2: (Sadiku, page 449)
Find the Thevenin equivalent circuit at terminals a-b.
Solution:
Use Current Divider Rule to solve for 𝑍𝑍 ;
𝑍𝑍 =
8 − 𝑍6
(5𝑍45𝑍°) = 5.59𝑍 − 18.43°𝑍
8 − 𝑍6 + 𝑍10
𝑍𝑍ℎ = 𝑍10 𝑍𝑍
𝑍𝑍ℎ = 𝑍10(5.59𝑍 − 18.43° )
𝑍𝑍ℎ = 55.9𝑍71.56° 𝑍
To find the 𝑍𝑍ℎ, open circuit all the current source, consider the circuit below;
𝑍𝑍ℎ = [(8 − 𝑍6) ∥ 𝑍10]
𝑍𝑍ℎ = 11.18𝑍26.56°𝑍
Page 25
Thevenin Equivalent Circuit:
26.56
71.56
Example 3: (Sadiku, page 428 )
Find the Thevenin equivalent circuit at terminals a-b.
Solution:
Use Nodal analysis to find the 𝑍𝑍ℎ :
𝑍−100𝑍20
𝑍
+ −𝑍4
6+𝑍2
1
1
+
)
6+𝑍2
−𝑍4
𝑍(
=0
=
100𝑍20
6+𝑍2
𝑍(0.15 + 𝑍0.2) = 15.81 + 𝑍0.432
𝑍 = 39.315 − 𝑍49.541 𝑍
𝑍 = 𝑍𝑍ℎ = 63.24𝑍 − 51.57° 𝑍
To find the 𝑍𝑍ℎ, short circuit the voltage source:
𝑍𝑍ℎ = [(6 + 𝑍2) ∥ −𝑍4] + 10
𝑍𝑍ℎ = 12.4 − 𝑍3.2 Ω
𝑍𝑍ℎ = 12.81𝑍 − 14.47°Ω
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Thevenin Equivalent Circuit:
Exercise 1
Find the Thevenin equivalent circuit at terminals a-b.
Exercise 2
Find the 𝑍𝑍ℎ in the given circuit at terminals a-b.
Exercise 3
Find the Thevenin equivalent circuit at terminals a-b.
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Exercise 4
Find the Thevenin equivalent circuit at terminals a-b.
Exercise 5
Find the Thevenin equivalent circuit at terminals a-b.
Exercise 6
Find the Thevenin equivalent circuit at terminals a-b.
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Exercise 7
Find the Thevenin equivalent circuit at terminals a-b. Take ω = 10 rad/s.
Exercise 8
Find the Thevenin equivalent circuit at terminals a-b.
Exercise 10
Using Thevenin’s Theorem, find the 𝑍𝑍 in the given circuit.
Exercise 11
Calculate the output impedance of the given circuit.
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Exercise 12
Find the Thevenin equivalent circuit at terminals a-b.
Exercise 13
Find the Thevenin Equivalent at terminals a-b.
Exercise 14
Find the Thevenin Equivalent at terminals a-b.
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Exercise 15
Find the Thevenin Equivalent at terminals a-b.
NORTON’S THEOREM
OBJECTIVES:
At the end of the lesson, the learners should be able;
●
●
●
to simplify complex circuits by replacing them with a single AC current source and
impedance using Norton’s Theorem.
to simplify in the analysis and understanding of circuit behavior in the frequency
domain.
include achieving impedance matching, improving circuit analysis efficiency, and
supporting circuit design and troubleshooting efforts in the context of AC signals.
INTRODUCTION:
Norton's theorem simplifies complex electrical circuits by replacing them with a single current
source and a resistor. It is based on the principle of superposition, where the total response is the sum
of individual responses caused by each source acting alone. The Norton current is found by calculating
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the short-circuit current, and the Norton resistance is determined by looking into the circuit. Using
Norton's theorem makes circuit analysis easier and more manageable.
Example 1.
Determine the Norton equivalent of the circuit as seen from the terminals a-b. Use the equivalent to
find 𝑍𝑍 .
Solution:
Impedances:
𝑍1 = 4 + 𝑍2
𝑍2 = 1 − 𝑍3
Find 𝑍𝑍 .
@ Mesh 2
8(𝑍2 − 𝑍1 ) + 𝑍1 + 𝑍2 (𝑍2 − 𝑍3 )
=0
8(𝑍2 − 𝑍1 ) + (4 + 𝑍2)𝑍2 + (1 − 𝑍3)(𝑍2 − 𝑍3 ) = 0
−8𝑍1 + (8 + 4 + 𝑍2 + 1 − 𝑍3)𝑍2 − (1 − 𝑍3)𝑍3 = 0
−8𝑍1 + (13 − 𝑍)𝑍2 − (1 − 𝑍3)𝑍3 = 0………..(1)
@ Super Mesh
−20𝑍0 + 8(𝑍1 − 𝑍2 ) + 𝑍2 (𝑍3 − 𝑍2 ) = 0
−20𝑍0 + 8(𝑍1 − 𝑍2 ) + (1 − 𝑍3)(𝑍3 − 𝑍2 ) = 0
−20 + 8(𝑍1 − 𝑍2 ) + (1 − 𝑍3)(𝑍3 − 𝑍2 ) = 0
8𝑍1 − (8 + 1 − 𝑍3)𝑍2 + (1 − 𝑍3)𝑍3 = 20
8𝑍1 − (9 − 𝑍3)𝑍2 + (1 − 𝑍3)𝑍3 = 20…….(2)
𝑍1 − 𝑍2 = 4𝑍 − 90
𝑍1 − 𝑍2 = −𝑍4…………(3)
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𝛥 =
= −8[0 − (−9 + 𝑍3)] − (13 − 𝑍)[(−1)(1 − 𝑍3) − 8] + (−1 + 𝑍3)[(−1)(−9 + 𝑍3)]
= −8(9 − 𝑍3) + (13 − 𝑍)(9 − 𝑍3) + (−1 + 𝑍3)(9 − 𝑍3)
= (−8 + 13 − 𝑍 − 1 + 𝑍3)(9 − 𝑍3)
= (4 + 𝑍2)(9 − 𝑍3)
= 42 + 𝑍6
𝛥3 =
= −8[0 − (−𝑍4)(−9 + 𝑍3)] − (13 − 𝑍)[(−1)(20) + 𝑍32)]
= −8(−12 − 𝑍36) − (13 − 𝑍)(−20 + 𝑍32)
= 96 + 𝑍288 + 228 − 𝑍436
= 324 − 𝑍148
𝑍𝑍 = 𝑍3
𝑍3
=
𝑍
324 − 𝑍148
=
42 + 𝑍6
= 7.0666 − 𝑍4.5333
𝑍𝑍 = 8.396𝑍 − 32.68°𝑍
Find 𝑍𝑍 .
𝑍𝑍 = (4 + 𝑍2) ∥ (8 + 1 − 𝑍3)
= (4 + 𝑍2) ∥ (9 − 𝑍3)
(4 + 𝑍2)(9 − 𝑍3)
=
(4 + 𝑍2) + (9 − 𝑍3)
42 + 𝑍6
=
13 − 𝑍
𝑍𝑍 = 3.176 + 𝑍0.706𝑍
Norton Equivalent Circuit:
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8.396∠32.68°
3.176+j
0.706Ω
𝑍𝑍
𝑍𝑍 = (
)𝑍
𝑍𝑍 + 10 − 𝑍5 𝑍
3.176 + 𝑍0.706
=(
) (8.396𝑍 − 32.68°)
3.176 + 𝑍0.706 + 10 − 𝑍5
3.176 + 𝑍0.706
=(
) (8.396𝑍 − 32.68°)
13.176 − 𝑍4.294
= (0.202 + 𝑍0.1195)(8.396𝑍 − 32.68°)
𝑍𝑍 = 1.9714𝑍 − 2.10°𝑍
(Sadiku, P.P 10.10, #430)
Example 2.
Find the Norton equivalent circuit at terminals a-b.
Find 𝑍𝑍 .
@ Mesh 1
𝑍1 ( 13 + 10 − 𝑍5) + 𝑍2 (−13 + 𝑍5) + 𝑍3 (−10) = 60𝑍45°
𝑍1 ( 23 − 𝑍5) + 𝑍2 (−13 + 𝑍5) + 𝑍3 (−10) = 60𝑍45°……….(1)
@ Mesh 2
𝑍1 (−13 + 𝑍5)+𝑍2 (13 + 12 − 𝑍5) = 0
𝑍1 (−13 + 𝑍5)+𝑍2 (25 − 𝑍5) = 0………….(2)
@ Mesh 3
𝑍1 (−10)+𝑍3 (10 + 8 + 𝑍6) = 0
𝑍1 (−10)+𝑍3 (18 + 𝑍6) = 0………..(3)
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= [(23 − 𝑍5)(25 − 𝑍5)(18 + 𝑍6) + (−13 + 𝑍5)(0)(−10) + (−10)(−13 + 𝑍5)(0)] − [(−10)
(25 − 𝑍5)(−10) + (23 − 𝑍5)(0)(0) + (−13 + 𝑍5)(−13 + 𝑍5)(18 + 𝑍6)]
= 5468 + 𝑍956
= [(23 − 𝑍5)(0)(18 + 𝑍6) + (60𝑍45°)(0)(−10) + (−10)(−13 + 𝑍5)(0)] − [(−10)
(0)(−10) + (23 − 𝑍5)(0)(0) + (60𝑍45°)(−13 + 𝑍5)(18 + 𝑍6)]
= 11709.6883 + 𝑍10691.45453
= [(23 − 𝑍5)(25 − 𝑍5)(0) + (−13 + 𝑍5)(0)(−10) + (60𝑍45°)(−13 + 𝑍5)(0)] − [(60𝑍45°)
(25 − 𝑍5)(−10) + (23 − 𝑍5)(0)(0) + (−13 + 𝑍5)(−13 + 𝑍5)(0)]
= 12727.92206 + 𝑍8485.281374
𝑍2
𝑍
11709.6883 + 𝑍10691.45453
=
5468 + 𝑍956
𝑍2 = 2.41 + 𝑍1.53
𝑍2 =
𝑍3
𝑍
12727.92206 + 𝑍8485.281374
=
5468 + 𝑍956
𝑍3 = 2.52 + 𝑍1.11
𝑍3 =
𝑍𝑍 = 𝑍3 − 𝑍2
= (2.52 + 𝑍1.11) − (2.41 + 𝑍1.53)
𝑍𝑍 = 0.11 − 𝑍0.42 = 0.44𝑍 − 75.18°𝑍
Find 𝑍𝑍 .
𝑍𝑍 = 10 ∥ (13 − 𝑍5) + 12 ∥ (8 + 𝑍6)
(10)(13 − 𝑍5)
(12)(8 + 𝑍6)
𝑍𝑍 =
+
(10) + (13 − 𝑍5) (12) + (8 + 𝑍6)
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𝑍𝑍 = 11.24 + 𝑍1.079 = 11.29𝑍5.48°Ω
Norton Equivalent Circuit
a
11.29∠5.48°Ω
b
(Sadiku, P 10.67, #451)
Example 3.
Find the Norton equivalent circuit at terminals a-b.
Find 𝑍𝑍 .
@ Mesh 1
𝑍1 = 4𝑍
@ Mesh 2
𝑍1 (−8)+𝑍2 (8 − 𝑍5 + 𝑍10) + 𝑍3 (−𝑍10) = 0
(4)(−8)+𝑍2 (8 − 𝑍5 + 𝑍10) + 𝑍3 (−𝑍10) = 0
𝑍2 (8 + 𝑍5) + 𝑍3 (−𝑍10) = 32……….(1)
Substitute 4𝑍 in 𝑍1
@ Mesh 3
𝑍2 (−𝑍10)+𝑍3 (𝑍10) = 0……….(2)
= (8 + 𝑍5)(𝑍10) − (−𝑍10)(−𝑍10)
= 50 + 𝑍80
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= (8 + 𝑍5)(0) − (32)(−𝑍10)
= 𝑍320
𝑍3
𝑍
𝑍320
=
50 + 𝑍80
𝑍𝑍 = 2.876 + 𝑍1.798 = 3.39𝑍32.01°
𝑍𝑍 =
Find 𝑍𝑍 .
𝑍𝑍 = 8 − 𝑍5 ∥ 𝑍10
= 8 − 𝑍5 ∥ 𝑍10
(8 − 𝑍5)(𝑍10)
=
8 − 𝑍5 + 𝑍10
50 + 𝑍80
=
8 + 𝑍5
𝑍𝑍 = 8.989 + 𝑍4.382 = 10𝑍25.99°Ω
Norton Equivalent Circuit:
a
3.39∠32.
01°A
10∠25.
99°Ω
b
(Sadiku, P 10.55, #449)
EXERCISES:
1. Find the Norton equivalent circuit at terminals a-b.
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2. Find the Norton equivalent circuit at terminals a-b.
3. Find the Norton
terminals a-b.
equivalent circuit at
4. Find the Norton equivalent circuit at terminals a-b.
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5. Find the Norton equivalent circuit at terminals a-b.
6. Find the Norton equivalent circuit at terminals c-d.
7. Find the Norton equivalent circuit at terminals a-b.
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8. Find the Norton equivalent circuit at terminals a-b.
9. Find the Norton equivalent circuit at terminals a-b.
10. Find the
circuit at
Norton equivalent
terminals a-b.
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11. Find the Norton equivalent circuit at terminals a-b.
12. Find the Norton equivalent circuit for the network external to the 7Ω capacitive reactance.
13. Find the Norton equivalent circuit at terminals a-b.
14. Find the Norton equivalent circuit at terminals a-b.
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15. Find the Norton equivalent circuit at terminals a-b.
Page 42