of Electric Circuits 7th Edition
By Charles Alexander,
Matthew Sadiku (All
Chapters 1-19, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse: 19-1
This is The Only Original and
Complete Test Bank for 7 th
Edition, All Other Files in the
Market are Fake/Old/Wrong
Edition.
,12/3/25, 4:15 AM Chapter 19
1. Award: 10.00 points Problems? Adjust credit for all students.
Obtain the z parameters for the network shown. Take R = 10 Ω.
A B
The z parameters for the network are [ ] Ω .
C D
The value of A is 16 ± 2% Ω.
The value of B is 2 ± 2% Ω.
The value of C is 2 ± 2% Ω.
The value of D is 3.3333 ± 2% Ω.
Explanation:
To get z11 and z21, consider the circuit shown.
V1
z11 = = 10 + (12 ∥ (8 + 4)) = 16 Ω
I1
1
Io = I1
2
V 2 = 4Io = 2I1
V2
z21 = = 2 Ω
I1
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,12/3/25, 4:15 AM Chapter 19
To get z22 and z12, consider the circuit given below.
V2
z22 = = 4 ∥ (8 + 12) = 3.3333 Ω
I2
4 1
Io = I2 = I2
4+20 6
′
V 1 = 12Io = 2I2
V1
z12 = = 2 Ω
I2
16 2
[z] = [ ] Ω
2 3.3333
The value of A is 16 Ω.
The value of B is 2 Ω.
The value of C is 2 Ω.
The value of D is 3.3333 Ω.
Hints References
Hint #1
Source: Fundamentals of Electric Circuits (Alexander, 7, ISBN 1260477649) > Chapter 19 Two-Port
Networks
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, 12/3/25, 4:15 AM Chapter 19
2. Award: 10.00 points Problems? Adjust credit for all students.
Find the impedance parameter equivalent of the given network where R = 3 Ω.
z11 = 8.200 ± 2% Ω
z21 = 0.022 ± 2% Ω
z12 = 0.022 ± 2% Ω
z22 = 8.200 ± 2% Ω
Explanation:
Consider the following circuit to get z11 and z21.
z11 = V1 / I1 = (3 + 3) + 3 || [(3 + 3) + 3 || (3 + 3 + 3)]
Solving this, we get z11 = 8.200 Ω.
3 ′
Io = Io
3+3+3+3
′ 3
Io = I1
3 + 8.200
3 3
Io = ( ) I1 = 0.022I1
3+3+3+3 3 + 8.200
V2 = Io = 0.022I1
z21 = V2 / I1 = z12 = 0.022 Ω
To get z22, consider the circuit given below.
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