FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
AN INFORMATION TO THE STUDENT!
Reference for the solutions is Fundamentals of Electric Circuits-4th Edition Alexander Sadiku.
This book has been prepared to provide elegant and clean solutions for application problems
and "problems" sections that are meticulously found at the end of each unit!
The solutions for the "problems" section are prepared for odd numbered questions!
While making the calculations, the significant figures were taken into consideration and the
necessary procedures were applied meticulously!
Some problems may not be solved because they are not needed!
In the practice problems and applications sections, just one problem will be solved at the end
of each unit!
As we know, some symbols or representations differ in Europe and America. The
representations of the referenced book will be applied in this book!
Recommendation!
If you feel inadequate in Linear Algebra, the course that underpins Circuit Theory, you should
qualify yourself in Linear Algebra before reviewing this book!
The recommended book for you to understand the basics of Linear Algebra is Linear Algebra: A
Modern Introduction by David Poole.
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1
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Chapter 1-Practice Problems
1.1
As it is known, the charge of one proton is 1.602 × 10−19 𝐶 .
This implies that, (4 × 106 )(1.602 × 10−19 𝐶) = 𝟔. 𝟒𝟎𝟖 × 𝟏𝟎−𝟏𝟑 𝑪
1.2
ⅆ𝑞
ⅈ(𝑡) = ⅆ𝑡 =
ⅆ(10−10ⅇ −2𝑡 )
ⅆ𝑡
= 20 ⋅ ⅇ −2𝑡
This implies that, ⅈ(0.5) = 7.36 × 10−3 𝐴 = 𝟕. 𝟑𝟔 𝒎𝑨
1.3
𝑡
ⅆ𝑞
ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
2
1
2
1
2
This implies that, ∫0 ⅈ(𝜓) ⅆ𝜓 = ∫𝑜 ⅈ(𝜓) ⅆ𝜓 + ∫1 ⅈ(𝜓) ⅆ𝜓 = ∫0 (2𝐴) ⅆ𝜓 + ∫1 (2𝜓 2 𝐴) ⅆ𝜓
2
3
So that, 𝑞 = (2𝜓)|𝜓=1
𝜓=0 + (3 𝑡 )|
1.4
𝜓=2
𝜓=1
2
2
3
= (2) ⋅ (1) − (2). (0) + 3 (23 ) − (13 ) = 𝟔. 𝟔𝟔𝟕𝑪
𝛥𝑤
It’s known that, 𝑣 = 𝛥𝑞
−30𝐽
−30𝐽
This implies that, (a) 𝑣𝑎𝑏 = 2𝐶 = −15𝑉 (b) 𝑣𝑎𝑏 = −6𝐶 = 𝟓𝑽
1.5
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
This implies that, (a) 𝑝(𝑡) = 2ⅈ ⋅ (5 𝑐𝑜𝑠(60𝜋𝑡)) = 50 𝑐𝑜𝑠 2 (60𝜋𝑡) ⇒ 𝑝(0.005) = 17.27𝑊
𝑡
𝑡
25
(b) 𝑣 = (10 + 5 ∫0 (ⅈ) ⅆ𝜓) = 10 + (5 ∫0 5 𝑐𝑜𝑠(60𝜋𝜓) ⅆ𝜓) = 10 + 60𝜋 𝑠ⅈ𝑛(60𝜋𝑡)
This implies that,
25
𝑝(𝑡) = 𝑣 ⋅ ⅈ = [10 + 60𝜋 𝑠ⅈ𝑛(60𝜋𝑡)] . [5 𝑐𝑜𝑠(60𝜋𝑡)] ⇒ 𝑝(0.005) = 𝟐𝟗. 𝟕𝑾
1.6
By using the definitions of power, voltage and current,
𝛥𝑞
𝛥𝑤
𝛥𝑞
𝛥𝑤
ⅈ = 𝛥𝑡 and 𝑣 = 𝛥𝑞 ⇒ 𝛥𝑡 = ⅈ = ⅈ⋅𝛥𝑣 It’s obvious that 𝑡0 = 0!
𝑤
60×103 𝐽
This implies that, 𝑡 = ⅈ⋅𝑣 = 15𝐴⋅240𝑣 = 𝟏𝟔. 𝟔𝟔𝟕𝒔
2
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1.7
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
Power with respect to voltage drop direction gives, 𝑝 = 𝑣 ⋅ ⅈ otherwise 𝑝 = −𝑣 ⋅ ⅈ
𝑣3 = 3𝐴 ⋅ 0.6𝐼 and 𝐼 = 5𝐴 ⇒ 𝑣3 = 𝟑𝑽
Therefore,
𝑝1 = −5𝑉 ⋅ 8𝐴 = −𝟒𝟎𝑾
𝑝2 = 2𝑉 ⋅ 8𝐴 = 𝟏𝟔𝑾
𝑝3 = 3𝑉 ⋅ 3𝐴 = 𝟗𝑾
𝑝4 = 3𝑉 ⋅ 5𝐴 = 𝟏𝟓𝑾
So that, negative signed values mean power is supplied, positive signed values mean power is absorbed!
1.8
Let n be the number of electrons and 𝑉0 as acceleration voltage!
ⅆ𝑞
ⅆ𝑛
This implies that, ⅈ = ⅆ𝑡 = ⅇ ⋅ ⅆ𝑡 = ⅇ
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
Now it’s clear-cut that, power in the beam equals to 𝑝𝑏ⅇ𝑎𝑚 = 𝑉0 ⋅ ⅈ
So that, 𝑝𝑏ⅇ𝑎𝑚 = 𝑉0 ⋅ ⅈ = (30 × 103 𝑉) ⋅ (1.602 × 10−19 𝐶) = 48 × 10−3 𝑊 = 𝟒𝟖𝒎𝑾
1.9
Minimum monthly charge = $12.00, First 100 kWh × $0.16/kWh = $16.00
Next 200 kWh × $0.10/kWh = $20.00, Remaining charge 100 kWh × $0.06/kWh = $6.00
This implies that, average cost = $54/[100+200+100] = 13.5 Cents/kWh
Chapter 1- Problems
1.1
It’s known that one electron charge is ⅇ = −𝟏. 𝟔𝟎𝟐 × 𝟏𝟎−𝟏𝟗
Therefore,
(a) (6.482 × 1017 ) ⋅ (−1.602 × 10−19 ) = −𝟎. 𝟏𝟎𝟑𝟖𝑪
(b) (1.24 × 1018 ) ⋅ (−1.602 × 10−19 ) = −𝟎. 𝟏𝟗𝟗𝑪
(c) (2.46 × 1017 ) ⋅ (−1.602 × 10−19 ) = −𝟑. 𝟗𝟒𝑪
(d) (1.628 × 1017 ) ⋅ (−1.602 × 10−19 ) = −𝟐𝟔. 𝟎𝟖𝑪
3
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1.3
𝑡
ⅆ𝑞
By the definition of current, ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
Therefore,
𝑡
𝑡
(a) 𝑞(𝑡) = 𝑞(0) + ∫0 ⅈ(𝜓) ⅆ𝜓 = 𝑞(0) + ∫0 (3𝐴) ⅆ𝜓 = 1𝐶 + 3𝑡|𝜓=𝑡
𝜓=0 = (1 + 3𝑡)𝐶
𝑡
𝑡
2
(b) 𝑞(𝑡) = 𝑞(0) + ∫0 ⅈ(𝜓) ⅆ𝜓 = 𝑞(0) + ∫0 (2𝜓 + 5) ⅆ𝜓 = 𝑞(0) + 𝜓 2 + 5𝜓|𝜓=𝑡
𝜓=0 = (𝑡 + 5𝑡)𝐶
𝑡
𝑡
𝜋
𝜋
(c) 𝑞(𝑡) = 𝑞(0) + ∫0 ⅈ(𝜓) ⅆ𝜓 = 𝑞(0) + ∫ 20 𝑐𝑜𝑠 (10𝜓 + 6 ) ⅆ𝜓 = (2 𝑠ⅈ𝑛 (10𝑡 + 𝑏 ) + 1) µ𝐶
0
𝑡
(d) 𝑞(𝑡) = 𝑞(0) + ∫0 ⅈ(𝜓) ⅆ𝜓
𝑡
= 𝑞(0) + ∫0 10 ⋅ ⅇ −30𝜓 ⋅ sin 40𝜓 ⅆ𝜓 = −ⅇ−𝟑𝟎𝒕 [𝟎. 𝟏𝟔 𝒄𝒐𝒔 𝟒𝟎𝒕 + 𝟎. 𝟏𝟐 𝒔𝒊𝒏 𝟒𝟎𝒕]𝑪
1.5
𝑡
ⅆ𝑞
By the definition of current, ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
10
10
𝜓2
1
This implies that, 𝑞 = ∫0 ⅈ(𝑡) ⅆ𝑡 = ∫ (2 𝜓) ⅆ𝜓 = 4 |
1.7
0
𝜓=10
𝜓=0
= 𝟐𝟓𝑪
ⅆ𝑞(𝑡)
By the definition of current, ⅈ(𝑡) = ⅆ𝑡
This implies that, the slope of the graph gives the current for the time interval!
25𝐴,
0 < 𝑡 < 2,
25𝐴,
6 < 𝑡 < 8.
−25𝐴,
2 < 𝑡 < 6,
So that graph can be drawn by using the data above!
1.9
ⅆ𝑞
𝑡
By the definition of current, ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
This implies that the area of the region between the line segments and the x-axis gives the charge for
the time interval!
So that,
1
∫0 ⅈ(𝑡) ⅆ𝑡 = (10) ⋅ (1) = 10𝐶
3
∫2 ⅈ(𝑡) ⅆ𝑡 = (5) ⋅ (1) = 5𝐶
4
2
1
∫1 ⅈ(𝑡) ⅆ𝑡 = (5 + 10) ⋅ (2) = 7.5𝐶
4
∫3 ⅈ(𝑡) ⅆ𝑡 = (5) ⋅ (1) = 5𝐶
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5
1
∫4 ⅈ(𝑡) ⅆ𝑡 = (5) ⋅ (2) = 2.5𝐶
This implies that,
10𝐶,
7.5𝐶,
0 < 𝑡 < 1,
1 < 𝑡 < 2,
5𝐶,
2 < 𝑡 < 3,
2.5𝐶,
4 < 𝑡 < 5.
5𝐶,
So that,
3 < 𝑡 < 4,
(a) 10C (b) 22.5C (c) 30C
1.11
𝑡
ⅆ𝑞
By the definition of current, ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
𝛥𝑞
This implies that, ⅈ = 𝛥𝑡 ⇒ 𝛥𝑞 = ⅈ ⋅ 𝛥𝑡
So that, 𝑞 = ⅈ ⋅ 𝑡 = (85 × 10−3 𝐴) ⋅ (12ℎ ⋅
By the definition of voltage, 𝑣 =
𝛥𝑤
This implies that, 𝑣 = 𝛥𝑞
ⅆ𝑤(𝑞)
ⅆ𝑞
3600𝑠
) = 3672𝐶 = 𝟑. 𝟔𝟕𝟐𝒌𝑪
1ℎ
It’s obvious that 𝑞0 and 𝑤0 equals to zero!
So that, 𝑤 = 𝑣 ⋅ 𝑞 = (3672𝐶) ⋅ (1.2𝑉) = 4406𝐽 = 𝟒. 𝟒𝟎𝟔𝒌𝑱
1.13
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
ⅆ𝑞
By the definition of current, ⅈ(𝑡) = ⅆ𝑡
So that, ⅈ(𝑡) =
ⅆ𝑞(𝑡)
ⅆ(10 𝑠ⅈ𝑛 4𝜋𝑡)
=
= 40𝜋 𝑐𝑜𝑠 4𝜋𝑡
ⅆ𝑡
ⅆ𝑡
(Note that this value is in mA!)
This implies that, 𝑝(𝑡) = ⅈ(𝑡) ⋅ 𝑣(𝑡) = (80𝜋 𝑐𝑜𝑠 2 (4𝜋𝑡)) ⇒ 𝑝(0.3) = 250.2𝑚𝑊
By the definition of power, 𝑝(𝑡) =
0.6
𝑡
ⅆ𝑤(𝑞)
⇒ 𝑤(𝑡) = ∫0 𝑝(𝜓) ⅆ𝜓
ⅆ𝑡
0.6
This implies that , 𝑤(0.6) = ∫0 𝑝(𝜓) ⅆ𝜓 = ∫0 80𝜋 𝑐𝑜𝑠 2 (4𝜋𝜓) ⅆ𝜓
= 𝟓 𝒔𝒊𝒏 (
5
𝟐𝟒𝝅
) + 𝟐𝟒𝝅 = 𝟕𝟖. 𝟑𝟒𝒎𝑱
𝟓
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1.15
𝑡
ⅆ𝑞
By the definition of current, ⅈ(𝑡) = ⅆ𝑡 ⇒ 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓
𝑡
2
1
ⅇ −4
This implies that, 𝑞(𝑡) = ∫0 ⅈ(𝜓) ⅆ𝜓 ⇒ 𝑞(2) = ∫0 3 ⋅ ⅇ −2𝑡 ⅆ𝜓 = 3 (2 − 2 ) = 1.473𝐶
By the definition of power, 𝑝 =
ⅆⅈ
Also, 𝑣(𝑡) = 5 ⅆ𝑡 =
5 ⅆ(3ⅇ −2𝑡 )
ⅆ𝑡
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
= 5((−2) ⋅ (3ⅇ −2𝑡 )) = −30ⅇ −2𝑡
This implies that, 𝑝(𝑡) = 𝑣(𝑡) ⋅ ⅈ(𝑡) = (−30ⅇ −2𝑡 ) ⋅ (3ⅇ −2𝑡 ) = −90ⅇ −4𝑡 𝑊
By the definition of power, 𝑝(𝑡) =
𝑡
𝑡
ⅆ𝑤(𝑞)
⇒ 𝑤(𝑡) = ∫0 𝑝(𝜓) ⅆ𝜓
ⅆ𝑡
3
1
This implies, 𝑤(𝑡) = ∫0 𝑝(𝜓) ⅆ𝜓 = ∫0 −90ⅇ −4𝜓 ⅆ𝜓 = −90 (4 −
1.17
ⅇ −12
) = −𝟐𝟐. 𝟓𝑱
4
As it known, ∑𝑝 = 𝑝𝑎𝑏𝑠𝑜𝑟𝑏ⅇⅆ + 𝑝ⅆⅇ𝑙ⅈ𝑣ⅇ𝑟ⅇⅆ = 𝟎!
This implies that, 𝑝1 + 𝑝2 + 𝑝3 + 𝑝4 + 𝑝5 = 0 ⇒ 𝑝3 = −(𝑃1 + 𝑝2 + 𝑝4 + 𝑝5 ) = −(−70𝑊) = 𝟕𝟎𝑾
1.19
As it known, ∑𝑝 = 𝑝𝑎𝑏𝑠𝑜𝑟𝑏ⅇⅆ + 𝑝ⅆⅇ𝑙ⅈ𝑣ⅇ𝑟ⅇⅆ = 𝟎!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
Note that voltage drop directions are signed positive!
Therefore,
∑𝑝 = −(4) ⋅ (9) + (1) ⋅ (9) + (𝐼) ⋅ (3) + (𝐼) ⋅ (6) = 0 ⇒ 9𝐼 = 27 ⇒ 𝐼 = 𝟑𝑨
1.21
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
ⅆ𝑞
𝛥𝑞
Also we’ve known that, ⅈ(𝑡) = ⅆ𝑡 ⇒ ⅈ = 𝛥𝑡
It’s obvious that, 𝑡0 and 𝑞0 equals to zero!
Rearranging the equations gives,
𝑞 =ⅈ⋅𝑡 =
𝑝
60𝑤
36005
⋅𝑡 =
⋅ (24ℎ ⋅
) = 43200𝐶 = 𝟒𝟑. 𝟐𝒌𝑪
𝑣
120𝑣
1ℎ
Let denote 𝑛ⅇ as number of electrons!
𝑞
43200
As it known, 𝑛ⅇ = |ⅇ | = |−1.602×10−19 | = 𝟐. 𝟔𝟗𝟕 × 𝟏𝟎𝟐𝟑
END!
6
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Chapter 2-Practice Problems
2.1
By the Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
𝑣
110𝑣
This implies that, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 10𝛺 = 𝟏𝟏𝑨
2.2
It’s obvious that, given current i equals to the current that supplied by the independent current source!
By the Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
This implies that, 𝑣 = ⅈ ⋅ 𝑅 = (2 × 10−3 𝐴) ⋅ (10 × 103 𝛺) = 𝟐𝟎𝑽
1
1
1
As it known, 𝑅 = 𝐺 ⇒ 𝐺 = 𝑅 = 10×103𝛺 = 10−4 𝑆 = 100µ𝑆
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
Note that voltage drop directions are signed positive!
So that, 𝑝 = 𝑣 ⋅ ⅈ = 20𝑉 ⋅ (2 × 10−3 𝐴) = 40 × 10−3 𝑊 = 𝟒𝟎𝒎𝑾
2.3
By the definition of power, 𝑝 =
𝑝
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
𝑝
=
⋅
=𝑣⋅ⅈ ⇒ⅈ =
𝑣
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
20 𝑐𝑜𝑠2 𝑡
This implies that, ⅈ = 𝑣 = 10 𝑐𝑜𝑠 𝑡 = (2 𝑐𝑜𝑠 𝑡)𝑚𝐴
𝑣
By the Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑅 = ⅈ
𝑣
10 𝑐𝑜𝑠 𝑡
This implies, 𝑅 = ⅈ = 2 𝑐𝑜𝑠 𝑡×10−3 = 5 × 103 𝛺 = 𝟓𝒌𝜴
2.4
By the definition of branch, number of single elements gives the number of branches!
So that, 4 branches and 1 independent voltage source gives, 4 + 1 = 𝟓!
By the definition of node, number of connections between branches gives the number of nodes!
This implies that there exist 3 different nodes!
2.5
Let denote KCL as Kirchhoff’s Current Law and KVL as Kirchhoff’s Voltage Law!
Assume that direction of voltage drops signed positive and direction of current i through the loop is
clockwise direction!
So by KVL, −10 + 𝑣1 − 8 − 𝑣2 = 𝟎
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣1 = 4ⅈ and 𝑣2 = −2ⅈ
7
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
This implies , 𝑣1 − 𝑣2 = 4ⅈ − (−2ⅈ) = 18 ⇒ ⅈ = 3𝐴 so, 𝑣1 = 4ⅈ = 4 ⋅ 3 = 12𝑉
𝑣2 = −2ⅈ = (−2) ⋅ (3) = −𝟔𝑽
2.6
Let denote KCL as Kirchhoff’s Current Law and KVL as Kirchhoff’s Voltage Law!
Assume that direction of voltage drops signed positive and direction of current i through the loop is
clockwise direction!
So by KVL, −35 + 𝑣𝑥 + 2𝑣𝑥 − 𝑣0 = 0
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣𝑥 = 10ⅈ and 𝑣0 = −5ⅈ
This implies that, 3𝑣𝑥 − 𝑣0 = 3 ⋅ (10ⅈ) − (−5ⅈ) = 35 ⇒ ⅈ = 1𝐴
So that, 𝑣𝑥 = 10ⅈ = 10 ⋅ 1 = 10𝑉 and 𝑣0 = −5ⅈ = (−5) ⋅ (1) = −𝟓𝑽
2.7
Let redraw the circuit with nodes and important currents!
Let denote currents leaving a node positive!
ⅈ
By KCL for node a and node b gives, −6 + ⅈ0 + ⅈ3 = 0 and −ⅈ3 + 40 + ⅈ1 = 0
Assume that direction of voltage drops signed positive!
It’s obvious that, 𝑣2𝛺 = 𝑣0 because they are connected in same pair of terminals!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣2𝛺 = 2ⅈ0 and 𝑣0 = 8ⅈ1
This implies that, 2ⅈ0 = 8ⅈ1 ⇒ ⅈ0 = 4ⅈ1 !
ⅈ
ⅈ
ⅈ
By substitution, 6 − ⅈ0 = 40 + ⅈ1 ⇒ 6 − ⅈ0 = 40 + 40 ⇒ ⅈ0 = 4𝐴
ⅈ
As it known, 𝑣0 = 8ⅈ1 = 8 ( 40 ) = 2ⅈ0 = 𝟖𝑽
2.8
Let denote currents leaving a node positive!
By KCL for upper middle node, −ⅈ1 + ⅈ2 + ⅈ3 = 𝟎
Assume that direction of voltage drops signed positive!
CONT . . .
8
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By KVL for both left and right closed paths, −5 + 𝑣1 + 𝑣2 = 𝟎 and −𝑣2 + 𝑣3 − 3 = 𝟎
Ohm’s Law gives, 𝑣1 = 2ⅈ1 , 𝑣2 = 8ⅈ2 and 𝑣3 = 4ⅈ3
By substitution, 2ⅈ1 + 8ⅈ2 = 𝟓 and −8ⅈ2 + 4ⅈ3 = 𝟑
Now we have 3 unknowns and 3 equations!
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ1
0
−1 1 1 ⅈ1
1.5
[2
8 0] . [ⅈ2 ] = [5] ⇒ [ⅈ2 ] = [0.25]
ⅈ3
3
0 −8 4 ⅈ3
1.25
This implies that, ⅈ1 = 1.5𝐴, ⅈ2 = 0.25𝐴 and ⅈ3 = 1.25𝐴
So that, 𝑣1 = 2ⅈ1 = 2 ⋅ (1.5𝐴) = 𝟑𝑽, 𝑣2 = 8ⅈ2 = 8 ⋅ (0.25𝐴) = 𝟐𝑽, 𝑣3 = 4ⅈ3 = 4 ⋅ (1.25𝐴) = 𝟓𝑽
2.9
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
It’s obvious that, 4𝛺 and 5𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 4𝛺 + 5𝛺 = 9𝛺
Now, 9𝛺 and 3𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 9𝛺 + 3𝛺 = 12𝛺
1
1
1
1
1
1
It’s clear cut that, 9𝛺 and 3𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 12 + 4 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 3𝛺
Now, 3𝛺 and 3𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 3𝛺 + 3𝛺 =
6𝛺
𝑒𝑞
It’s clear- cut that, 6𝛺 and 6𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 6 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 3𝛺
𝑒𝑞
It’s obvious that, , 2𝛺, 3𝛺 and 1𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 2𝛺 + 3𝛺 + 1𝛺 = 𝟔𝜴
This implies, the equivalent resistance of the circuit is 𝟔𝜴
2.10
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
1
−1
for parallel connections
1
1
1
1
It’s clear that, 20𝛺 and 5𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 20 + 5 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4𝛺
Now, 4𝛺 and 1𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 4𝛺 + 1𝛺 = 5𝛺
1
𝑒𝑞
It’s clear that, 5𝛺 and 20𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 20 + 5 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4𝛺
9
𝑒𝑞
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
CONT . . .
Now, 4𝛺 and 2𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 4𝛺 + 2𝛺 = 𝟔𝜴
1
1
1
It’s clear that, 6𝛺 and 9𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 9 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟑. 𝟔𝜴
1
1
𝑒𝑞
1
Now, 3.6𝛺 and 18𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 18 + 3.6 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟑𝜴
𝑒𝑞
Finally, 3𝛺 and 8𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 3𝛺 + 8𝛺 = 𝟏𝟏𝜴
2.11
Let denote equivalent resistance as 𝑅ⅇ𝑞 and equivalent counductance as 𝐺ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
This implies that, 𝐺ⅇ𝑞 = (∑
∞
1
)
𝐺
𝑛=1 𝑛
−1
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
for the series connections and 𝐺ⅇ𝑞 = ∑∞
𝑛=1 𝐺𝑛 !
It’s obvious that, 8𝑆 and 4𝑆 resistors are connected parallel so 𝐺ⅇ𝑞 for them is 8𝑆 + 4𝑆 = 12𝑆
1
1
1
Now, , 6𝑆 and 12𝑆 resistors are connected in series so 𝐺ⅇ𝑞 for them is 6 + 12 = 𝐺 ⇒ 𝐺ⅇ = 4𝑆
𝑒
It’s obvious that, 4𝑆 and 2𝑆 resistors are connected parallel so 𝐺ⅇ𝑞 for them is 4𝑆 + 2𝑆 = 6𝑆
1
1
1
Now, 6𝑆 and 12𝑆 resistors are connected in series so 𝐺ⅇ𝑞 for them is 6 + 12 = 𝐺 ⇒ 𝐺ⅇ = 𝟒𝑺
2.12
𝑒
Let redraw the circuit with nodes and important currents!
Note that Voltage and Current Division will make things easier but the basic operations are shown to
make everything clear!
Let denote currents leaving a node positive!
So by KCL for node b gives, −ⅈ3 − ⅈ1 + ⅈ0 + ⅈ2 = 0
10
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 𝑣1 = 𝑣6𝛺 and 𝑣2 = 𝑣40𝛺 because they are both connected to the same pair of
terminals!( if there is any confusion you can verify it using Kirchhoff's Voltage Law!)
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣1 = 𝑣6𝛺 = 12ⅈ1 = 6ⅈ3 and 𝑣2 = 𝑣10𝛺 = 40ⅈ2 = 10ⅈ0
Assume that direction of voltage drops signed positive!
By KVL for left closed path with clockwise direction gives, −15 + 𝑣6𝛺 + 𝑣10𝛺 = 0
By substitution, 12ⅈ1 + 40ⅈ2 = 𝟏𝟓𝑽
So rearranging the equations gives,
−ⅈ1 + ⅈ2 − ⅈ3 + ⅈ0 = 𝟎
12ⅈ1 + 0ⅈ2 − 6ⅈ3 + 0ⅈ0 = 0
0ⅈ1 + 40ⅈ2 + 0ⅈ3 − 10ⅈ0 = 0
12ⅈ1 + 40ⅈ2 + 0ⅈ3 + 0ⅈ0 = 15
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ1
ⅈ1
−1 1 −1
1
0
0.4167
ⅈ
ⅈ
12 0 −6
0
0
0.25
[
] [ 2] = [ ] ⇒ [ 2] = [
]
ⅈ3
0 40 0 −10 ⅈ3
0
0.8333
ⅈ0
ⅈ0
12 40 0
0
15
1
This implies that, ⅈ1 = 416.7𝑚𝐴, ⅈ2 = 250𝑚𝐴, ⅈ3 = 833.3𝑚𝐴 and ⅈ0 = 1𝐴
Now by substitution, 𝑣1 = 𝑣6𝛺 = 12ⅈ1 = 5𝑉, 𝑣2 = 𝑣10𝛺 = 40ⅈ2 = 10𝑉
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
Note that direction of voltage drops signed positive!
This implies, 𝑝1 = 𝑣1 ⋅ ⅈ1 = 2083𝑚𝑊 = 𝟐. 𝟎𝟖𝟑𝑾 and 𝑝2 = 𝑣2 ⋅ ⅈ2 = 2500𝑚𝑊 = 𝟐. 𝟓𝑾
2.13
Let redraw the circuit with nodes and important currents!
Note that Voltage and Current Division will make things easier but the basic operations are shown to
make everything clear!
CONT . . .
11
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote currents leaving a node positive!
By KCL for both nodes a and b gives, ⅈ1 − 0.01 + ⅈ3 = 0 and −ⅈ3 + ⅈ0 + ⅈ2 = 0
It’s obvious that, 𝑣0 = 𝑣2 = 𝑣10𝑚𝐴 because they are connected to the same pair of terminals!( if there is
any confusion you can verify it using Kirchhoff's Voltage Law!)
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣0 = 𝑣2 = 5000ⅈ0 = 20000ⅈ2 , 𝑣1𝑘𝛺 = 1000ⅈ1 and 𝑣1 = 3000ⅈ1
Assume that direction of voltage drops signed positive!
By KCL for both leftmost closed path gives, −𝑣1 − 𝑣1𝑘𝛺 + 𝑣0 = 0
By substitution, −3000ⅈ1 − 1000ⅈ1 + 5000ⅈ0 = 𝟎
So rearranging the equations gives
ⅈ1 + 0ⅈ2 + ⅈ3 + 0ⅈ0 = 0.01
0ⅈ1 + ⅈ2 − ⅈ3 + ⅈ0 = 0
0ⅈ1 − 20000ⅈ2 + 0ⅈ3 + 5000ⅈ0 = 0
−4000ⅈ1 + 0ⅈ2 + 0ⅈ3 + 5000ⅈ0 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
0
[
0
−4000
0
1
−20000
0
1
−1
0
0
ⅈ1
ⅈ1
0
0.01
5 × 10−3
−3
ⅈ
ⅈ
1
0
] [ 2] = [
] ⇒ [ 2 ] = [1 × 10−3 ]
ⅈ3
5000 ⅈ3
0
5 × 10
ⅈ0
5000 ⅈ0
0
4 × 10−3
This implies that, ⅈ1 = 5𝑚𝐴, ⅈ2 = 1𝑚𝐴, ⅈ3 = 5𝑚𝐴 and ⅈ0 = 4𝑚𝐴
Now by substitution, 𝑣0 = 𝑣2 = 5000ⅈ0 = 20𝑉 and 𝑣1 = 3000ⅈ1 = 15𝑉
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
Note that direction of voltage drops signed positive!
This implies, 𝑝3𝑘𝛺 = 𝑣1 ⋅ ⅈ1 = 75 × 10−3 𝑊 = 75𝑚𝑊 and 𝑝20𝑘𝛺 = 𝑣2⋅ⅈ2 = 20 × 10−3 𝑊 = 𝟐𝟎𝒎𝑾
2.14
Let redraw the circuit with the 𝛥 and 𝑌 form nested! (This form is called Superposition of 𝛥 and 𝑌
networks!)
CONT . . .
12
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By setting 𝑅𝑎𝑐 (𝑌) = 𝑅𝑎𝑐 (𝛥), 𝑅𝑎𝑏 (𝑌) = 𝑅𝑎𝑏 (𝛥) and 𝑅𝑏𝑐 (𝑌) = 𝑅𝑏𝑐 (𝛥),
𝑅 (𝑅𝑎+𝑅 )
𝑅 (𝑅 +𝑅 )
𝑅 (𝑅 +𝑅 )
𝑏
𝑎 𝑏
𝑐
𝑏
𝑐
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 = 𝑅𝑎+𝑅
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
+𝑅
+𝑅𝑐
𝑏
𝑐
𝑅 𝑅
𝑎
𝑏
𝑅 𝑅
𝑅 𝑅
𝑐
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
By using multiplication and division,
𝑅𝑎 =
𝑎
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑏
𝑐
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
, 𝑅𝑏 = 1 2 𝑅2 3 3 1 and 𝑅𝑐 = 1 2 𝑅2 3 3 1
𝑅1
2
3
Now, 𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1 = (10) ⋅ (20) + (20) ⋅ (40) + (40) ⋅ (10) = 1400
This implies that,
𝑅𝑎 =
𝑅𝑏 =
𝑅𝑐 =
2.15
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
1400
= 10 = 𝟏𝟒𝟎𝜴,
𝑅1
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
1400
= 20 = 𝟕𝟎𝜴,
𝑅2
𝑅1 𝑅2 +𝑅2 𝑅3+𝑅3𝑅1
1400
=
= 𝟑𝟓𝜴
𝑅3
40
Let redraw the part of the circuit with Superposition that will be converted from wye to delta!
Let denote, 20𝛺 as 𝑅3 , 50𝛺 as 𝑅2 and 10𝛺 as 𝑅1 !
By setting 𝑅𝑎𝑐 (𝑌) = 𝑅𝑎𝑐 (𝛥), 𝑅𝑎𝑏 (𝑌) = 𝑅𝑎𝑏 (𝛥) and 𝑅𝑏𝑐 (𝑌) = 𝑅𝑏𝑐 (𝛥),
𝑅 (𝑅𝑎+𝑅 )
𝑅 (𝑅 +𝑅 )
𝑅 (𝑅 +𝑅 )
𝑏
𝑎 𝑏
𝑐
𝑏
𝑐
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 = 𝑅𝑎+𝑅
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
+𝑅
+𝑅𝑐
𝑏
𝑐
𝑎
𝑏
𝑅 𝑅𝑐
𝑅 𝑅
𝑅 𝑅
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
𝑎
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑐
𝑏
By using multiplication and division,
𝑅𝑎 =
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
, 𝑅𝑏 = 1 2 𝑅2 3 3 1 and 𝑅𝑐 = 1 2 𝑅2 3 3 1
𝑅1
2
3
Now, 𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1 = (10) ⋅ (50) + (50) ⋅ (20) + (20) ⋅ (10) = 1700
This implies that,
𝑅𝑎 =
13
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
1700
= 10 = 𝟏𝟕𝟎𝜴,
𝑅1
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑅𝑏 =
𝑅𝑐 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
1700
= 50 = 𝟑𝟒𝜴,
𝑅2
𝑅1 𝑅2 +𝑅2 𝑅3+𝑅3𝑅1
1700
= 20 = 𝟖𝟓𝜴
𝑅3
So that the new circuit becomes,
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
1
−1
for parallel connections
1
1
1
1
It’s clear that, 24𝛺 and 34𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 24 + 34 = 𝑅
⇒ 𝑅ⅇ𝑞 = 𝟏𝟒. 𝟎𝟕𝜴
𝑒𝑞
1
It’s obvious that , 30𝛺 and 170𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 30 + 170 = 𝑅
𝑒𝑞
⇒ 𝑅ⅇ𝑞 = 𝟐𝟓. 𝟓𝜴
Now, 14.07𝛺 and 25.5𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 14.07𝛺 + 25.5𝛺 = 39.6𝛺
1
1
1
It’s clear that, 85𝛺 and 39.6𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 85 + 39.6 = 𝑅
⇒ 𝑅ⅇ𝑞 = 𝟐𝟕𝜴
𝑒𝑞
Finally, 27𝛺 and 13𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 27𝛺 + 13𝛺 = 40𝛺
This implies, the equivalent resistance of the circuit is 40𝛺!
Assume that direction of voltage drops signed positive!
By KVL for closed path in clockwise direction gives, −100𝑉 + 𝑣40𝛺 = 0 ⇒ 𝑣40𝛺 = 100𝑉
𝑣
100
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 40 = 𝟐. 𝟓𝑨
2.16
(a) By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
Power with respect to voltage drop direction gives, 𝑝 = 𝑣 ⋅ ⅈ otherwise 𝑝 = −𝑣 ⋅ ⅈ
As it known, ∑𝑝 = 𝑝𝑎𝑏𝑠𝑜𝑟𝑏ⅇⅆ + 𝑝ⅆⅇ𝑙ⅈ𝑣ⅇ𝑟ⅇⅆ = 0
14
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote total power absorbed by bulbs as 𝑝𝑏𝑢𝑙𝑏𝑠 !
This implies that, 𝑝𝑏𝑢𝑙𝑏𝑠 − 𝑝110𝑣 = 0 ⇒ 𝑝110𝑣 = 𝑝𝑏𝑢𝑙𝑏𝑠
So that, 𝑝110𝑣 = 𝑃𝑏 = (40) ⋅ (10) = 400𝑊
𝑝
400
Now, 𝑝 = 𝑣 ⋅ ⅈ ⇒ ⅈ = 𝑣 ⇒ ⅈ110𝑣 = 110 = 3.64𝐴
Now, the source current must be divided equally into 10 parts. As a result, an equal amount of current
must pass through each bulb!
ⅈ
= 0.364𝐴
This implies,the current for each bulb “ ⅈ𝑏 ” is, ⅈ𝑏 = 110𝑣
10
(b) By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ
ⅆ𝑡
Power with respect to voltage drop direction gives, 𝑝 = 𝑣 ⋅ ⅈ otherwise 𝑝 = −𝑣 ⋅ ⅈ
As it known, ∑𝑝 = 𝑝𝑎𝑏𝑠𝑜𝑟𝑏ⅇⅆ + 𝑝ⅆⅇ𝑙ⅈ𝑣ⅇ𝑟ⅇⅆ = 0
Let denote total power absorbed by bulbs as 𝑝𝑏𝑢𝑙𝑏𝑠 !
This implies that, 𝑝𝑏𝑢𝑙𝑏𝑠 − 𝑝110𝑣 = 0 ⇒ 𝑝110𝑣 = 𝑝𝑏𝑢𝑙𝑏𝑠
So that, 𝑝110𝑣 = 𝑃𝑏 = (40) ⋅ (10) = 400𝑊
𝑝
400
Now, 𝑝 = 𝑣 ⋅ ⅈ ⇒ ⅈ = 𝑣 ⇒ ⅈ110𝑣 = 110 = 𝟑. 𝟔𝟒𝑨
Because of series connection, each bulb will be supplied by the same amount of current which
corresponds “source current”!
Chapter 2- Problems
2.3
𝑙
By the definition of resistance, 𝑅 = 𝑝 𝐴 where 𝑝 denotes resistivity, 𝑙 denotes length and 𝐴 denotes
cross-sectional area!
𝑙
𝑙
𝑅
𝜋
This implies that, 𝑅 = 𝑝 𝐴 = 𝑝 𝜋𝑟2 ⇒ 𝑟 = √𝑝 ⋅ 𝑙
𝑅
𝜋
So that, 𝑟 = √𝑝 ⋅ 𝑙 = 𝟏𝟖𝟒. 𝟑𝒎𝒎
2.11
Assume that direction of voltage drops signed positive!
By KVL for both left and right closed paths with the clockwise direction gives,
−𝑣1 + 1 + 5 = 0 ⇒ 𝑣1 = 𝟔𝑽 and −5 + 2 + 𝑣2 = 0 ⇒ 𝑣2 = 𝟑𝑽
15
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2.13
Let denote currents leaving a node positive!
By KCL for middle first, middle second, middle third and middle last gives,
−𝐼1 − 𝐼2 + 2 = 0,
−7 + 𝐼3 − 𝐼4 = 0,
𝐼2 + 3 + 7 = 0,
−2 + 𝐼4 + 4 = 0
It’s obvious that, 𝐼2 = −𝟏𝟎𝑨 and 𝐼4 = −𝟐𝑨, also by using substitution, 𝐼1 = 𝟏𝟐𝑨 and 𝐼3 = 𝟓𝑨
2.15
Assume that direction of voltage drops signed positive!
By KVL for both left and right closed paths with the clockwise direction gives,
−12 + 𝑣 + 2 = 0 ⇒ 10𝑉 and −2 + 8 + 3ⅈ𝑥 = 0 ⇒ ⅈ𝑥 = −𝟐𝑨
2.17
Assume that direction of voltage drops signed positive!
By KVL for both upper and lower triangular closed paths with the clockwise directions gives,
−24 + 𝑣1 − 𝑣2 = 0 and 𝑣2 + 𝑣3 + 12 = 0
It’s obvious that, 𝑣3 = 10𝑉 because they are connected to the same pair of terminals!( if there is any
confusion you can verify it using Kirchhoff's Voltage Law!)
By substitution, 𝑣2 = −𝟐𝟐𝑽 and 𝑣1 = 𝟐𝑽
2.19
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
Source power with respect to passive sign convention gives, 𝑝𝑠𝑜𝑢𝑟𝑐ⅇ = −𝑣𝑠𝑜𝑢𝑟𝑐ⅇ ⋅ ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
This implies that,
𝑝20𝑉 = (−20) ⋅ (𝐼) = −20𝐼,
𝑝10𝑉 = −(−10) ⋅ (𝐼) = 10𝐼,
𝑝−4𝑉 = −(−4) ⋅ (𝐼) = 4𝐼
Assume that direction of voltage drops signed positive!
By KVL for closed path with the clockwise direction gives,
−20 + 10 + 𝑣3𝛺 − (−4) = 0 ⇒ 𝑣3𝛺 = 6𝑉
𝑣
6
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 ⇒ 𝐼 = 3 = 2𝐴
This implies, 𝑝20𝑉 = −𝟒𝟎𝑾, 𝑝10𝑉 = 20𝑊, 𝑝−4𝑉 = 𝟖𝑾
As it known, ∑𝑝 = 𝑝𝑎𝑏𝑠𝑜𝑟𝑏ⅇⅆ + 𝑝ⅆⅇ𝑙ⅈ𝑣ⅇ𝑟ⅇⅆ = 0
This implies, −40 + 20 + 8 + 𝑝3 = 0 ⇒ 𝑝3 = 𝟏𝟐𝑾
16
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2.21
Assume that direction of voltage drops signed positive!
By KVL for closed path with the clockwise direction gives,
−15 + 𝑣1𝛺 + 2𝑣𝑥 + 𝑣𝑥 + 𝑣2𝛺 = 0
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
This implies that, 𝑣1𝛺 = ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ , 𝑉2𝛺 = 2ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ and 𝑉𝑥 = 5ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
By substitution, 𝑣𝑥 =
15−(𝑣1𝛺 +𝑣2𝛺 )
15−(ⅈ𝑠𝑜𝑢𝑟𝑐𝑒 +2ⅈ𝑠𝑜𝑢𝑟𝑐𝑒 )
⇒ 𝑉𝑥 = 5ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ =
⇒ ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ = 0.833𝐴
3
3
This implies, 𝑉𝑥 = 5ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ = 𝟒. 𝟏𝟕𝑽
2.23
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑛=1 𝑅𝑛
1
1
−1
for parallel connections
1
It’s clear that, 3𝛺 and 6𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 3 + 6 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 2𝛺
Now, 2𝛺 and 4𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 2𝛺 + 4𝛺 = 6𝛺
1
𝑒𝑞
1
1
It’s clear that, 8𝛺 and 12𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 8 + 12 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4.8𝛺
𝑒𝑞
Now, 1.2𝛺 and 4.8𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 1.2𝛺 + 4.8𝛺 = 6𝛺
1
1
1
It’s clear that, 6𝛺 and 6𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 6 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 3𝛺
Finally, 1𝛺 and 3𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 1𝛺 + 3𝛺 = 4𝛺
𝑒𝑞
Let denote the current passing through the 𝑣𝑥 as ⅈ𝑥 !
𝑅
Now by Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
Here 𝑅ⅇ𝑞 corresponds the equivalent resistance of the circuit which can be calculated as,
1
1
1
+ 4 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 1.3𝛺
2
𝑒𝑞
Also 𝑅𝑥 corresponds the resistance that is combined parallel with 2𝛺,that is 4𝛺!
𝑅
1.3
So by substitution, ⅈ𝑥 = 𝑅𝑒 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ = 4 6𝐴 = 2.0𝐴
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
𝑥
This implies that, 𝑣𝑥 = ⅈ𝑥 ⋅ (1) = 2𝑉
Now let apply the same process to 2 rightmost closed paths, the current ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ passing through the
1.2𝛺 will be 1𝐴!
CONT . . .
17
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
Here 𝑅ⅇ𝑞 corresponds the equivalent resistance of the circuit which can be calculated as,
1
1
1
+ 12 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4.8𝛺
8
𝑒𝑞
Also 𝑅𝑥 corresponds the resistance that is combined parallel with 8𝛺,that is 12𝛺!
𝑅
4.8
So by substitution, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ = 12 1𝐴 = 0.4𝐴
𝑥
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
This implies, 𝑝12𝛺 = (0.4)2 ⋅ (12) = 𝟏. 𝟗𝟐𝑾
2.25
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
1
)
𝑅
𝑛=1 𝑛
As it known, 𝑅ⅇ𝑞 = (∑
−1
for parallel connections of resistors!
Here 𝑅ⅇ𝑞 corresponds the equivalent resistance of the middle and the rightmost closed paths! which
can be calculated as,
1
1
1
+ 20 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟒𝒌𝜴
5
𝑒𝑞
Now, Current Division equation is applied for the part of the circuit that covers the middle and the
rightmost closed paths!
ⅈ20𝑘𝛺 =
4
0.01𝑉0
20
It’s obvious that, 𝑣5𝑚𝐴 = 𝑉0 because they are connected to same pair of terminals!( if there is any
confusion you can verify it using Kirchhoff's Voltage Law!)
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑉0 = 𝑣5𝑚𝐴 = (5 × 10−3 ) ⋅ (10 × 103 ) = 50𝑉
By substitution,
𝑉0 = 50𝑉 ⇒ ⅈ20𝑘𝛺 = 0.1𝐴,
𝑣20𝑘𝛺 = (0.1) ⋅ (20 × 103 ) = 2000𝑉 = 2𝑘𝑉,
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
This implies that, 𝑝20𝑘𝛺 = 𝑣20𝑘𝛺 ⋅ ⅈ20𝑘𝛺 = (ⅈ20𝑘𝛺 )2 ⋅ 𝑅20𝑘𝛺
Now it’s clear that, 𝑝20𝑘𝛺 = (2000) ⋅ (0.1) = 200𝑊 = 𝟎. 𝟐𝒌𝑾
2.27
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
18
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 4𝛺 and 6𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 4𝛺 + 6𝛺 = 10𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
4
By substitution, 𝑣0 = 10 16𝑉 = 𝟔. 𝟒𝑽
2.29
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
Let apply the process from the rightmost to leftmost resistance!
It’s clear that, 1𝛺 and 1𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 1𝛺 + 1𝛺 = 2𝛺
It’s obvious that, 2𝛺 and 1𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 0.667𝛺
2
1
𝑅𝑒𝑞
Now, 0.667𝛺 and 1𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 0.667𝛺 + 1𝛺 = 1.667𝛺
It’s obvious that, 1.667𝛺 and 1𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 0.625𝛺
1.667
1
𝑅𝑒𝑞
It’s clear that, 1𝛺 and 0.625𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 1𝛺 + 0.625𝛺 = 𝟏. 𝟔𝟐𝟓𝜴
2.31
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
It’s obvious that, 2𝛺, 1𝛺 𝑎𝑛ⅆ 4𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
1
+ 1 + 4 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 0.571𝛺
2
𝑒𝑞
It’s clear that, 3𝛺 and 0.571𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
𝑅ⅇ𝑞 = 3𝛺 + 0.571𝛺 = 3.571𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
3
𝑒𝑞
By substitution, 𝑣3𝛺 = 3.571 40𝑉 = 33.6𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
19
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
This implies, ⅈ1 =
33.6
= 11.2𝐴
3
As it known, , 2𝛺, 1𝛺 𝑎𝑛ⅆ 4𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 𝑅ⅇ𝑞 = 0.571𝛺!
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
Here ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ corresponds the source current which is equal to, ⅈ1 = 11.2𝐴
This implies,
ⅈ2 =
0.571
11.2𝐴 = 1.60𝐴
4
ⅈ5 =
0.571
11.2𝐴 = 3.20𝐴
2
ⅈ4 =
0.571
11.2𝐴 = 6.40𝐴
1
Let denote currents leaving a node positive!
By KCL for the node that connects 1𝛺 and 2𝛺 gives,
−ⅈ3 + ⅈ4 + ⅈ5 = 0 ⇒ ⅈ3 = ⅈ4 + ⅈ5 = 𝟗. 𝟔𝟎𝑨
2.33
Let denote equivalent resistance as 𝑅ⅇ𝑞 and equivalent conductance as 𝐺ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
This implies that, 𝐺ⅇ𝑞 = ∑∞
𝑛=1 𝐺𝑛 for the parallel connections and 𝐺ⅇ𝑞 = (∑
connections of resistors!
1
−1
for parallel connections
∞
1
)
𝐺
𝑛=1 𝑛
1
−1
for series
1
It’s obvious that, 6𝑆 and 3𝑆 are connected in series so so 𝐺ⅇ𝑞 for them is 6 + 3 = 𝐺 ⇒ 𝐺ⅇ𝑞 = 2𝑆
𝑒𝑞
It’s clear that, 2𝑆 and 2𝑆 are connected in parallel so 𝐺ⅇ𝑞 for them is 2𝑆 + 2𝑆 = 4𝑆
1
1
1
Now, 4𝑆 and 4𝑆 are connected in series so 𝐺ⅇ𝑞 for them is 4 + 4 = 𝐺 ⇒ 𝐺ⅇ𝑞 = 2𝑆
𝑒𝑞
Finally, 2𝑆 and 1𝑆 are connected in parallel so 𝐺ⅇ𝑞 for them is 2𝑆 + 1𝑆 = 3𝑆
𝑅
𝐺
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ ⇒ 𝐺 𝑥 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
2
This implies, ⅈ = 3 9𝐴 = 6𝐴
𝑥
𝑒𝑞
Let denote currents leaving a node positive and ⅈ1 as the current passing through the 3𝑆 resistor!
By KCL for the node that connects 4𝑆 and 1𝑆 gives, −9 + ⅈ + ⅈ1 = 0 ⇒ ⅈ1 = 9 − ⅈ = 3𝐴
ⅈ
3
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑣 = 𝐺 = 1 = 3𝑉
20
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2.35
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
It’s obvious that, 70𝛺 𝑎𝑛ⅆ 30𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+
=
⇒ 𝑅ⅇ𝑞 = 21𝛺
70 30 𝑅ⅇ𝑞
1
1
1
Now, 20𝛺 𝑎𝑛ⅆ 5𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 20 + 5 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4𝛺
𝑒𝑞
It’s obvious that, 𝑣70𝛺 = 𝑣30𝛺 and 𝑣20𝛺 = 𝑉0 because they are connected to same pair of terminals!( if
there is any confusion you can verify it using Kirchhoff's Voltage Law!)
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
21
4
This implies, 𝑣30𝛺 = 𝑣70𝛺 = 25 50 = 42𝑉 and 𝑣20𝛺 = 𝑉0 = 25 50 = 8𝑉
Let denote ⅈ1 as current passing through the 30𝛺 and ⅈ2 as current passing through the 5𝛺 resistor!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Now, 𝑣30𝛺 = 30ⅈ1 ⇒ ⅈ1 = 𝟏. 𝟒𝑨 and 𝑣0 = 5ⅈ2 ⇒ ⅈ2 = 𝟏. 𝟔𝑨
Let denote currents leaving a node positive!
By KCL for rightmost middle node gives, −𝐼0 − ⅈ1 + ⅈ2 = 0 ⇒ 𝐼0 = ⅈ2 − ⅈ1 = 𝟎. 𝟐𝑨
2.37
Let denote the current that flows around the circuit as ⅈ𝑠 !
Assume that direction of voltage drops signed positive!
By KVL for closed path with the clockwise direction gives,
−20 + 10 + 𝑣10𝛺 − 30 = 0 ⇒ 𝑣10𝛺 = 40𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
This implies, 𝑣10𝛺 = 10ⅈ𝑠 ⇒ ⅈ𝑠 = 4𝐴
Also, 𝑣𝑅 = ⅈ𝑠 ⋅ 𝑅 ⇒ 𝑅 = 𝟐. 𝟓𝜴
2.39
(a) Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
CONT . . .
21
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 2𝑘𝛺 𝑎𝑛ⅆ 1𝑘𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
2
1
+ =
1
1
𝑅𝑒𝑞
⇒ 𝑅ⅇ𝑞 = 𝟎. 𝟔𝟔𝟕𝒌𝜴
Now, 0.667𝑘𝛺 and 2𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 0.667𝑘𝛺 + 2𝑘𝛺 = 2.667𝛺
It’s clear that, 2.667𝑘𝛺 𝑎𝑛ⅆ 1𝑘𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 1 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 0.7273𝑘𝛺 = 727.3𝛺
2.667
𝑒𝑞
(b) It’s clear that, 12𝑘𝛺 𝑎𝑛ⅆ 12𝑘𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 6𝑘𝛺
12
𝑅𝑒𝑞
12
Now, 6𝑘𝛺 and 6𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 6𝑘𝛺 + 6𝑘𝛺 = 12𝑘𝛺
1
1
1
Finally, 12𝑘𝛺 𝑎𝑛ⅆ 4𝑘𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 12 + 4 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟑𝒌𝜴
2.41
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑛=1 𝑅𝑛
𝑒𝑞
−1
for parallel connections
It’s obvious that, 12𝛺, 12𝛺 𝑎𝑛ⅆ 12𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
1
+
+
=
⇒ 𝑅ⅇ𝑞 = 4𝛺
12 12 12 𝑅ⅇ𝑞
Now, 4𝛺 and 𝑅 are connected in series so 𝑅ⅇ𝑞 for them is 4𝛺 + 𝑅 !
It’s clear that, 30𝛺 𝑎𝑛ⅆ 60𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 20𝛺
30
60
𝑅𝑒𝑞
Now, 10𝛺 and 20𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝛺 + 20𝛺 = 30𝛺
Finally, 4𝛺 + 𝑅 𝑎𝑛ⅆ 30𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
4𝛺 + 𝑅 + 30𝛺 = 50𝛺 ⇒ 𝑅 = 16𝛺
2.43
(a) Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
CONT . . .
22
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s clear that, 10𝛺 𝑎𝑛ⅆ 40𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 40 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 8𝛺
10
𝑒𝑞
Now, 5𝛺 𝑎𝑛ⅆ 20𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 20 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 4𝛺
5
𝑒𝑞
Finally, 8𝛺 𝑎𝑛ⅆ 4𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 8𝛺 + 4𝛺 = 𝟏𝟐𝜴
(b) It’s clear that, 60𝛺, 20𝛺 𝑎𝑛ⅆ 30𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
1
+
+
=
⇒ 𝑅ⅇ𝑞 = 10𝛺
60 20 30 𝑅ⅇ𝑞
Now, 10𝛺 and 10𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝛺 + 10𝛺 = 𝟐𝟎𝜴
1
1
1
Finally, 80𝛺 𝑎𝑛ⅆ 20𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 80 + 20 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟏𝟔𝜴
2.45
(a) Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
𝑒𝑞
−1
for parallel connections
It’s obvious that, 50𝛺 𝑎𝑛ⅆ 5𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 50𝛺 + 5𝛺 = 55𝛺
It’s clear that, 10𝛺, 40𝛺, 20𝛺 and 30𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
1
1
+
+
+
=
⇒ 𝑅ⅇ𝑞 = 𝟒. 𝟖𝜴
10 40 20 30 𝑅ⅇ𝑞
It’s clear that, 55𝛺 𝑎𝑛ⅆ 4.8𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 55𝛺 + 4.8𝛺 = 59.8𝛺
(b) It’s clear that, 12𝛺 𝑎𝑛ⅆ 60𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 10𝛺
12
60
𝑅𝑒𝑞
Now, 10𝛺 𝑎𝑛ⅆ 20𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝛺 + 20𝛺 = 30𝛺
It’s obvious that, 30𝛺 𝑎𝑛ⅆ 30𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 30 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟏𝟓𝜴
30
𝑒𝑞
Now, 15𝛺 𝑎𝑛ⅆ 10𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 15𝛺 + 10𝛺 = 25𝛺
It’s clear that, 25𝛺 𝑎𝑛ⅆ 25𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
23
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1
1
1
+ 25 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 12.5𝛺
25
𝑒𝑞
Finally, 5𝛺, 12.5𝛺 and 15𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
5𝛺 + 12.5𝛺 + 15𝛺 = 32.5𝛺
2.47
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
It’s obvious that, 5𝛺 𝑎𝑛ⅆ 20𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ =
⇒ 𝑅ⅇ𝑞 = 4𝛺
5
20
𝑅𝑒𝑞
1
1
1
Likewise, 6𝛺 𝑎𝑛ⅆ 3𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 3 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 2𝛺
𝑒𝑞
Finally, 10𝛺, 4𝛺 , 2𝛺 and 8𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
10𝛺 + 4𝛺 + 2𝛺 + 8𝛺 = 𝟐𝟒𝜴
2.49
(a) By setting 𝑅𝑎𝑐 (𝑌) = 𝑅𝑎𝑐 (𝛥), 𝑅𝑎𝑏 (𝑌) = 𝑅𝑎𝑏 (𝛥) and 𝑅𝑏𝑐 (𝑌) = 𝑅𝑏𝑐 (𝛥),
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 =
𝑅𝑏 (𝑅𝑎+𝑅𝑐)
𝑅 (𝑅 +𝑅𝑏 )
𝑅𝑎 (𝑅𝑏 +𝑅𝑐 )
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
𝑅𝑎+𝑅𝑏 +𝑅𝑐
𝑎
𝑏
𝑏 +𝑅𝑐
𝑅 𝑅
𝑅 𝑅
𝑅 𝑅
𝑐
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
𝑎
Therefore,
𝑅 𝑅
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑐
(12)⋅(12)
𝑐
𝑅1 = 𝑅 +𝑅𝑏 +𝑅
= 12+12+12 = 𝟒𝜴
𝑎
𝑏
𝑅 𝑅
𝑐
(12)⋅(12)
𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 = 12+12+12 = 𝟒𝜴
𝑎
𝑏
𝑅 𝑅
(12)⋅(12)
𝑎 𝑏
𝑅3 = 𝑅 +𝑅
= 12+12+12 = 𝟒𝜴
+𝑅
𝑎
𝑏
𝑐
(b) By applying the same process,
𝑅1 =
𝑅𝑏 𝑅𝑐
𝑅𝑎 +𝑅𝑏 +𝑅𝑐
𝑅 𝑅
(30)⋅(60)
=
10+30+60
=
10+30+60
(60)⋅(10)
= 𝟏𝟖𝜴
𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 = 10+30+60 = 𝟔𝜴
𝑅3 =
2.51
𝑎
𝑏
𝑅𝑎 𝑅𝑏
𝑅𝑎 +𝑅𝑏 +𝑅𝑐
(10)⋅(30)
= 𝟑𝜴
Let redraw the part of the circuit with Superposition that will be converted from wye to delta!
CONT . . .
24
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote, 10𝛺 as 𝑅3 , 20𝛺 as 𝑅2 and 20𝛺 as 𝑅1 !
By setting 𝑅𝑎𝑐 (𝑌) = 𝑅𝑎𝑐 (𝛥), 𝑅𝑎𝑏 (𝑌) = 𝑅𝑎𝑏 (𝛥) and 𝑅𝑏𝑐 (𝑌) = 𝑅𝑏𝑐 (𝛥),
𝑅 (𝑅𝑎+𝑅 )
𝑅 (𝑅 +𝑅 )
𝑅 (𝑅 +𝑅 )
𝑏
𝑎 𝑏
𝑐
𝑏
𝑐
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 = 𝑅𝑎+𝑅
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
+𝑅
+𝑅𝑐
𝑏
𝑐
𝑅 𝑅
𝑎
𝑏
𝑅 𝑅
𝑅 𝑅
𝑐
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
By using multiplication and division,
𝑅𝑎 =
𝑎
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑏
𝑐
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
, 𝑅𝑏 = 1 2 𝑅2 3 3 1 and 𝑅𝑐 = 1 2 𝑅2 3 3 1
𝑅1
2
3
Now, 𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1 = (20) ⋅ (20) + (20) ⋅ (10) + (10) ⋅ (20) = 𝟖𝟎𝟎
This implies that,
𝑅𝑎 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
800
= 20 = 40𝛺,
𝑅1
𝑅𝑐 =
𝑅1 𝑅2 +𝑅2 𝑅3+𝑅3𝑅1
800
= 10 = 80𝛺
𝑅3
𝑅𝑏 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
800
= 20 = 40𝛺,
𝑅2
So that the new circuit becomes,
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
It’s obvious that, 10𝛺 𝑎𝑛ⅆ 40𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
25
−1
for parallel connections
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1
1
1
+ 40 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟖𝜴
10
𝑒𝑞
Likewise, 10𝛺 𝑎𝑛ⅆ 40𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
10
+
1
40
=
1
𝑅𝑒𝑞
⇒ 𝑅ⅇ𝑞 = 8𝛺
It’s obvious that, 8𝛺 𝑎𝑛ⅆ 8𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
8𝛺 + 8𝛺 = 16𝛺
It’s clear that, 30𝛺, 16𝛺 𝑎𝑛ⅆ 80𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
1
+
+
=
⇒ 𝑅ⅇ𝑞 = 𝟗. 𝟐𝜴
30 16 80 𝑅ⅇ𝑞
2.53
Let redraw the part of the circuit with Superposition that will be converted from wye to delta!
Let denote, 10𝛺 as 𝑅3 , 60𝛺 as 𝑅2 and 50𝛺 as 𝑅1 !
𝑅 (𝑅𝑎+𝑅 )
𝑅 (𝑅 +𝑅 )
𝑅 (𝑅 +𝑅 )
𝑏
𝑎 𝑏
𝑐
𝑏
𝑐
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 = 𝑅𝑎+𝑅
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
+𝑅
+𝑅𝑐
𝑏
𝑐
𝑎
𝑏
𝑅 𝑅𝑐
𝑅 𝑅
𝑅 𝑅
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
𝑎
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑐
𝑏
By using multiplication and division,
𝑅𝑎 =
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
, 𝑅𝑏 = 1 2 𝑅2 3 3 1 and 𝑅𝑐 = 1 2 𝑅2 3 3 1
𝑅1
2
3
Now, 𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1 = (50) ⋅ (60) + (60) ⋅ (10) + (10) ⋅ (50) = 4100
This implies that,
𝑅𝑎 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
4100
= 50 = 82𝛺,
𝑅1
𝑅𝑐 =
𝑅1 𝑅2 +𝑅2 𝑅3+𝑅3𝑅1
4100
= 10 = 410𝛺
𝑅3
𝑅𝑏 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
4100
= 60 = 68.33𝛺,
𝑅2
So that the new circuit becomes,
CONT . . .
26
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑛=1 𝑅𝑛
−1
for parallel connections
It’s obvious that, 68.33𝛺 𝑎𝑛ⅆ 40𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 40 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 25.23𝛺
68.33
𝑒𝑞
Likewise, 30𝛺 𝑎𝑛ⅆ 82𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 82 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 21.96𝛺
30
𝑒𝑞
It’s obvious that, 25.23𝛺 𝑎𝑛ⅆ 21.96𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
25.23𝛺 + 21.96𝛺 = 47.19𝛺
It’s obvious that, 47.19𝛺 𝑎𝑛ⅆ 410𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+
=
⇒ 𝑅ⅇ𝑞 = 42.32𝛺
47.19𝛺
410
𝑅𝑒𝑞
Finally, 20𝛺, 42.32𝛺 and 80𝛺 are connected in series so 𝑅ⅇ𝑞 for them is
20𝛺 + 42.32𝛺 + 80𝛺 = 𝟏𝟒𝟐. 𝟑𝟐𝜴
2.55
Let redraw the part of the circuit with Superposition that will be converted from wye to delta!
Let denote, 40𝛺 as 𝑅3 , 20𝛺 as 𝑅2 and 10𝛺 as 𝑅1 !
By setting 𝑅𝑎𝑐 (𝑌) = 𝑅𝑎𝑐 (𝛥), 𝑅𝑎𝑏 (𝑌) = 𝑅𝑎𝑏 (𝛥) and 𝑅𝑏𝑐 (𝑌) = 𝑅𝑏𝑐 (𝛥),
𝑅 (𝑅𝑎+𝑅 )
𝑅 (𝑅 +𝑅 )
𝑅 (𝑅 +𝑅 )
𝑏
𝑎 𝑏
𝑐
𝑏
𝑐
𝑅𝑎𝑐 = 𝑅1 + 𝑅3 = 𝑅𝑎+𝑅
, 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 = 𝑅𝑐 +𝑅𝑎 +𝑅𝑐
and 𝑅𝑏𝑐 = 𝑅2 + 𝑅3 = 𝑅𝑎+𝑅
+𝑅
+𝑅𝑐
𝑏
27
𝑐
𝑎
𝑏
𝑏
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑅 𝑅
𝑅 𝑅
𝑅 𝑅
𝑐
By subtraction and additions, 𝑅1 = 𝑅 +𝑅𝑏 +𝑅
, 𝑅2 = 𝑅 +𝑅𝑐 𝑎+𝑅𝑐 and 𝑅3 = 𝑅 +𝑅𝑎 𝑏+𝑅
By using multiplication and division,
𝑅𝑎 =
𝑎
𝑏
𝑐
𝑎
𝑏
𝑎
𝑏
𝑐
𝑅1 𝑅2+𝑅2𝑅3 +𝑅3 𝑅1
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
𝑅 𝑅 +𝑅 𝑅 +𝑅 𝑅
, 𝑅𝑏 = 1 2 𝑅2 3 3 1 and 𝑅𝑐 = 1 2 𝑅2 3 3 1
𝑅1
2
3
Now, 𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1 = (10) ⋅ (20) + (20) ⋅ (40) + (40) ⋅ (10) = 1400
This implies that,
𝑅𝑎 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
1400
= 10 = 140𝛺,
𝑅1
𝑅𝑐 =
𝑅1 𝑅2 +𝑅2 𝑅3+𝑅3𝑅1
1400
= 40 = 35𝛺
𝑅3
𝑅𝑏 =
𝑅1 𝑅2 +𝑅2 𝑅3 +𝑅3 𝑅1
1400
= 20 = 70𝛺,
𝑅2
So that the new circuit becomes,
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
It’s obvious that, 60𝛺 𝑎𝑛ⅆ 140𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 140 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 42𝛺
60
𝑒𝑞
Now, 20𝛺 𝑎𝑛ⅆ 50𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 20𝛺 + 50𝛺 = 70𝛺
Likewise, 70𝛺 𝑎𝑛ⅆ 70𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 70 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 35𝛺
70
𝑒𝑞
It’s obvious that, 42𝛺 𝑎𝑛ⅆ 35are connected in series so 𝑅ⅇ𝑞 for them is
42𝛺 + 35𝛺 = 77𝛺
Finally, , 77𝛺 𝑎𝑛ⅆ 35𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is
1
1
1
+ 35 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 24.0625𝛺
77𝛺
𝑒𝑞
28
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that 24V= 𝑣𝑅𝑒𝑞 because they are connected to same pair of terminals!( if there is any
confusion you can verify it using Kirchhoff's Voltage Law!)
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
24
This implies that, 𝐼0 = 24.0625 = 0.9974𝐴 = 997.4𝑚𝐴
Chapter 3-Practice Problems
3.1
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 −𝑣
By KCL for both a and b node gives, −1 + 21 + 1 6 2 = 0 and
Rearranging the equations gives,
1
2
1
6
1
6
1
6
1
6
𝑣
𝑣2 −𝑣1
+ 72 + 4 = 0
6
1
7
𝑣1 ( + ) + 𝑣2 (− ) = 1, 𝑣1 (− ) + 𝑣2 ( + ) = −4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
4
6
[ 1
−6
1
−6
𝑣1
𝑣1
1
−2
13 ] [ 𝑣2 ] = [ −4 ] ⇒ [ 𝑣2 ] = [−14]
42
This implies that, 𝑣1 = −𝟐𝑽 and 𝑣2 = −𝟏𝟒𝑽
3.2
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 1, 2 and 3 gives,
𝑣 −𝑣
𝑣 −𝑣
−10 + 1 2 3 + 1 3 2 = 0,
𝑣
𝑣2 −𝑣1
+ 42 − 4ⅈ𝑥 = 0
3
𝑣
𝑣3 −𝑣1
+ 4ⅈ𝑥 + 63 = 0
2
𝒗
It’s obvious that, ⅈ𝑥 = 𝟒𝟐!
29
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Now we have 4 equations and 4 unknowns!
Rearranging the equations gives,
1 1
1
1
𝑣1 ( + ) + 𝑣2 (− ) + 𝑣3 (− ) + 0ⅈ𝑥 = 10
2 3
3
2
1
1 1
𝑣1 (− ) + 𝑣2 ( + ) + 0𝑣3 + ⅈ𝑥 (−4) = 0
3
3 4
1
1 1
𝑣1 (− ) + 0𝑣2 + 𝑣3 ( + ) + ⅈ𝑥 (4) = 0
2
2 6
1
0𝑣1 + 𝑣2 (− ) + 0𝑣3 + ⅈ𝑥 = 0
4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
5
6
1
−
3
1
−
2
[ 0
1
1
−
0
3
2
𝑣1
𝑣1
7
10
80
0 −4 𝑣
𝑣
2
0
2
−64
12
[𝑣 ] = [ ] ⇒ [𝑣 ] = [
]
4
0
3
156
3
0
4 ⅈ
ⅈ𝑥
0
−16
𝑥
6
1
0
1]
−
4
−
This implies that, 𝑣1 = 80𝑉, 𝑣2 = −𝟔𝟒𝑽 and 𝑣3 = 𝟏𝟓𝟔𝑽
3.3
Let denote the lower essential node as reference node!
It’s obvious that 9𝑉 source connects two non-reference nodes!
This implies that, 9𝑉 source is a supernode!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and 𝑣ⅈ as voltage across the terminals of 2𝛺 resistor!
By KCL for supernode gives,
𝑣
𝑣
𝑣
𝑣−21
+ 3 + 2𝑖 + 6𝑖 = 0
4
Also it’s clear that, 𝑣ⅈ − 𝑣 = 9
Rearranging the equations gives,
1 1
1 1
21
𝑣 ( + ) + 𝑣ⅈ ( + ) =
4 3
2 6
4
𝑣(−1) + 𝑣ⅈ (1) = 9
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
30
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
7 4 𝑣
21
𝑣
−0.6
[ 12 6] [𝑣 ] = [ 4 ] ⇒ [𝑣 ] = [
]
8.4
ⅈ
ⅈ
9
−1 1
This implies that, 𝑣 = −𝟎. 𝟔𝑽 and 𝑣ⅈ = 𝟖. 𝟒𝑽
𝑣
Using Ohm’s Law, 𝑣ⅈ = 2ⅈ ⇒ ⅈ = 2𝑖 = 𝟒. 𝟐𝑨
3.4
It’s obvious that 10𝑉 independent voltage and 5ⅈ dependent voltage source connects two non-reference
nodes!
This implies that, 10𝑉 and 5ⅈ sources create supernodes!(Note that we can combine them as single
larger supernode!)
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
𝑣
By KCL for larger supernode gives, 21 + 42 + 33 = 0
𝑣
It’s clear that, 𝑣1 − 𝑣2 = 10, 𝑣3 − 𝑣2 = 5ⅈ and ⅈ = 21!
Rearraning the equations gives,
1
1
1
𝑣1 ( ) + 𝑣2 ( ) + 𝑣3 ( ) = 0
2
4
3
𝑣1 − 𝑣2 + 0𝑣3 = 10
𝑣
2
−5
2
0𝑣1 + 𝑣3 − 𝑣2 = 5 1 ⇒ 𝑣1 ( ) − 𝑣2 + 𝑣3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
2
1
5
[− 2
1 1
𝑣1
0
3.043
4 3 𝑣1
−1 0 [𝑣2 ] = [10] ⇒ [𝑣2 ] = [−6.956]
𝑣3
𝑣3
0
0.6521
−1 1]
This implies that, 𝑣1 = 𝟑. 𝟎𝟒𝟑𝑽, 𝑣2 = −𝟔. 𝟗𝟓𝟔𝑽 and 𝑣3 = 𝟎. 𝟔𝟓𝟐𝟐𝑽
3.5
Assume that direction of voltage drops signed positive!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
By KVL for each mesh gives,
CONT . . .
31
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
−36 + 2ⅈ1 + 12(ⅈ1 − ⅈ2 ) + 4ⅈ1 = 0
9ⅈ2 + 24 + 3ⅈ2 + 12(ⅈ2 − ⅈ1 ) = 0
Rearranging the equations gives,
18ⅈ1 − 12ⅈ2 = 36
−12ⅈ1 + 24ⅈ2 = −24
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ
2
36
18 −12 ⅈ1
] ⇒ [ 1] = [ ]
][ ] = [
[
ⅈ2
0
−24
−12
24 ⅈ2
This implies that, ⅈ1 = 𝟐𝑨, ⅈ2 = 𝟎𝑨
3.6
Assume that direction of voltage drops signed positive!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
By KVL for each mesh gives,
−20 + 4(ⅈ1 − ⅈ3 ) + 2(ⅈ1 − ⅈ2 ) = 0
8(ⅈ2 − ⅈ3 ) − 10ⅈ3 + 2(ⅈ2 − ⅈ1 ) = 0
6ⅈ3 + 8(ⅈ3 − ⅈ2 ) + 4(ⅈ3 − ⅈ1 ) = 0
Also It’s obvious that, 𝐼0 = ⅈ3
By substitution, second equation becomes,
Rearranging the equations gives,
6ⅈ1 − 2ⅈ2 − 4ⅈ3 = 20, −2ⅈ1 + 10ⅈ2 − 18ⅈ3 = 0 and −4ⅈ1 − 8ⅈ2 + 18ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
6
[ −2
−4
−2
10
−8
ⅈ1
−4 ⅈ1
20
−3.214
−18 ] [ⅈ2 ] = [ 0 ] ⇒ [ⅈ2 ] = [ −9.642 ]
ⅈ3
18 ⅈ3
0
−5
This implies that, ⅈ3 = 𝐼0 = −𝟓𝑨
3.7
Assume that direction of voltage drops signed positive!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, independent current source causes a supermesh!(Two meshes have a current source
in common!)
By KVL for supermesh gives, −6 + 2(ⅈ1 − ⅈ3 ) + 4(ⅈ2 − ⅈ3 ) + 8ⅈ2 = 0
By KVL for mesh ⅈ3 gives, 2ⅈ3 + 4(ⅈ3 − ⅈ2 ) + 2(ⅈ3 − ⅈ1 ) = 0
32
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Also it’s obvious that, ⅈ1 − ⅈ2 = 3
Rearranging the equations gives,
2ⅈ1 + 12ⅈ2 − 6ⅈ3 = 6, −2ⅈ1 − 4ⅈ2 + 8ⅈ3 = 0 and ⅈ1 − ⅈ2 + 0ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
2
[ −2
1
12
−4
−1
ⅈ1
−6 ⅈ1
6
3.474
8] [ⅈ2 ] = [0] ⇒ [ⅈ2 ] = [0.4737]
ⅈ3
0 ⅈ3
3
1.1052
This implies that, ⅈ1 = 𝟑. 𝟒𝟕𝟒𝑨, ⅈ2 = 𝟎. 𝟒𝟕𝟑𝟕𝑨 and ⅈ3 = 𝟏. 𝟏𝟎𝟓𝟐𝑨
3.8
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 − 𝑣3 +
𝑣1 − 𝑣2 𝑣1
+
=0
5
10
𝑣2 − 𝑣1
−1−2=0
5
𝑣3 − 𝑣4
𝑣3 − 𝑣1 + 1 +
=0
4
𝑣4 − 𝑣3 𝑣4
+ −3=0
2
4
Rearranging the equations gives,
1 1
1
𝑣1 (1 + + ) + 𝑣2 (− ) − 𝑣3 + 0𝑣4 = 0
5 10
5
1
1
𝑣1 (− ) + 𝑣2 ( ) + 0𝑣3 + 0𝑣4 = 3
5
5
1
1
−𝑣1 + 0𝑣2 + 𝑣3 (1 + ) + 𝑣4 (− ) = −1
4
4
1
1 1
0𝑣1 + 0𝑣2 + 𝑣3 (− ) + 𝑣4 ( + ) = 3
4
2 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
𝑣1
1.3 −0.2
−1
0
𝟎
𝑣2
−0.2
0.2
0
0
𝟑
[
][ ] = [ ]
−1
0
1.25 −0.25 𝑣3
−𝟏
0
0 −0.25
0.75 𝑣4
𝟑
So that, node voltages can be found by solving the matrix above!
33
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3.9
Assume that direction of voltage drops signed positive!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
By KVL for each mesh gives,
−24 + 50ⅈ1 + 40(ⅈ1 − ⅈ2 ) + 80(ⅈ1 − ⅈ4 ) = 0
30(ⅈ2 − ⅈ3 ) + 10(ⅈ2 − ⅈ4 ) + 40(ⅈ2 − ⅈ1 ) = 0
12 + 20(ⅈ3 − ⅈ5 ) + 30(ⅈ3 − ⅈ2 ) = 0
−10 + 80(ⅈ4 − ⅈ1 ) + 10(ⅈ4 − ⅈ3 ) = 0
20(ⅈ5 − ⅈ3 ) + 60ⅈ5 + 10 = 0
Rearranging the equations gives,
170ⅈ1 − 40ⅈ2 + 0ⅈ3 − 80ⅈ4 + 0ⅈ5 = 24
−40ⅈ1 + 80ⅈ2 − 30ⅈ3 − 10ⅈ4 + 0ⅈ5 = 0
0ⅈ1 − 30ⅈ2 + 50ⅈ3 + 0ⅈ4 − 20ⅈ5 = −12
−80ⅈ1 − 10ⅈ2 + 0ⅈ3 + 90ⅈ4 + 0ⅈ5 = 10
0ⅈ1 + 0ⅈ2 − 20ⅈ3 + 0ⅈ4 + 80ⅈ5 = −10
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ1
170 −40
0
−80
0
24
ⅈ1
−40 80 −30 −10
0
0
0
−30 50
0
−20 ⅈ1 = −12
ⅈ1
−80 −10
0
90
0
10
[ 0
]
0
−20
0
80 [ⅈ1 ] [ −10]
So that, mesh currents can be found by solving the matrix above!
3.10
Note that PSpice must be used but in order to make everything clear, regular process will be applied!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
𝑣
𝑣
𝑣 −𝑣
By KCL for node voltages gives, 301 + 601 + 2 = 0 and −2 + 502 + 252 + 21003 = 0
And also it’s clear that, 𝑣3 = 200𝑉
This implies that, 𝑣1 = −40𝑉 and 𝑣2 = 57.14𝑉
34
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3.11
Let redraw the circuit with mesh currents!
It’s obvious that independent current source and dependent current source are in common with
different meshes,
So that they’re both results in supermeshes! (Note that we can think them as single large supermesh!)
Assume that direction of voltage drops signed positive!
By KVL for large supermesh gives, 4ⅈ𝑏 + 2ⅈⅆ + ⅈ𝑐 − ⅈ𝑎 = 0
By KVL for mesh ⅈ𝑎 gives, −10 + 2ⅈ𝑎 + ⅈ𝑎 − ⅈ𝑐 = 0
It’s clear that, ⅈ𝑐 − ⅈ𝑏 = 2
We can clearly see that, ⅈ1 = ⅈ𝑏 ⇒ ⅈ𝑐 − ⅈⅆ = ⅈ1 = ⅈ𝑏 ⇒ ⅈ𝑐 − ⅈⅆ − ⅈ𝑏 = 0
Rearranging the equations gives,
−ⅈ𝑎 + 4ⅈ𝑏 + ⅈ𝑐 + 2ⅈⅆ = 0
3ⅈ𝑎 + 0ⅈ𝑏 − ⅈ𝑐 + 0ⅈⅆ = 10
0ⅈ𝑎 − ⅈ𝑏 + ⅈ𝑐 + 0ⅈⅆ = 2
0ⅈ𝑎 − ⅈ𝑏 + ⅈ𝑐 − ⅈⅆ = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ𝑎
−1 4
1
2 ⅈ𝑎
0
3.857
ⅈ𝑏
3
0 −1 0 ⅈ𝑏
10
−0.4286
[
][ ] = [ ] ⇒ [ ] = [
]
ⅈ𝑐
0 −1 1
0 ⅈ𝑐
2
1.571
ⅈⅆ
0 −1 1 −1 ⅈⅆ
0
2
This implies that, ⅈ𝑎 = 3.8571𝐴, ⅈ𝑏 = −0.4286𝐴, ⅈ𝑐 = 1.5714𝐴 and ⅈⅆ = 2𝐴
Now it’s clear that, ⅈ1 = ⅈ𝑏 , ⅈ2 = ⅈ𝑎 − ⅈ𝑐 and ⅈ3 = ⅈⅆ
So that, ⅈ1 = −𝟎 ⋅ 𝟒𝟐𝟖𝟔𝑨, ⅈ2 = 2.286𝐴 and ⅈ3 = 𝟐𝑨
35
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Chapter 3- Problems
3.3
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
𝑣
𝑣
By KCL for node 𝑣0 gives, −10 + 100 + 200 + 300 + 2 + 600 = 0 ⇒ 𝑣0 = 40𝑉
This implies that, 𝐼1 = 𝟒𝑨, 𝐼2 = 𝟐𝑨, 𝐼3 = 𝟏. 𝟑𝟑𝟑𝟑𝑨 and 𝐼4 = 𝟎. 𝟔𝟔𝟔𝟕𝑨
3.5
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and lower middle node as reference node!
It’s clear that, upper middle node voltage equals to 𝑣0 because they are connected to same pair of
terminals! ( if there is any confusion you can verify it using Kirchhoff's Voltage Law!)
𝑣 −30
𝑣 −20
𝑣
0
0
0
+ 5000
+ 4000
= 0 ⇒ 𝑣0 = 𝟐𝟎𝑽
By KVL for upper middle node gives, 2000
3.7
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and second lower middle node as reference node!
𝑉
𝑉
By KCL for second middle upper node gives, −2 + 10𝑥 + 20𝑥 + 0.2𝑉𝑥 = 0 => 𝑉𝑥 = 𝟓. 𝟕𝟏𝟒𝑽
3.9
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and lower middle node as reference node!
Let denote 𝑣150𝛺 as 𝑣1 and 𝑣50𝛺 as 𝑣𝑠 !
36
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By KCL for node 𝑣𝑠 gives,
𝑣𝑠 −12
𝑣
𝑣1
+ 50𝑠 + 150
=0
250
12−𝑣
It’s clear that, 60𝐼𝑏 = 𝑣𝑠 − 𝑣1 and 𝐼𝑏 = 250 𝑠
Rearranging the equations gives,
𝑣1 (
1
1
1
12
) + 𝑣𝑠 (
+ ) + 0𝐼𝑏 =
150
250 50
250
𝑣1 − 𝑣𝑠 + 60𝐼𝑏 = 0
0𝑣1 + 𝑣𝑠 (
1
12
) + 𝐼𝑏 =
250
250
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
6
12
1 𝑣1
𝑣1
−0.2975
150 250
250
𝑣
1
−1 60 [ 𝑠 ] = 0 ⇒ [ 𝑣𝑠 ] = [ 2.0826 ]
𝐼𝑏
𝐼𝑏
1
12
0.03967
1]
[ 0
[
]
250
250
This implies that, 𝐼𝑏 = 0.03967𝐴 = 𝟑𝟗. 𝟔𝟕𝒎𝑨
3.11
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 +12
By KCL for upper middle node gives, 𝑣0 − 36 + 20 + 0 4
By the definition of power, 𝑝 =
= 0 ⇒ 𝑣0 = 18.86𝑉
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
𝑣2
=
⋅
=𝑣⋅ⅈ =
𝑅
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
𝑣2
This implies that, 𝑝1𝛺 = (36 − 𝑣0 )2 = 𝟐𝟗𝟑. 𝟖𝑾, 𝑝2𝛺 = 20 = 𝟏𝟕𝟕. 𝟖𝑾 and 𝑣4𝛺 =
3.13
(𝑣0 +12)2
4
= 𝟐𝟑𝟖. 𝟏𝑾
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣2 gives,
𝑣
𝑣
𝑣2 +2−𝑣1
+ 42 − 3 = 0
2
𝑣 +2−𝑣1
Also it’s clear that, 81 = 2 2
37
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Rearranging the equations gives,
𝑣1 (
−1
1 1
) + 𝑣2 ( + ) = 2
2
2 4
1 1
−1
𝑣1 ( + ) + 𝑣2 ( ) = 1
8 2
2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
−1 3
4 ] [𝑣1 ] = [2] ⇒ [𝑣1 ] = [8]
[2
𝑣2
5 −1 𝑣2
1
8
8
2
This implies that, 𝑣1 = 𝑣2 = 𝟖𝑽
3.15
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and lower middle node as reference node!
Let denote first,second and middle upper nodes as, 𝑣1 , 𝑣2 and 𝑣3
It’s obvious that 10𝑉 independent voltage source connects two non-reference nodes!
This implies that, 10𝑉source creates a supernode!
By KCL for supernode gives, 6𝑣1 + 2 + 5𝑣2 + 3(𝑣2 − 𝑣3 ) = 0
By KCL for node 𝑣3 gives, −2 + 3(𝑣3 − 𝑣2 ) − 4 = 0
Also it’s clear that, 𝑣1 − 𝑣2 = 10
Rearranging the equations gives,
6𝑣1 + 8𝑣2 − 3𝑣3 = −2
0𝑣1 − 3𝑣2 + 3𝑣3 = 6
𝑣1 − 𝑣2 + 0𝑣3 = 10
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
6
[0
1
𝑣1
8 −3 𝑣1
−2
4.409
𝑣
𝑣
−3
3 ] [ 2 ] = [ 6 ] ⇒ [ 2 ] = [ −5.909]
𝑣3
−1
0 𝑣3
10
−3.909
This implies that, 𝑣1 = 4.909𝑉, 𝑣2 = −5.091𝑉 and 𝑣3 = −3.091𝑉
So that, ⅈ0 = 6𝑣1 = 𝟐𝟗. 𝟒𝟓𝑨
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
𝑣2
=
⋅
=
𝑣
⋅
ⅈ
=
= 𝑣2 ⋅ 𝐺
𝑅
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
This implies that, 𝑝6𝑆 = 6𝑣12 = 𝟏𝟒𝟒. 𝟔𝑾, 𝑝5𝑆 = 5𝑣22 = 𝟏𝟐𝟗. 𝟔𝑾 and 𝑝3𝑆 = 3(𝑣3 − 𝑣2 )2 = 𝟏𝟐𝑾
38
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3.17
Let redraw the circuit with some important voltages and a reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for the node voltages 2 and 3 gives,
𝑣1 −𝑣2
= ⅈ0
4
It’s clear that, 𝑣1 = 60𝑉 and
𝑣 −𝑣
𝑣
𝑣 −𝑣
𝑣2 −𝑣1
𝑣 −𝑣
+ 2 2 3 + 82 = 0 and 310 1 + 3 2 2 − 3ⅈ0 = 0
4
Rearranging the equations gives,
1 1 1
1
𝑣2 ( + + ) + 𝑣3 (− ) + 0ⅈ0 = 15
4 2 8
2
1
1 1
𝑣2 (− ) + 𝑣3 ( + ) − 3ⅈ0 = 6
2
10 2
1
𝑣2 (− ) + 0𝑣3 − ⅈ0 = −15
4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
8
1
−
2
1
−
[ 4
1
2
6
10
−
0
0
𝑣2
𝑣2
15
𝑣
𝑣3 ] = [
=
⇒
[
]
[
]
[
3
6
−3
ⅈ0
ⅈ0
−15
−1]
53.08
62.88]
1.73
This implies that, ⅈ0 = 𝟏. 𝟕𝟑𝑨
3.19
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
39
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 − 𝑣3 𝑣1 − 𝑣2 𝑣1
+
+ =0
2
8
4
𝑣2 − 𝑣1 𝑣2 𝑣2 − 𝑣3
+ +
=0
8
2
4
−5 + 3 +
−3 +
𝑣3 − 𝑣1 𝑣3 − 𝑣2 𝑣3 − 12
+
+
=0
8
2
4
Rearranging the equations gives,
7𝑣1 − 𝑣2 − 4𝑣3 = 16
−𝑣1 + 7𝑣2 − 2𝑣3 = 0
−4𝑣1 − 2𝑣2 + 7𝑣3 = 36
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
[ −1
−4
−1
7
−2
𝑣1
−4 𝑣1
16
10
𝑣
𝑣
−2 ] [ 2 ] = [ 0 ] ⇒ [ 2 ] = [ 4.933 ]
𝑣3
7 𝑣3
36
12.267
This implies that, 𝑣1 = 𝟏𝟎𝑽, 𝑣2 = 𝟒. 𝟗𝟑𝟑𝑽 and 𝑣3 = 𝟏𝟐. 𝟐𝟔𝟕𝑽
3.21
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 +3𝑣 −𝑣
1
2
0
2
By KCL for node voltages gives, −0.003 + 4000
+ 1 2000
= 0 and
Also it’s clear that, 𝑣2 = 𝑣0 !
𝑣2 −𝑣1
𝑣 −3𝑣0 −𝑣1
𝑣2
+ 2 2000
+ 1000
=0
4000
Rearranging the equations gives,
3𝑣1 + 3𝑣0 = 12 and −3𝑣1 + 𝑣0 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
3
−3
𝑣1
1
12
3 𝑣1
] [𝑣 ] = [ ] ⇒ [𝑣 ] = [ ]
3
0
1 0
0
This implies that, 𝑣1 = 1𝑉 and 𝑣2 = 𝟑𝑽
3.23
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
CONT . . .
40
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive, lower left middle node as reference node and 𝑣16𝛺 as 𝑣1 !
𝑉
𝑉 −2𝑉𝑜 −𝑣1
=0
4
By KCL for node 𝑉0 gives, 𝑉0 − 30 + 20 + 0
By KCL for node 𝑣1 gives,
𝑣
𝑣1 +2𝑉0 −𝑉0
+ 161 − 3 = 0
4
Rearranging the equations gives, 5𝑉0 − 𝑣1 = 120 and 4𝑉0 + 4𝑣1 = 48
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
5
[4
1
4
1
4 ] [𝑉0 ] = [30] ⇒ [𝑉0 ] = [ 22.34 ]
𝑣1
5 𝑣1
−8.275
3
16
−
This implies that, 𝑉0 = 𝟐𝟐. 𝟑𝟒𝑽
3.25
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for the node voltages gives,
𝑣1 − 𝑣4
+ 𝑣1 − 𝑣2 = 0
20
𝑣2 𝑣2 − 𝑣3
𝑣2 − 𝑣1 + +
=0
8
10
𝑣3 − 𝑣2 𝑣3 − 𝑣4 𝑣3
+
+
=0
10
10
20
𝑣4 − 𝑣1 𝑣4 − 𝑣3 𝑣4
+
+
=0
20
10
30
−4 +
Rearranging the equations gives,
𝑣1 (
1
1
+ 1) − 𝑣2 + 0𝑣3 + 𝑣4 (− ) = 4
20
20
−𝑣1 + 𝑣2 (
1 1
1
+ + 1) + 𝑣3 (− ) + 0𝑣4 = 0
10 8
10
0𝑣1 + 𝑣2 (−
𝑣1 (−
1
1
1
1
1
) + 𝑣3 ( +
+ ) + 𝑣4 (− ) = 0
10
10 10 20
10
1
1
1
1
1
) + 0𝑣2 + 𝑣3 (− ) + 𝑣4 ( +
+ )=0
20
10
20 10 30
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
41
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
21
20
−1
0
1
[− 20
−1
49
40
1
−
10
0
0
0
𝑣1
1
ⅈ𝑎
4
25.52
0 𝑣
ⅈ𝑏
0
2
22.05
10
[ ]=[ ]⇒[ ]=[
]
ⅈ𝑐
1
1 𝑣3
0
14.84
−
ⅈⅆ
0
15.06
4
10 𝑣4
1
11
−
10
60]
−
This implies that, 𝑣1 = 25.52𝑉, 𝑣2 = 22.05𝑉, 𝑣3 = 14.84𝑉 and 𝑣4 = 15.06𝑉
3.27
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and lower middle node as reference node!
By KCL for node voltages gives,
−2 + 4(𝑣1 − 𝑣3 ) + 3ⅈ0 + 𝑣1 − 𝑣2 + 2𝑣1 = 0
𝑣2 − 𝑣1 + 4𝑣2 + 𝑣2 − 𝑣3 = 0
4(𝑣3 − 𝑣1 ) − 3ⅈ0 + 𝑣3 − 𝑣2 + 2𝑣3 − 4 = 0
It’s clear that, ⅈ0 = 4𝑣2
Rearranging the equations with substitution gives,
7𝑣1 + 11𝑣2 − 4𝑣3 = 2
−𝑣1 + 6𝑣2 − 𝑣3 = 0
−4𝑣1 − 13𝑣2 + 7𝑣3 = 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
[ −1
−4
11
6
−13
𝑣1
−4 𝑣1
2
0.625
𝑣
𝑣
−1 ] [ 2 ] = [ 0] ⇒ [ 2 ] = [ 0.375]
𝑣3
7 𝑣3
4
1.625
This implies that, 𝑣1 = 0.625𝑉 = 625𝑚𝑉, 𝑣2 = 𝟎. 𝟑𝟕𝟓𝑽 = 𝟑𝟕𝟓𝒎𝑽 and 𝑣3 = 𝟏. 𝟔𝟐𝟓𝑽
3.29
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for the node voltages gives,
CONT . . .
42
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5 + 𝑣1 − 𝑣4 + 2𝑣1 + 𝑣1 − 𝑣2 = 0
𝑣2 − 𝑣1 + 2𝑣2 + 4(𝑣2 − 𝑣3 ) = 0
4(𝑣3 − 𝑣2 ) + 𝑣3 − 𝑣4 − 6 = 0
𝑣4 − 𝑣1 − 2 + 3𝑣4 + 𝑣4 − 𝑣3 = 0
Rearranging the equations gives,
4𝑣1 − 𝑣2 + 0𝑣3 − 𝑣4 = −5
−𝑣1 + 7𝑣2 − 4𝑣3 + 0𝑣4 = 0
0𝑣1 − 4𝑣2 + 5𝑣3 − 𝑣4 = 6
−𝑣1 + 0𝑣2 − 𝑣3 + 5𝑣4 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
4
−1
[
0
−1
−1
7
−4
0
0
−4
5
−1
This implies that,
ⅈ𝑎
−1 𝑣1
−0.7708
−5
ⅈ𝑏
0 𝑣2
1.209
0
][ ] = [ ] ⇒ [ ] = [
]
ⅈ𝑐
−1 𝑣3
2.309
6
ⅈⅆ
5 𝑣4
0.7076
2
𝑣1 = −0.7708𝑉 = −𝟕𝟕𝟎. 𝟖𝒎𝑽, 𝑣2 = 𝟏. 𝟐𝟎𝟗𝑽, 𝑣3 = 𝟐. 𝟑𝟎𝟗𝑽 and 𝑣4 = 𝟎. 𝟕𝟎𝟕𝟔𝑽 = 𝟕𝟎𝟕. 𝟔𝒎𝑽
3.31
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
It’s obvious that, 4𝐼0 independent voltage source connects two non-reference nodes!
This implies that, source creates a supernode!
𝑣
By KCL for supernode gives, −1 + 𝑣1 − 𝑣3 + 41 + 𝑣2 − 2𝑣0 = 0
It’s clear that, 𝑣1 − 𝑣3 = 𝑣0 , 𝑣2 − 𝑣1 = 4𝐼0 and 𝑣3 = 4𝐼0
𝑣
𝑣 −10
By KCL for node 𝑣3 gives, 2𝑣0 + 𝑣3 − 𝑣1 + 43 + 3 2
Rearraning the equations with substitution gives,
=0
1
𝑣1 (1 + ) + 𝑣2 − 2𝑣0 − 4𝐼0 = 1
4
𝑣1 + 0𝑣2 − 𝑣0 − 4𝐼0 = 0
−𝑣1 + 𝑣2 + 0𝑣0 − 4𝐼0 = 0
−𝑣1 + 0𝑣2 + 2𝑣0 + 7𝐼0 = 5
43
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1.25
1
[
−1
−1
1
0
1
0
−2
−1
0
2
𝑣1
4.970
1
−4 𝑣1
𝑣
𝑣
2
2
0
4.848
−4
]
] [ ] = [ ] ⇒ [𝑣 ] = [
0
5.091
0
−4 𝑣0
𝐼0
5
−0.03
7 𝐼0
This implies that, 𝑣1 = 4.970𝑉, 𝑣2 = 4.848𝑉, 𝑣0 = 5.091𝑉 and 𝑣1 = −0.03𝐴
So that, 𝑣3 = 4𝐼0 = −𝟎. 𝟏𝟐𝑽 = −𝟏𝟐𝟎𝒎𝑽
3.39
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for each mesh gives, −10 + 4ⅈ1 − 2𝐼𝑥 + 6(ⅈ1 − ⅈ2 ) = 0 and 2ⅈ2 + 12 + 6(ⅈ2 − ⅈ1 ) = 0
Also it’s clear that, 𝐼𝑥 = ⅈ1 − ⅈ2 (If there’s any confusion, you can use Kirchhoff’s Current Law!)
Rearranging the equations gives,
10ⅈ1 − 6ⅈ2 − 2𝐼𝑥 = 10
−6ⅈ1 + 8ⅈ2 + 0𝐼𝑥 = −12
−ⅈ1 + ⅈ2 + 𝐼𝑥 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
10 −6
[ −6
8
−1
1
ⅈ1
−2 ⅈ1
10
0.80
0 ] [ⅈ2 ] = [−12] ⇒ [ⅈ2 ] = [ −0.90 ]
𝐼𝑥
1 𝐼𝑥
0
1.70
This implies that, ⅈ1 = 𝟎. 𝟖𝟎𝑨 = 𝟖𝟎𝟎𝒎𝑨 and ⅈ2 = −𝟎. 𝟗𝟎𝑨 = −𝟗𝟎𝟎𝒎𝑨
3.41
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for each mesh gives,
10ⅈ1 − 6 + 2(ⅈ1 − ⅈ2 ) = 0
2(ⅈ2 − ⅈ1 ) + ⅈ2 − ⅈ3 + 8 + 4ⅈ2 = 0
6 + 5ⅈ3 − 8 + ⅈ3 − ⅈ2 = 0
Rearranging the equations gives,
12ⅈ1 − 2ⅈ2 + 0ⅈ3 = 6
−2ⅈ1 + 7ⅈ2 − ⅈ3 = −8
0ⅈ1 − ⅈ2 + 6ⅈ3 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
44
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
12
[ −2
0
−2
7
−1
ⅈ1
0 ⅈ1
6
0.3291
−1] [ⅈ2 ] = [−8] ⇒ [ⅈ2 ] = [ −1.026 ]
ⅈ3
6 ⅈ3
2
0.1624
This implies that, ⅈ1 = 0.3291𝐴 = 329.1𝑚𝐴, ⅈ2 = −1.026𝐴 and ⅈ3 = 0.1624𝐴 = 162.4𝑚𝐴
As it known, ⅈ = ⅈ3 − ⅈ2 = 𝟏. 𝟏𝟖𝟖𝑨
3.43
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Let redraw the circuit with mesh currents!
By KVL for each mesh gives,
−80 + 20(ⅈ1 − ⅈ2 ) + 30(ⅈ1 − ⅈ3 ) + 20ⅈ1 = 0
−80 + 20ⅈ2 + 30(ⅈ2 − ⅈ3 ) + 20(ⅈ2 − ⅈ1 ) = 0
30ⅈ3 + 30(ⅈ3 − ⅈ1 ) + 30(ⅈ3 − ⅈ2 ) = 0
Rearranging the equations gives,
70ⅈ1 − 20ⅈ2 − 30ⅈ3 = 80
−20ⅈ1 + 70ⅈ2 − 30ⅈ3 = 80
−30ⅈ1 − 30ⅈ2 + 90ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
70
[ −20
−30
−20
70
−30
ⅈ1
−30 ⅈ1
80
2.6667
−30 ] [ⅈ2 ] = [ 80] ⇒ [ⅈ2 ] = [ 2.6667 ]
ⅈ3
90 ⅈ3
0
1.7778
This implies that, ⅈ3 = 𝟏. 𝟕𝟕𝟕𝟖𝑨
It’s clear that, ⅈ0 = ⅈ3 = 𝟏. 𝟕𝟕𝟖𝑨
Now, 𝑣𝑎𝑏 = 30ⅈ0 = 𝟓𝟑. 𝟑𝟑𝑽
Let redraw the circuit with mesh currents!
CONT . . .
45
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s clear that, independent current source is in common with two different meshes, this implies that,
there exist a supermesh!
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for supermesh gives, 4ⅈ1 + 8ⅈ2 + 6(ⅈ2 − ⅈ4 ) + 2(ⅈ1 − ⅈ3 ) = 0
By KVL for bottom meshes gives, −30 + 2(ⅈ3 − ⅈ1 ) + 3(ⅈ3 − ⅈ4 ) = 0
6(ⅈ4 − ⅈ2 ) + ⅈ4 + 3(ⅈ4 − ⅈ3 ) = 0
It’s clear that, ⅈ2 − ⅈ1 = 4
Rearranging the equations gives,
6ⅈ1 + 14ⅈ2 − 2ⅈ3 − 6ⅈ4 = 0
−2ⅈ1 + 0ⅈ2 + 5ⅈ3 − 3ⅈ4 = 30
0ⅈ1 − 6ⅈ2 − 3ⅈ3 + 10ⅈ4 = 0
−ⅈ1 + ⅈ2 + 0ⅈ3 + 0ⅈ4 = 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
6
−2
[
0
−1
14
0
−6
1
−2
5
−3
0
ⅈ1
−6 ⅈ1
0
0.5530
ⅈ2
−3 ⅈ2
30
3.447
][ ] = [ ] ⇒ [ ] = [
]
ⅈ3
10 ⅈ3
0
8.561
ⅈ4
0 ⅈ4
4
4.636
This implies that, ⅈ3 = 𝟖. 𝟓𝟔𝟏𝑨
It’s clear that, ⅈ = ⅈ3 = 𝟖. 𝟓𝟔𝟏𝑨
46
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3.49
Let redraw the circuit with mesh currents!
It’s clear that, dependent current source is in common with two different meshes, this implies that,
there exist a supermesh!
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for supermesh gives, ⅈ1 − ⅈ3 + 2(ⅈ2 − ⅈ3 ) + 16 + 2ⅈ1 = 0
By KVL for upper mesh gives, 3ⅈ3 + 2(ⅈ3 − ⅈ2 ) + ⅈ3 − ⅈ1 = 0
It’s clear that, ⅈ2 − ⅈ1 = 2ⅈ0 and ⅈ0 = −ⅈ1
Rearranging the equations with substitution gives,
3ⅈ1 + 2ⅈ2 − 3ⅈ3 = −16
−ⅈ1 − 2ⅈ2 + 6ⅈ3 = 0
ⅈ1 + ⅈ2 + 0ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
2
[−1 −2
1
1
ⅈ1
−3 ⅈ1
−16
−10.67
ⅈ
ⅈ
6 ] [ 2 ] = [ 0 ] ⇒ [ 2 ] = [ 10.67 ]
ⅈ3
0 ⅈ3
0
1.78
This implies that, ⅈ1 = −10.67𝐴, ⅈ2 = 10.67𝐴 and ⅈ3 = 1.78𝐴
So that, ⅈ0 = −ⅈ1 = 10.67𝐴
Let denote currents leaving a node positive!
This implies that,
47
𝑣𝑜 −16
= ⅈ2 − ⅈ3 ⇒ 𝑣𝑜 = 𝟑𝟑. 𝟕𝟖𝑽
2
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3.51
Let redraw the circuit with mesh currents!
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for bottom meshes gives,
−40 + ⅈ1 + 2(ⅈ1 − ⅈ2 ) + 4(ⅈ1 − ⅈ3 ) = 0
8ⅈ3 − 20 + 4(ⅈ3 − ⅈ1 ) = 0
It’s clear that, ⅈ2 = 𝟓𝑨
Rearranging the equations with substitution gives,
7ⅈ1 − 4ⅈ3 = 50
−4ⅈ1 + 12ⅈ3 = 20
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
ⅈ
10
7 −4 ⅈ1
50
] [ ] = [ ] ⇒ [ 1] = [ ]
[
ⅈ3
5
−4 12 ⅈ3
20
This implies that, ⅈ1 = 10𝐴 and ⅈ3 = 5𝐴
𝑣
It’s clear that, 40 = ⅈ1 − ⅈ3 ⇒ 𝑣0 = 4(ⅈ1 − ⅈ3 ) = 𝟐𝟎𝑽
3.53
Assume that direction of voltage drops signed positive!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
By KVL for each mesh gives,
−12 + 1000(𝐼1 − 𝐼3 ) + 3000(𝐼1 − 𝐼2 ) = 𝟎
4000(𝐼2 − 𝐼4 ) + 3000(𝐼2 − 𝐼1 ) = 𝟎
6000(𝐼3 − 𝐼5 ) + 8000(𝐼3 − 𝐼4 ) + 1000(𝐼3 − 𝐼1 ) = 𝟎
48
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2000𝐼5 + 8000(𝐼5 − 𝐼4 ) + 6000(𝐼5 − 𝐼3 ) = 𝟎
It’s clear that, 𝐼4 = 3𝑚𝐴
Rearranging the equations with substitution gives,
4𝐼1 − 3𝐼2 − 1𝐼3 + 0𝐼5 = 12
−3𝐼1 + 7𝐼2 + 0𝐼3 + 0𝐼5 = −12
𝐼1 + 0𝐼2 − 15𝐼3 + 6𝐼5 = 24
0𝐼1 + 0𝐼2 − 6𝐼3 + 16𝐼5 = 24
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
4 −3
−3
7
[
1
0
0
0
−1
0
−15
−6
𝐼1
0 𝐼1
1.6196
12
𝐼2
0 𝐼2
−12
−1.0202
][ ] = [
]⇒[ ]=[
]
𝐼3
6 𝐼3
24
−2.4612
𝐼5
16 𝐼5
24
−2.4230
This implies that, ⅈ1 = 1.6196𝑚𝐴, 𝐼2 = 𝟏. 𝟔𝟏𝟗𝟔𝒎𝑨 , 𝐼3 = 𝟏. 𝟔𝟏𝟗𝟔𝒎𝑨 and 𝐼5 = 𝟏. 𝟔𝟏𝟗𝟔𝒎𝑨
3.55
Let redraw the circuit with mesh currents!
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
It’s clear that, independent current source is in common with two different meshes, this implies that,
there exist a supermesh!
By KVL for supermesh gives, 10 + 2ⅈ𝑐 + 4(ⅈ𝑐 − ⅈⅆ ) + 6(ⅈ𝑏 − ⅈ𝑎 ) = 0
By KVL for bottom mesh gives, 4(ⅈⅆ − ⅈ𝑐 ) − 8 + 12(ⅈⅆ − ⅈ𝑎 ) = 0
It’s clear that, ⅈ𝑎 = 4𝐴 and ⅈ𝑏 − ⅈ𝑐 = 1
Rearranging the equations with substitution gives,
6ⅈ𝑏 + 6ⅈ𝑐 − 4ⅈⅆ = 14
0ⅈ𝑏 − 4ⅈ𝑐 + 16ⅈⅆ = 56
ⅈ𝑏 − ⅈ𝑐 + 0ⅈⅆ = 1
49
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
6
[0
1
6
−4
−1
ⅈ𝑏
−4 ⅈ𝑏
14
3
16] [ ⅈ𝑐 ] = [ 56] ⇒ [ ⅈ𝑐 ] = [2]
ⅈⅆ
0 ⅈⅆ
1
4
This implies that, ⅈ𝑏 = 3𝐴, ⅈ𝑐 = 2𝐴 and ⅈⅆ = 4𝐴
It’s clear that, ⅈ1 = ⅈ𝑏 − ⅈ𝑎 = −𝟏𝑨 , ⅈ2 = ⅈⅆ − ⅈ𝑎 = 𝟎𝑨 and ⅈ3 = ⅈⅆ − ⅈ𝑐 = 𝟐𝑨
3.55
It’s clear that 𝑅 and 3𝑘𝛺 resistors are connected in parallel!
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
1
1 −1
This implies that, 𝑅ⅇ𝑞 for 𝑅 and 3𝑘𝛺 resistors is 𝑅ⅇ𝑞 = (3 + 𝑅)
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
3⋅𝑅
= 3+𝑅
It’s clear that, 𝑅ⅇ𝑞 for 𝑅 and 3𝑘𝛺 is connected in series with 4𝑘𝛺 resistor!
3⋅𝑅
This implies, 𝑅ⅇ𝑞 = 3+𝑅 + 4
It’s clear that 100 = 𝑣𝑅𝑒𝑞 because they are both connected to the same pair of terminals!( if there is any
confusion you can verify it using Kirchhoff's Voltage Law!)
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ 𝑅 = ⅈ
100
This implies that, 𝑅ⅇ𝑞 = 18×10−3 = 𝟓. 𝟓𝟓𝟔𝒌𝜴
3⋅𝑅
As it known, 5.556 − 4 = 3+𝑅 ⇒ 𝑅 = 𝟑. 𝟐𝟑𝒌𝜴
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
1.56
𝑒𝑞
4
This implies that, 𝑉1 = 5556 100𝑉 = 𝟐𝟖𝑽 and 𝑉2 = 5.556 100𝑉 = 𝟕𝟐𝑽
3.61
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
Let denote the mesh currents for the consecutive closed paths as ⅈ1 , ⅈ2 and ⅈ3 !
It’s clear that, ⅈ1 = ⅈ𝑠 , ⅈ3 = ⅈ0 and 𝑣0 = 30(ⅈ1 − ⅈ2 )
By KVL for other meshes gives, 20ⅈ2 − 5𝑣0 + 30(ⅈ2 − ⅈ1 ) = 0 and 10ⅈ3 + 40ⅈ3 + 5𝑣0 = 0
Rearranging the equations with substitution gives,
CONT . . .
50
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
−180ⅈ𝑠 + 200ⅈ2 = 0
−150ⅈ2 + 50ⅈ0 + 150ⅈ𝑠 = 0
Now multiply both equations with the numbers needed!
−540ⅈ𝑠 + 600ⅈ2 = 0
−600ⅈ2 + 200ⅈ0 + 600ⅈ𝑠 = 0
Now in order to get a new equation add them up!
60ⅈ𝑠 = −200ⅈ0
ⅈ
This implies that, current gain becomes ⅈ0 = −𝟎. 𝟑
3.63
𝑠
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
Let denote the mesh currents for the first and last closed paths as ⅈ1 and ⅈ2 !
It’s clear that, both the independent and dependent current source are in common with two different
meshes, this implies that, there exist a supermesh!(Note that super mesh can be applied as single larger
supermesh!)
By KVL for larger supermesh gives, 10ⅈ1 + 5ⅈ2 + 4ⅈ𝑥 − 50 = 0
Let denote currents leaving a node positive!
𝑣
By KCL for upper second middle node gives, −ⅈ𝑥 − 3 − 4𝑥 + ⅈ2 = 0
It’s clear that, ⅈ1 = ⅈ𝑥 and 𝑣𝑥 = 2(ⅈ1 − ⅈ2 )
Rearranging the equations with substitution gives,
14ⅈ𝑥 + 5ⅈ2 + 0𝑣𝑥 = 50
−ⅈ𝑥 + ⅈ2 −
𝑣𝑥
=3
4
−2ⅈ𝑥 + 2ⅈ2 + 𝑣𝑥 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
14
[−1
−2
5
1
2
0
ⅈ𝑥
ⅈ𝑥
50
2.105
1
− 4 ] [ ⅈ2 ] = [ 3 ] ⇒ [ ⅈ2 ] = [4.105]
𝑣𝑥
0
−4
1 𝑣𝑥
This implies that, ⅈ𝑥 = 𝟐. 𝟏𝟎𝟓𝑨 and 𝑣𝑥 = −𝟒𝑽
3.65
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
CONT . . .
51
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By KVL for each mesh gives,
−12 + 5ⅈ1 + ⅈ1 − ⅈ4 + 6(ⅈ1 − ⅈ2 ) = 0
ⅈ2 − ⅈ4 + ⅈ2 − ⅈ5 + 8(ⅈ2 − ⅈ3 ) + 6(ⅈ2 − ⅈ1 ) = 0
ⅈ3 − ⅈ5 + 6ⅈ3 − 9 + 8(ⅈ3 − ⅈ2 ) = 0
−6 + 3ⅈ4 + 2(ⅈ4 − ⅈ5 ) + ⅈ4 − ⅈ2 + ⅈ4 − ⅈ1 = 0
4ⅈ5 − 10 + ⅈ5 − ⅈ3 + ⅈ5 − ⅈ2 + 2(ⅈ5 − ⅈ4 ) = 0
Rearranging the equations gives,
12ⅈ1 − 6ⅈ2 + 0ⅈ3 − ⅈ4 + 0ⅈ5 = 12
−6ⅈ1 + 16ⅈ2 − 8ⅈ3 − ⅈ4 − ⅈ5 = 0
0ⅈ1 − 8ⅈ2 + 15ⅈ3 + 0ⅈ4 − ⅈ5 = 9
−ⅈ1 − ⅈ2 + 0ⅈ3 + 7ⅈ4 − 2ⅈ5 = 6
0ⅈ1 − ⅈ2 − ⅈ3 − 2ⅈ4 + 8ⅈ5 = 10
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
12
−6
0
−1
[ 0
−6
16
−8
−1
−1
0
−8
15
0
−1
−1
−1
0
7
−2
ⅈ1
0 ⅈ1
12
2.17
ⅈ
ⅈ
−1 2
0
1.99
2
−1 ⅈ3 = 9 = ⅈ3 = 1.81
ⅈ4
6
−2 ⅈ4
2.09
8 ] [ⅈ5 ] [10] [ⅈ5 ] [2.25]
This implies that, ⅈ1 = 𝟐. 𝟏𝟕𝑨 , ⅈ2 = 𝟏. 𝟗𝟗𝑨, ⅈ3 = 𝟏. 𝟖𝟏𝑨, ⅈ4 = 2.09𝐴 and ⅈ5 = 𝟐. 𝟐𝟓𝑨
3.67
Let denote 𝑣10𝛺 as 𝑣1 , 𝑣5𝛺 as 𝑣2 and 𝑣4𝐴 as 𝑣3
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
−3𝑉0 + 2 +
𝑣1 − 𝑣2 𝑣1
+
=0
4
10
𝑣
𝑣 −𝑣
𝑣2 −𝑣1
+ 52 + 2 2 3 = 0
4
𝑣3 − 𝑣2
−2−4=0
2
It’s clear that, 𝑣2 − 𝑣3 = 𝑉0
52
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣2 (− ) + 0𝑣3 − 3𝑉0 = −2
4 10
4
1
1 1 1
1
𝑣1 (− ) + 𝑣2 ( + + ) + 𝑣3 (− ) − 3𝑉0 = 0
4
4 5 2
2
1
1
0𝑣1 + 𝑣2 (− ) + 𝑣3 ( ) + 0𝑉0 = 6
2
2
0𝑣1 + 𝑣2 − 𝑣3 − 𝑉0 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
7
20
1
−
4
1
4
19
20
1
−
2
1
−
0
0
0
−3
𝑣1
𝑣1
−2
−258.95
𝑣
−3 [ 2 ] = [ 0 ] ⇒ [𝑣2 ] = [−210.53]
𝑣3
𝑣3
6
−198.53
𝑉
𝑉0
0
0
−12
0
1
−
2
1
2
−1
−1]
This implies that, 𝑉0 = −𝟏𝟐𝑽
3.69
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 − 𝑣3 𝑣1 − 𝑣2
𝑣1
+
+
=0
1000
4000
2000
𝑣2 − 𝑣1
𝑣2 − 𝑣3
𝑣2
− 0.005 +
+
=0
4000
4000
2000
𝑣3 − 𝑣2
𝑣3 − 𝑣1
+ 0.005 +
− 0.010 = 0
4000
1000
−0.02 +
Rearranging the equations gives,
1
4
1
2
1
4
𝑣1 (1 + + ) + 𝑣2 (− ) + 𝑣3 (−1) = 𝟐𝟎
1
1
1
1
1
𝑣1 (− 4) + 𝑣2 (4 + 4 + 2) + 𝑣3 (− 4) = 𝟓
1
1
−𝑣1 + 𝑣2 (− 4) + 𝑣3 (4 + 1) = 𝟓
53
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1.75
[−0.25
−1
−0.25
1
−0.25
20
−1 ⅈ𝑥
−0.25 ] [ ⅈ2 ] = [ 5 ]
5
1.25 𝑣𝑥
So that, currents can be found by solving the matrix above!
3.71
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
−10 + 5(ⅈ1 − ⅈ3 ) + 4(ⅈ1 − ⅈ2 ) = 0
ⅈ2 − ⅈ3 + 2ⅈ2 + 5 + 4(ⅈ2 − ⅈ1 ) = 0
3ⅈ3 + ⅈ3 − ⅈ2 + 5(ⅈ3 − ⅈ1 ) = 0
Rearranging the equations gives,
9ⅈ1 − 4ⅈ2 − 5ⅈ3 = 10
−4ⅈ1 + 7ⅈ2 − ⅈ3 = −5
−5ⅈ1 − ⅈ2 + 9ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
9
[−4
−5
−4
7
−1
ⅈ1
10
−5 ⅈ1
2.085
ⅈ
ⅈ
−1 ] [ 2 ] = [−5] ⇒ [ 2 ] = [0.6533]
ⅈ3
0
9 ⅈ3
1.231
This implies that, ⅈ1 = 𝟐. 𝟎𝟖𝟓𝑨, ⅈ2 = 0.6533𝐴 = 𝟔𝟓𝟑. 𝟑𝒎𝑨 and ⅈ3 = 𝟏. 𝟐𝟑𝟏𝑨
3.73
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
−6 + 2ⅈ1 + 3(ⅈ1 − ⅈ2 ) + 4(ⅈ1 − ⅈ3 ) = 0
5ⅈ2 − 4 + 3(ⅈ2 − ⅈ1 ) = 0
4(ⅈ3 − ⅈ1 ) + ⅈ3 − ⅈ4 − 2 + ⅈ3 = 0
ⅈ4 + 3 + ⅈ4 − ⅈ3 = 0
Rearranging the equations gives,
9ⅈ1 − 3ⅈ2 − 4ⅈ3 + 0ⅈ4 = 6
−3ⅈ1 + 8ⅈ2 + 0ⅈ3 + 0ⅈ4 = 4
54
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
−4ⅈ1 + 0ⅈ2 + 6ⅈ3 − ⅈ4 = 2
0ⅈ1 + 0ⅈ2 − ⅈ3 + 2ⅈ4 = −3
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
9
−3
[
−4
0
−3
8
0
0
−4
0
6
−1
ⅈ1
0 ⅈ1
6
ⅈ
ⅈ
0 2
4
] [ ] = [ ] ⇒ [ 2]
ⅈ3
−1 ⅈ3
2
ⅈ
ⅈ4
2 4
−3
So that, currents can be found by solving the matrix above!
3.77
Note that PSpice must be used but in order to make everything clear, regular process will be applied!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 = 𝑣 ⋅ 𝐺
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑉 −𝑉
𝑉
By KCL for node voltages gives, −5 + 2ⅈ𝑥 + 1 5 2 + 21 = 0 and
𝑉
It’s clear that, ⅈ𝑥 = 21
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣2 (− ) + 2ⅈ𝑥 = 5
5 2
5
1
1
𝑣1 (− ) + 𝑣2 ( + 1) − 2ⅈ𝑥 = −2
5
5
1
𝑣1 (− ) + 0𝑣2 + ⅈ𝑥 = 0
2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
0.7
[−0.2
−0.5
−0.2
1.2
0
𝑉1
3.111
2 𝑉1
5
−2 ] [𝑉2 ] = [−2] ⇒ [𝑉2 ] = [ 1.444 ]
ⅈ𝑥
1.556
1 ⅈ𝑥
0
This implies that, 𝑉1 = 3.111𝑉 and 𝑉2 = 1.444𝑉
END!
55
𝑉2 −𝑉1
5
− 2ⅈ𝑥 + 2 + 𝑉2 = 0
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Chapter 4-Practice Problems
4.1
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 =
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
Let denote 𝑣4𝛺 as 𝑣1 !
𝑣
𝑣 −𝑣
By KCL for upper middle node gives, −ⅈ𝑠 + 41 + 112 0 = 0
It’s clear that,
𝑣
𝑣1 −𝑣0
= 80
12
When ⅈ𝑠 = 15𝐴, rearranging the equations with substitution gives,
1 1
1
𝑣1 ( + ) + 𝑣0 (− ) = 15
4 12
12
𝑣1 (
1
1 1
) + 𝑣0 (− − ) = 0
12
12 8
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
1
−
12] [𝑣1 ] = [15] ⇒ [𝑣1 ] = [50]
[3
𝑣0
1
5 𝑣0
20
0
−
12
24
This implies that, 𝑣0 = 𝟐𝟎𝑽
So that when ⅈ𝑠 = 𝟑𝟎𝑨
1
1
−
12] [𝑣1 ] = [30] ⇒ [𝑣1 ] = [100]
[3
𝑣0
1
5 𝑣0
40
0
−
12
24
This implies that, 𝑣0 = 𝟒𝟎𝑽
4.2
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅
Assume that direction of voltage drops signed positive!
Let denote the mesh currents for the first and second closed paths as ⅈ1 and ⅈ2 !
Assuming 𝑉0 as 1𝑉 gives, 1 = 8ⅈ2 ⇒ ⅈ2 = 0.125𝐴 = 125𝑚𝐴
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
56
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
This implies that, 𝑅ⅇ𝑞 for 12𝛺 and 8𝛺 is 𝑅ⅇ𝑞 = 12𝛺 + 8𝛺 = 20𝛺
Now, 𝑣5𝛺 = 20ⅈ2 = 𝟐. 𝟓𝑽
Assuming 𝑉0 as 1𝑉 gives independent voltage source voltage as 2.5𝑉, so that actual value of source
30𝑉 will give 𝑉0 = 𝟏𝟐𝑽
4.3
Let 𝑣0 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 4𝐴 current and 10𝑉 voltage
sources,respectively!
To obtain 𝑣1 set voltage source as zero!
Let denote 𝑣4𝐴 as 𝑣𝑥 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 =
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for upper middle node gives,
𝑣
𝑣 −𝑣
It’s clear that, 21 = 𝑥 3 1
𝑣
𝑣𝑥 −𝑣1
− 4 + 5𝑥 = 0
3
Rearranging the equations gives,
1
1 1
𝑣1 (− ) + 𝑣𝑥 ( + ) = 4
3
3 5
1 1
1
𝑣1 ( + ) + 𝑣𝑥 (− ) = 0
2 3
3
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
[ 3
5
6
−
8
15 ] [𝑣1 ] = [4] ⇒ [𝑣1 ] = [ 4 ]
𝑣𝑥
1 𝑣𝑥
0
10
−
3
This implies that, 𝑣1 = 𝟒𝑽
To obtain 𝑣2 set current source as zero!
Let denote the current that flows in the circuit as ⅈ𝑠 !
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, 2ⅈ𝑠 + 3ⅈ𝑠 + 5ⅈ𝑠 + 10 = 0 ⇒ ⅈ𝑠 = −𝟏𝑨
This implies that, 𝑣2 = −2ⅈ𝑠 = 𝟐𝑉
So that, 𝑣0 = 𝑣1 + 𝑣2 = 4𝑉 + 2𝑉 = 𝟔𝑽
57
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.4
Let 𝑣𝑥 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 20𝑉 voltage and 4𝐴 current
sources, respectively!
To obtain 𝑣1 set current source as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅 =
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣1 gives,
This implies that, 𝑣1 = 𝟓𝑽
𝑣1 −20
𝑣
+ 41 − 0.1𝑣1 = 0
20
To obtain 𝑣2 set voltage source as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
By KCL for node 𝑣2 gives, 202 − 4 + 42 − 0.1𝑣2 = 0
This implies that, 𝑣2 = 𝟐𝟎𝑽
So that, 𝑣𝑥 = 𝑣1 + 𝑣2 = 5𝑉 + 20𝑉 = 𝟐𝟓𝑽
4.5
Let 𝐼 = ⅈ1 + ⅈ2 + ⅈ3 where ⅈ1 , ⅈ2 and ⅈ3 are the contributions due to the 16𝑉 voltage, 4𝐴 and 12𝑉
current sources, respectively!
To obtain ⅈ1 set 4𝐴 and 12𝑉 sources as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −16 + 6ⅈ1 + 2ⅈ1 + 8ⅈ1 = 0
This implies that, ⅈ1 = 1𝐴
To obtain ⅈ2 set 16𝑉 and 12𝑉 sources as zero!
Let redraw the circuit with mesh currents!
CONT . . .
58
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that direction of voltage drops signed positive!
It’s clear that, independent current source is in common with two different meshes, this implies that,
there exist a supermesh!
By KVL for supermesh gives, 6ⅈ𝑎 + 2ⅈ𝑏 + 8ⅈ𝑎 = 0
It’s clear that, ⅈ𝑎 − ⅈ𝑏 = 𝟒
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
14
[
1
ⅈ
0
2 ⅈ𝑎
0.5
]
] [ ] = [ ] ⇒ [ 𝑎] = [
ⅈ𝑏
4
−1 ⅈ𝑏
−3.5
This implies that, ⅈ𝑎 = ⅈ2 = 0.5𝐴
To obtain ⅈ3 set 16𝑉 and 4𝐴 sources as zero!
By KVL for closed path gives, 6ⅈ3 + 2ⅈ3 + 8ⅈ3 + 12 = 0
This implies that, ⅈ3 = −𝟎. 𝟕𝟓𝑨
So that, 𝐼 = ⅈ1 + ⅈ2 + ⅈ3 = 1𝐴 + 0.5𝐴 − 0.75𝐴 = 𝟎. 𝟕𝟓𝑨
4.6
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
1
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s clear that, 1𝛺 and 4𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 1𝛺 + 4𝛺 = 5𝛺
1
1
1
Now, 6𝛺 and 3𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 3 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 2𝛺
𝑒𝑞
It’s obvious that, 5𝐴 source and 2𝛺 resistor is connected in parallel so that, they can be replaced with
10𝑉 source and 2𝛺 resistor connected in series!
Now, it’s clear that, 10𝑉 and 5𝑉 sources connected in series with same polarity!
This implies that, we can assume that situation as single 15𝑉 source!
It’s obvious that, 15𝑉 source and 2𝛺 resistor are connected in series so that, they can be replaced with
7.5𝐴 source and 2𝛺 resistor connected in parallel!
CONT . . .
59
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote source current as ⅈ𝑠 !
Now, in order to create single source current, just add two independent current sources!
This implies that, ⅈ𝑆 = 7.5 + 3 = 10.5𝐴
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
It’s clear that, 2𝛺, 7𝛺 and 5𝛺 resistor are connected in parallel so that,
𝑅ⅇ𝑞 for them is
This implies, ⅈ0 =
1
1
1
1
+ + =
⇒ 𝑅ⅇ𝑞 = 𝟏. 𝟏𝟖𝟔𝜴
2
7
5
𝑅𝑒𝑞
4.7
1.186
10.5 = 𝟏. 𝟕𝟖𝑨
7
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
1
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, 2ⅈ𝑥 source and 5𝛺 resistor is connected in series so that, they can be replaced with
0.4𝐴 source and 5𝛺 resistor connected in parallel!(Note that direction must be downward!)
Let denote source current as ⅈ𝑠 !
Now, in order to create single source current, just add two independent current sources!
This implies that, ⅈ𝑆 = 0.024 − 0.4ⅈ𝑥
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
1
1
1
It’s clear that, 10𝛺 and 5𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 10 + 5 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 𝟑. 𝟑𝟑𝟑𝜴
This implies, ⅈ𝑥 =
4.8
3.333
(0.024 − 0.4ⅈ𝑥) ⇒ ⅈ𝑥 = 7.056 × 10−3 𝐴 = 𝟕. 𝟎𝟓𝟔𝒎𝑨
10
𝑒𝑞
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣4𝛺 directly and voltage drop across
the terminals of 1𝛺 resistor is zero!
Let denote 𝑣3𝐴 as 𝑣1 !
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
60
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By KCL for node voltages gives,
𝑣1 − 18
𝑣1 − 𝑣𝑇ℎ
−3+
=0
6
6
𝑣𝑇ℎ − 𝑣1 𝑣𝑇ℎ
+
=0
4
6
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣𝑇ℎ (− ) = 6
6 6
6
1
1 1
𝑣1 (− ) + 𝑣𝑇ℎ ( + ) = 0
6
6 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
[ 3
1
−
6
1
6] [ 𝑣1 ] = [6] ⇒ [ 𝑣1 ] = [ 22.5]
𝑣𝑇ℎ
5 𝑣𝑇ℎ
0
9
12
−
This implies that, 𝑣𝑇ℎ = 9𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 4𝛺 resistor becomes zero!
Let denote 𝑣3𝐴 as 𝑣1 !
𝑣
𝑣1 −18
− 3 + 61 = 0 ⇒ 𝑣1 = 𝟏𝟖𝑽
6
By KCL for node 𝑣1 gives,
Now it’s clear that, voltage across the first 6𝛺 resistor is zero!
Let denote currents leaving a node positive!
By KCL for node a gives, −3 + ⅈ𝑠 = 0 ⇒ ⅈ𝑠 = 𝟑𝑨
𝑣
9
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 3 = 𝟑𝜴
𝑠
So that the Thevenin equivalent circuit is the given below!
Here, 𝑅𝐿 corresponds to 1𝛺 resistor!
By KVL for closed path gives, −9 + 3𝐼 + 𝐼 = 0 ⇒ 𝐼 = 𝟐. 𝟐𝟓𝑨
61
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.9
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣4𝛺 directly!
Let denote 𝑣1.5𝐼𝑥 as 𝑣1 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 − 6
𝑣1 − 𝑣𝑇ℎ
− 1.5𝐼𝑥 +
=0
5
3
𝑣𝑇ℎ − 𝑣1 𝑣𝑇ℎ
+
=0
3
4
𝑣 −𝑣
Also, it’s clear that, 𝐼𝑥 = 1 3 𝑇ℎ
Rearranging the equations gives,
1 1
1
6
𝑣1 ( + ) + 𝑣𝑇ℎ (− ) − 1.5𝐼𝑥 =
5 3
3
5
1
1 1
𝑣1 (− ) + 𝑣𝑇ℎ ( + ) + 0𝐼𝑥 = 0
3
3 4
1
1
𝑣1 (− ) + 𝑣𝑇ℎ ( ) + 𝐼𝑥 = 0
3
3
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
8
15
1
−
3
1
[ −3
1
3
7
12
1
3
−
−1.5
6
𝑣1
𝑣1
9.33
0 [𝑣𝑇ℎ ] = [ 5] ⇒ [𝑣𝑇ℎ ] = [ 5.33 ]
0
𝐼𝑥
𝐼𝑥
1.33
0
1]
This implies that, 𝑣𝑇ℎ = 𝟓. 𝟑𝟑𝑽
Let create a short-circuit for the terminals a and b in order to find short-circuit current ⅈ𝑠𝑐 !
This implies that, voltage drop across the 4𝛺 resistor becomes zero!
Let denote 𝑣1.5𝐼𝑥 as 𝑣1 !
By KCL for node voltages gives,
𝑣1 − 6
𝑣1
− 1.5𝐼𝑥 + = 0
5
3
62
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝒗
Also, it’s clear that, 𝐼𝑥 = 𝟑𝟏 !
It’s obvious that, ⅈ𝑠𝑐 = 𝐼𝑥
Rearranging the equations with substitution gives,
1 1
6
𝑣1 ( + ) − 1.5ⅈ𝑠 =
5 3
5
1
𝑣1 (− ) + ⅈ𝑠 = 𝟎
3
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
8
[ 15
1
−
3
6
− 1.5 𝑣
𝑣1
1
36
]
] [ ⅈ ] = [ 5] ⇒ [ ⅈ ] = [
12
𝑠
𝑠
0
1
This implies that, ⅈ𝑠 = 12𝐴
𝑣
5.33
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 12 = 𝟎. 𝟒𝟒𝜴
𝑠
So that the Thevenin equivalent circuit is the given below!
4.10
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣15𝛺 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑇ℎ gives,
𝑣
𝑣𝑇ℎ +4𝑣𝑥 −𝑣𝑥
𝑣
+ 𝑇ℎ = 0
10
15
𝑣 +4𝑣 −𝑣𝑥
It’s clear that, 5𝑥 = 𝑇 10𝑥
Rearranging the equations gives,
CONT . . .
63
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣𝑇ℎ (
1
1
3
+ ) + 𝑣𝑥 ( ) = 0
10 15
10
𝑣𝑇ℎ (−
1
1 3
) + 𝑣𝑥 ( − ) = 0
10
5 10
1
[ 6
1
−
10
3
10] [𝑣𝑇ℎ ] = [0] ⇒ [𝑣𝑇ℎ ] = [0]
𝑣𝑥
𝑣𝑥
1
0
0
−
10
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
This implies that, 𝑣𝑇ℎ = 𝟎𝑽
Now in order to find the 𝑅𝑇ℎ , let redraw the circuit with test a test source and mesh currents!
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
5ⅈ1 + 10ⅈ1 + 4𝑣𝑥 + 15(ⅈ1 − ⅈ2 ) = 0
𝑣𝑇 + 15(ⅈ2 − ⅈ1 ) = 0
It’s clear that, 5ⅈ1 = −𝑣𝑥
Rearranging the equations with substitution gives,
10ⅈ1 − 15ⅈ2 + 0𝑣𝑇 = 0 ⇒ 10ⅈ1 = 𝟏𝟓𝒊𝟐
𝑣𝑇 = 15ⅈ1 − 15ⅈ2 ⇒ 𝑣𝑇 = 𝟕. 𝟓𝒊𝟐
It’s clear-cut that, ⅈ𝑇 = −ⅈ2
𝑣
It’s obvious that, 𝑅𝑇ℎ = ⅈ 𝑇 = −𝟕. 𝟓𝜴
𝑇
So that the Thevenin equivalent circuit is the given below!
CONT . . .
64
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.11
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣6𝛺 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
Let denote 𝑣4𝐴 as 𝑣1 !
By KCL for node voltages gives,
𝑣1 − 𝑣𝑇ℎ
𝑣1 − 15
−4+
=0
3
3
𝑣𝑇ℎ − 𝑣1 𝑣𝑇ℎ
+
=0
6
3
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣𝑇ℎ (− ) = 9
3 3
3
1
1 1
𝑣1 (− ) + 𝑣𝑇ℎ ( + ) = 0
3
3 6
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
2
[ 3
1
−
3
−
1
3] [ 𝑣1 ] = [9] ⇒ [ 𝑣1 ] = [20.25]
𝑣𝑇ℎ
1 𝑣𝑇ℎ
0
13.5
2
This implies that, 𝑣𝑇ℎ = 𝟏𝟑. 𝟓𝑽
Let create a short-circuit for the terminals a and b in order to find short-circuit current ⅈ𝑠𝑐 !
This implies that, voltage drop across the 6𝛺 resistor becomes zero!
65
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote 𝑣4𝐴 as 𝑣1 !
By KCL for node voltage 𝑣1 gives,
This implies, 𝑣1 = 𝟏𝟑. 𝟓𝑽
It’s clear that, ⅈ𝑠𝑐 = ⅈ3𝛺 =
𝑣
𝑣1 −15
− 4 + 31 = 0
3
13.5−0
= 𝟒. 𝟓𝑨
3
𝑣
13.5
As it known, 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ 𝑇 = 4.5 = 𝟑𝜴 and 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟒. 𝟓𝑨
𝑠𝑐
So that, the Norton equivalent circuit is the given below!
4.12
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣𝑥 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
Let denote 𝑣10𝐴 as 𝑣1 !
It’s obvious that dependent voltage source connects two non-reference nodes!
This implies that, source creates a supernode!
𝑣
𝑣
By KCL for supernode gives, 61 − 10 + 2𝑥 = 0
It’s clear that, 𝑣1 − 𝑣𝑥 = 2𝑣𝑥
Rearranging the equations gives,
1
1
𝑣1 ( ) + 𝑣𝑥 ( ) = 𝟏𝟎
6
2
𝑣1 − 3𝑣𝑥 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
[6
1
66
1 𝑣
𝑣1
10
1
30
2] [ ] = [ ] ⇒ [ ] = [ ]
−3
𝑣𝑥
0
𝑣𝑥
10
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
This implies that, 𝑣𝑥 = 𝑣𝑇ℎ = 𝟏𝟎𝑽
Let create a short-circuit for the terminals a and b in order to find short-circuit current ⅈ𝑠𝑐 !
This implies that, voltage drop across the 2𝛺 resistor becomes zero!
So that, voltage drop across the dependent voltage source becomes zero!
This implies that, ⅈ𝑠𝑐 = 10𝐴
𝑣
As it known, 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 1𝛺 and 𝐼𝑁 = ⅈ𝑠 = 𝟏𝟎𝑨
𝑠𝑐
So that, the Norton equivalent circuit is the given below!
4.13
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
This implies that, voltage drop across the 4𝛺 resistor becomes zero!
So that the node voltage for the middle node is equal to 𝑣𝑇ℎ !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for upper middle node gives,
It’s clear that, 𝑣𝑥 = 9 − 𝑣𝑇ℎ
𝑣𝑇 −9
+ 𝑣𝑇 − 3𝑣𝑥 = 0
2
Rearranging the equations gives,
1
9
𝑣𝑇 (2 + 1) − 3𝑣𝑥 = 2
𝑣𝑇 + 𝑣𝑥 = 9
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
CONT . . .
67
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3
[2
1
9
𝑣𝑇ℎ
7
−3 𝑣𝑇ℎ
] [ 𝑣 ] = [2] ⇒ [ 𝑣 ] = [ ]
𝑥
2
𝑥
9
1
This implies that, 𝑣𝑇ℎ = 𝟕𝑽
Let create a short-circuit for the terminals a and b in order to find short-circuit current ⅈ𝑠𝑐 !
Let denote the upper middle node voltage as 𝑣1 !
By KCL for node 𝑣1 gives,
It’s clear that, 𝑣𝑥 = 9 − 𝑣1
𝑣1 −9
𝑣
+ 𝑣1 − 3𝑣𝑥 + 1 = 0
2
4
Rearranging the equations gives,
1
1
9
𝑣1 (2 + 1 + 4) − 3𝑣𝑥 = 2
𝑣1 + 𝑣𝑥 = 9
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
[4
1
9
𝑣1
6.632
−3 𝑣1
] [𝑣 ] = [ 2 ] ⇒ [𝑣 ] = [
]
𝑥
𝑥
2.368
1
9
This implies that, 𝑣1 = 6.632𝑉
𝑣
It’s clear that, ⅈ𝑠𝑐 = 41 = 1.658𝐴
𝑣
As it known, 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 4.22𝛺 and 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟏. 𝟔𝟓𝟖𝑨
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 4.22𝛺 corresponds to 𝑅𝑇ℎ !
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝒗
𝟐
Now, 𝑝 = (𝑹 𝑻𝒉
) ⋅ 𝑹𝑳
+𝑹
𝑻𝒉
68
𝑳
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑉𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 0
] = 𝑉𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
𝑣
2
𝑣
2
𝑣2
So that, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ = 𝟐. 𝟗𝟎𝑾
+𝑅
𝑇ℎ
4.15
𝐿
𝑇ℎ
𝑇ℎ
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
Now, 𝑝 = (𝑅 𝑇ℎ
) ⋅ 𝑅𝐿
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑣𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 0
] = 𝑣𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
It’s clear that, 𝑣𝑇ℎ is equal to open-circuit voltage 1𝑉and 𝑅𝑇ℎ = 𝟐𝒌𝜴!
𝑣
2
𝑣
2
𝑣2
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ = 0.125𝑊 = 𝟏𝟐𝟓𝒎𝑾
This implies, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
+𝑅
4.16
𝑇ℎ
𝐿
𝑇ℎ
𝑇ℎ
Let redraw the circuit with the load resistor!
Here 𝑅𝐿 corresponds to 20𝛺 resistance!
69
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
20
By substitution, 𝑣𝑅𝐿 = 𝑅 9𝑉 = 8𝑉 ⇒ 𝑅ⅇ𝑞 = 22.5𝛺 ⇒ 𝑅𝑇ℎ = 2.5𝛺
𝑒𝑞
Now, for the circuit given above, 𝑅𝑇ℎ corresponds to 2.5𝛺!
As it known, new value of 𝑅𝐿 is 10𝛺
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿 = 𝟏𝟐. 𝟓𝜴
10
By substitution, 𝑣𝑅𝐿 = 12.5 9𝑉 = 𝟕. 𝟐𝑽
4.17
Let redraw the circuit first!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Because of the bridge is balanced 𝑣1 = 𝑣2 !
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑅
𝑒𝑞
𝑅
2
𝑣 = 𝑣2 = 𝑥 𝑣
This implies that, 𝑣1 = 𝑅 +𝑅
𝑅 +𝑅
1
2
𝑥
3
𝑅
𝑅
2
It’s clear that, no current flows through the galvanometer when, 𝑅 +𝑅
= 𝑥 ⇒ 𝑅2 𝑅3 = 𝑅1 𝑅𝑥
𝑅 +𝑅
𝑅
This implies, 𝑅𝑥 = 𝑅3 𝑅2
1
2
𝑥
3
1
By substitution, 𝑅𝑥 = 𝟑. 𝟐𝒌𝜴
4.18
Let redraw the circuit as open-circuit!
CONT . . .
70
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
1
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
It’s clear that, 60𝛺 and 40𝛺 are connected in series so that 𝑅ⅇ𝑞 for them is 60𝛺 + 40𝛺 = 100𝛺
Also, 20𝛺 and 30𝛺 are connected in series so that 𝑅ⅇ𝑞 for them is 20𝛺 + 30𝛺 = 50𝛺
It’s obvious that, 𝑣50𝛺 = 𝑣100𝛺 = 16𝑉 because they are both connected to the same pair of terminals!(
if there is any confusion you
can verify it using Kirchhoff's Voltage Law!)
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
40
𝑒𝑞
30
This implies that, 𝑣40𝛺 = 100 16𝑉 = 6.4𝑉 and 𝑣30𝛺 = 50 16𝑉 = 𝟗. 𝟔𝑽
It’s clear that, 𝑣𝑇ℎ = 𝑣𝑜𝑐 = 6.4𝑉 − 9.6𝑉 = −𝟑. 𝟐𝑽
Let redraw the circuit in order to find Thevenin resistance!
CONT . . .
71
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1
1
1
It’s clear that, 60𝛺 and 40𝛺 are connected int parallel so that 𝑅ⅇ𝑞 for them is, 60 + 40 = 𝑅 ⇒ 𝑅ⅇ𝑞 =
24𝛺
1
1
𝑒𝑞
1
Now, 20𝛺 and 30𝛺 are connected int parallel so that 𝑅ⅇ𝑞 for them is, 20 + 30 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 12𝛺
𝑒𝑞
Finally, 24𝛺 and 12𝛺 are connected int series so that 𝑅ⅇ𝑞 for them is 24𝛺 + 12𝛺 = 𝟑𝟔𝜴
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to 14𝛺!
It’s clear that, 𝐼 is equal to the current that flows through the galvanometer!
Assume that direction of voltage drops signed positive and direction of current 𝐼 through the loop is
clockwise direction!
By KVL for closed path gives, 3.2 + 36𝐼 + 14𝐼 = 0 ⇒ 𝐼 = −0.064𝐴 = −𝟔𝟒𝒎𝑨
Note that, negative sign indicates that the current flows in the direction opposite to the one assumed,
that is, from terminal b to a!
Chapter 4-Problems
4.1
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
By KCL for upper middle node gives, 𝑣1 − 1 + 81 + 81 = 0 ⇒ 𝑣1 = 𝟎. 𝟖𝑽
𝑣
It’s clear that, ⅈ0 = 81 = 𝟎. 𝟏𝑨
72
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By the linearity property, 𝑘ⅈ𝑅 = 𝑘𝑣
This implies that, if the input voltage becomes 10𝑉, 𝑘 becomes 10!
This implies that, new value of ⅈ0 is, 𝑘ⅈ0 = 𝟏𝑨
4.3
(a) Let redraw the circuit with mesh currents!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
−𝑣𝑠 + ⅈ1 − ⅈ2 + ⅈ1 − ⅈ3 = 0
ⅈ2 + ⅈ2 − ⅈ3 + ⅈ2 − ⅈ1 = 0
ⅈ3 − ⅈ2 + ⅈ3 + ⅈ3 − ⅈ1 = 0
Rearranging the equations with substitution gives,
2ⅈ1 − ⅈ2 − ⅈ3 = 1
−ⅈ1 + 3ⅈ2 − ⅈ3 = 0
−ⅈ1 − ⅈ2 + 3ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
2
[−1
−1
−1
3
−2
ⅈ1
1
1
−1 ⅈ1
−1 ] [ⅈ2 ] = [0] ⇒ [ⅈ2 ] = [ 0.5 ]
ⅈ3
0.5
0
3 ⅈ3
This implies that, ⅈ1 = 1𝐴, ⅈ2 = 0.5𝐴 and ⅈ3 = 0.5𝐴
It’s clear that, ⅈ0 = ⅈ3 = 𝟎. 𝟓𝑨 𝑣0 = ⅈ0 ⇒ 𝑣0 =
𝟎. 𝟓𝑽
(b) By the linearity property, 𝑘ⅈ𝑅 = 𝑘𝑣
This implies that, if the source voltage becomes 10𝑉, 𝑘 becomes 10!
So that, new value of 𝑣0 becomes 𝑘𝑣0 = 𝟓𝑽 and ⅈ0 = 𝟓𝑨
73
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
(c) By KVL for each mesh gives,
By KVL for each mesh gives,
−𝑣𝑠 + 10(ⅈ1 − ⅈ2 ) + 10(ⅈ1 − ⅈ3 ) = 0
10ⅈ2 + 10(ⅈ2 − ⅈ3 ) + 10(ⅈ2 − ⅈ1 ) = 0
10(ⅈ3 − ⅈ2 ) + 10ⅈ3 + 10(ⅈ3 − ⅈ1 = 0
Rearranging the equations with substitution gives,
20ⅈ1 − 10ⅈ2 − 10ⅈ3 = 10
−10ⅈ1 + 30ⅈ2 − 10ⅈ3 = 0
−10ⅈ1 − 10ⅈ2 + 30ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
20
[−10
−10
−10
30
−20
ⅈ1
−10 ⅈ1
10
1
−10 ] [ⅈ2 ] = [ 0 ] ⇒ [ⅈ2 ] = [ 0.5 ]
ⅈ3
30 ⅈ3
0
0.5
This implies that, ⅈ1 = 1𝐴, ⅈ2 = 0.5𝐴 and ⅈ3 = 0.5𝐴
It’s clear that, ⅈ0 = ⅈ3 = 𝟎. 𝟓𝑨 and 𝑣0 = ⅈ0 ⇒ 𝑣0 = 𝟎. 𝟓𝑽
4.5
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
Let denote the voltage drop across the first 6𝛺 resistor as 𝑣1 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and source voltage as 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ !
By KCL for node voltages gives,
𝑣1 − 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ 𝑣1 𝑣1 − 1
+ +
=0
3
2
6
1 − 𝑣1 1 1
+ + =0
6 6
3
It’s clear that, 𝑣1 = 2𝑉
By using substitution, 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ = 3.333𝑉
By the linearity property, 𝑘ⅈ𝑅 = 𝑘𝑣
74
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
This implies that, if the source voltage becomes 15𝑉, 𝑘 becomes 4.5!
So that, new value of 𝑣0 becomes 𝑘𝑣0 = 𝟒. 𝟓𝑽
4.7
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
Let denote the voltage drop across the first 3𝛺 resistor as 𝑣1 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive and source voltage as 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ !
𝑣
𝑣 −1
By KCL for node voltage gives, 𝑣1 − 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ + 31 + 14 = 0
It’s clear that,
1
𝑣1 −1
= 2 ⇒ 𝑣1 = 3𝑉
4
By using substitution, 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ = 4.5𝑉
By the linearity property, 𝑘ⅈ𝑅 = 𝑘𝑣
This implies that, if the source voltage becomes 4𝑉, 𝑘 becomes 0.8889!
So that, new value of 𝑉0 becomes 𝑘𝑉0 = 0.8889𝑉 = 𝟖𝟖𝟖. 𝟗𝒎𝑽
4.9
Let 𝑣0 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 6𝐴 current and 18𝑉 voltage
sources, respectively!
To obtain 𝑣1 set voltage source as zero!
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
resistors!
1
𝑛=1 𝑅𝑛
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
1
1
1
It’s clear that, 2𝛺 and 2𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 2 + 2 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 1𝛺
Now, 1𝛺 and 1𝛺 are connected int series so that 𝑅ⅇ𝑞 for them is 1𝛺 + 1𝛺 = 2𝛺
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
1
1
𝑒𝑞
1
It’s clear that, 2𝛺 and 4𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 2 + 4 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 1.333𝛺
Let denote the current flows through the 𝑣0 as ⅈ0
75
𝑒𝑞
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1.333
6𝐴 = 𝟑. 𝟗𝟗𝟗𝑨
2
This implies that, ⅈ0 =
So that, 𝑣1 = ⅈ0 = 4
To obtain 𝑣2 set current source as zero!
1
1
1
It’s clear that, 2𝛺 and 2𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 2 + 2 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 1𝛺
𝑒𝑞
Now, 1𝛺, 1𝛺 and 4𝛺 are connected int series so that 𝑅ⅇ𝑞 for them is 1𝛺 + 1𝛺 + 4𝛺 = 6𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
1
By substitution, 𝑣2 = 6 18𝑉 = 3𝑉
𝑒𝑞
This implies, 𝑣0 = 𝑣1 + 𝑣2 = 3.999𝑉 + 3𝑉 = 𝟔. 𝟗𝟗𝟗𝑽
4.11
Let denote 𝑣40𝛺 as 𝑣𝑎, 𝑣4ⅈ0 as 𝑣𝑏 and lower node as reference node!
Let 𝑣0 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 6𝐴 current and 30𝑉 voltage
sources, respectively!
To obtain 𝑣1 set voltage source as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
−6 +
𝑣𝑎 𝑣𝑎 − 𝑣𝑏
+
=0
40
10
𝑣𝑏 − 𝑣𝑎
𝑣𝑏
− 4ⅈ1 +
=0
10
20
𝑣 −𝑣
It’s clear that, ⅈ1 = 𝑎10 𝑏
Rearranging the equations gives,
1
1
1
+ ) + 𝑣𝑏 (− ) + 0ⅈ1 = 6
40
10
10
𝑣𝑎 (
𝑣𝑎 (−
𝑣𝑎 (−
1
1
1
) + 𝑣𝑏 ( + ) − 4ⅈ1 = 0
10
10
20
1
1
) + 𝑣𝑏 ( ) + ⅈ1 = 0
10
10
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
76
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
1
8
1
− 10
1
[− 10
1
− 10
3
20
1
10
0
6
𝑣𝑎
𝑣𝑎
176
3
−4 [𝑣𝑏 ] = [ ] ⇒ [𝑣𝑏 ] = [ 160 ]
2
ⅈ1
ⅈ1
1.6
0
1]
This implies that, ⅈ1 = 1.6𝐴
It’s clear that, 𝑣1 = 10ⅈ1 ⇒ 𝑣1 = 16𝑉
To obtain 𝑣2 set current source as zero!
By using same process rearranged equations are given below!
1
1
1
+ ) + 𝑣𝑏 (− ) + 0ⅈ2 = 0
40
10
10
𝑣𝑎 (
𝑣𝑎 (−
1
1
1
3
) + 𝑣𝑏 ( + ) − 4ⅈ2 = −
10
10
20
2
1
1
𝑣𝑎 (− 10) + 𝑣𝑏 (10) + ⅈ2 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
8
1
−
10
1
[− 10
1
− 10
3
20
1
10
0
0
𝑣𝑎
𝑣𝑎
−8
3
𝑣
𝑣
−4 [ 𝑏 ] = [− ] ⇒ [ 𝑏 ] = [−10]
2
ⅈ1
ⅈ2
0.2
0
1]
This implies that, ⅈ1 = 0.2𝐴
It’s clear that, 𝑣2 = 10ⅈ2 ⇒ 𝑣1 = 2𝑉
So that, 𝑣0 = 𝑣1 + 𝑣2 = 16𝑉 + 2𝑉 = 𝟏𝟖𝑽
𝑣
This implies, ⅈ0 = 100 = 𝟏. 𝟖𝑨
4.13
Let 𝑣0 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the independent current sources and
12𝑉 voltage source, respectively!
To obtain 𝑣1 set voltage source as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
Let denote 𝑣10𝛺 as 𝑣𝑎 !
CONT . . .
77
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By KCL for node voltage gives,
𝑣𝑎 − 𝑣1 𝑣𝑎
+
=0
8
10
𝑣1 𝑣1 − 𝑣𝑎
−4 + +
=0
5
8
−2 + 4 +
Rearranging the equations gives,
1
1 1
𝑣1 (− ) + 𝑣𝑎 ( + ) = −2
8
8 10
1 1
1
𝑣1 ( + ) + 𝑣𝑎 (− ) = 4
5 8
8
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1 9
𝑣1
𝑣1
−2
11.304
[ 8 40 ] [𝑣 ] = [ ] ⇒ [𝑣 ] = [
]
13
1 𝑎
𝑎
4
−2.609
−
40
8
−
This implies that, 𝑣1 = 11.304𝑉
To obtain 𝑣2 set current sources as zero!
Let denote the current that flows through the circuit as ⅈ𝑠 !
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, 10ⅈ𝑠 + 8ⅈ𝑠 + 12 + 𝑣2 = 0
It’s clear that, 𝑣2 = 5ⅈ𝑠
Rearranging the equations gives,
18ⅈ𝑠 + 𝑣2 = −12
−5ⅈ𝑠 + 𝑣2 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
18
−5
1] [ ⅈ𝑠 ] = [−12] ⇒ [ ⅈ𝑠 ] = [ −0.5217]
𝑣2
0
−2.609
1 𝑣2
This implies that, 𝑣2 = −𝟐. 𝟔𝟎𝟗𝑽
So that, 𝑣0 = 𝑣1 + 𝑣2 = 11.304𝑉 − 2.609𝑉 = 𝟖. 𝟔𝟗𝟓𝑽
4.15
Let ⅈ = ⅈ1 + ⅈ2 where ⅈ1 and ⅈ2 are the contributions due to the 0.2𝐴 source and independent voltage
sources, respectively!
To obtain 𝑣1 set voltage sources as zero!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
78
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that direction of voltage drops signed positive
Let redraw the circuit with mesh currents!
It’s obvious that, independent current source causes a supermesh!(Two meshes have a current source
in common!)
By KVL for supermesh gives, ⅈ𝑏 − ⅈ𝑎 + 4ⅈ𝑐 + 3(ⅈ𝑐 − ⅈ𝑎 ) = 0
By KVL for mesh ⅈ𝑎 gives,
2ⅈ𝑎 + ⅈ𝑎 − ⅈ𝑏 + 3(ⅈ𝑎 − ⅈ𝑐 ) = 0
It’s clear that, ⅈ𝑏 − ⅈ𝑐 = 2
Rearranging the equations gives,
−4ⅈ𝑎 + ⅈ𝑏 + 7ⅈ𝑐 = 0
6ⅈ𝑎 − ⅈ𝑏 − 3ⅈ𝑐 = 0
0ⅈ𝑎 + ⅈ𝑏 − ⅈ𝑐 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
−4
[ 6
0
1
−1
1
ⅈ𝑎
7 ⅈ𝑎
0.25
0
−3] [ⅈ𝑏 ] = [0] ⇒ [ⅈ𝑏 ] = [ 1.875 ]
ⅈ𝑐
2
−1 ⅈ𝑐
−0.125
This implies that, ⅈ𝑎 = 0.25𝐴, ⅈ𝑏 = 1.875𝐴 and ⅈ𝑐 = −0.125𝐴
It’s clear that, ⅈ1 = ⅈ𝑎 − ⅈ𝑐 = 0.375𝐴 = 375𝑚𝐴
To obtain 𝑣2 set current source as zero!
Now let redraw the circuit with mesh currents!
By KVL for each mesh gives,
CONT . . .
79
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2ⅈ𝑎 − 20 + ⅈ𝑎 − ⅈ𝑏 + 3(ⅈ𝑎 − ⅈ𝑏 ) = 0
4ⅈ𝑏 − 16 + 3(ⅈ𝑏 − ⅈ𝑎 ) + ⅈ𝑏 − ⅈ𝑎 = 0
6ⅈ𝑎 − 4ⅈ𝑏 = 20
−4ⅈ𝑎 + 8ⅈ𝑏 = 16
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
6 −4] [ⅈ𝑎 ] = [20] ⇒ [ⅈ𝑎 ] = [ 7 ]
ⅈ𝑏
5.5
16
−4
8 ⅈ𝑏
This implies that, ⅈ𝑎 = 7𝐴 and ⅈ𝑏 = 5.5𝐴
It’s clear that, ⅈ2 = ⅈ𝑎 − ⅈ𝑏 = 1.5𝐴
So that, ⅈ = ⅈ1 + ⅈ2 = 0.375𝐴 + 1.5𝐴 = 𝟏. 𝟖𝟕𝟓𝑨
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
This implies, 𝑝3𝛺 = 3ⅈ 2 = 𝟏𝟎. 𝟓𝟓𝑾
4.17
Let 𝑣𝑥 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 6𝐴 source and independent voltage
sources, respectively!
To obtain 𝑣1 set voltage sources as zero!
Let denote 𝑣60𝛺 as 𝑣𝑎 and 𝑣6𝐴 as 𝑣𝑏 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣𝑎 𝑣𝑎 𝑣𝑎 − 𝑣𝑏
+
+
=0
10
30 60
𝑣𝑏 𝑣𝑏
𝑣𝑏 − 𝑣𝑎
−6+
+
=0
30 20
10
Rearranging the equations gives,
1
1
1
1
𝑣𝑎 ( +
+ ) + 𝑣𝑏 (− ) = 0
30 60 10
10
𝑣𝑎 (−
1
1
1
1
) + 𝑣𝑏 ( +
+ )=6
10
10 30 20
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
80
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
3
1
−
𝑣𝑎
𝑣𝑎
0
34.286
20
10
[
] [ ] = [ ] ⇒ [𝑣 ] = [
]
1
11 𝑣𝑏
6
𝑏
51.429
−
10
60
This implies that, 𝑣𝑎 = 34.286𝑉 , 𝑣𝑏 = 51.429𝑉 and 𝑣1 = 𝑣𝑎 − 𝑣𝑏 = −17.143𝑉
To obtain 𝑣2 set current source as zero!
Let denote 𝑣60𝛺 as 𝑣𝑎 and 𝑣30𝛺 as 𝑣𝑏 !
By KCL for node voltages gives,
𝑣𝑎 − 90 𝑣𝑎 𝑣𝑎 − 𝑣𝑏
+
+
=0
60
10
30
𝑣𝑏 − 𝑣𝑎 𝑣𝑏 𝑣𝑏 − 40
+
+
=0
20
10
30
Rearranging the equations gives,
𝑣𝑎 (
1
1
1
1
+
+ ) + 𝑣𝑏 (− ) = 3
30 60 10
10
𝑣𝑎 (−
1
1
1
1
) + 𝑣𝑏 ( +
+ )=2
10
10 30 20
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
−
3
20
1
10
−
1
𝑣𝑎
3
42.857
10 𝑣𝑎
11 ] [𝑣𝑏 ] = [2] ⇒ [𝑣𝑏 ] = [34.286]
60
This implies that, 𝑣𝑎 = 42.857𝑉, 𝑣𝑏 = 34.286𝑉 and 𝑣2 = 𝑣𝑎 − 𝑣𝑏 = 𝟖. 𝟓𝟕𝟏𝑽
So that, 𝑣𝑥 = 𝑣1 + 𝑣2 = −17.143𝑉 + 8.571𝑉 = −𝟖. 𝟓𝟕𝟐𝑽
4.19
Let 𝑣𝑥 = 𝑣1 + 𝑣2 where 𝑣1 and 𝑣2 are the contributions due to the 6𝐴 and 4𝐴 sources, respectively!
To obtain 𝑣1 set 4𝐴 source as zero!
Let redraw the circuit with mesh currents!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
81
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that direction of voltage drops signed positive!
It’s obvious that, independent current source causes a supermesh!(Two meshes have a current source
in common!)
By KVL for supermesh gives, 2ⅈ𝑎 + 8ⅈ𝑏 + 4ⅈ1 = 0
It’s clear that, ⅈ1 = −ⅈ𝑎 and ⅈ𝑏 − ⅈ𝑎 = 6
Rearranging the equations gives,
2ⅈ𝑎 + 8ⅈ𝑏 + 4ⅈ1 = 0
ⅈ𝑎 + 0ⅈ𝑏 + ⅈ1 = 0
−ⅈ𝑎 + ⅈ𝑏 + 0ⅈ1 = 6
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
2
[ 1
−1
8
0
1
ⅈ𝑎
−8
4 ⅈ𝑎
0
1] [ⅈ𝑏 ] = [0] ⇒ [ⅈ𝑏 ] = [−2]
ⅈ1
6
8
0 ⅈ1
This implies that, ⅈ𝑎 = −8𝐴, ⅈ𝑏 = −2𝐴 and ⅈ1 = 8𝐴
It’s clear that, 𝑣1 = 8ⅈ𝑏 = −𝟏𝟔𝑽
To obtain 𝑣2 set 6𝐴 source as zero!
Let redraw the circuit with mesh currents!
It’s obvious that, independent current source causes a supermesh!(Two meshes have a current source
in common!)
By KVL for supermesh gives, 2ⅈ𝑎 + 8ⅈ𝑏 + 4ⅈ1 = 0
It’s clear that, ⅈ2 = −ⅈ𝑎 and ⅈ𝑏 − ⅈ𝑎 = 4
Rearranging the equations gives,
2ⅈ𝑎 + 8ⅈ𝑏 + 4ⅈ1 = 0
ⅈ𝑎 + 0ⅈ𝑏 + ⅈ2 = 0
−ⅈ𝑎 + ⅈ𝑏 + 0ⅈ2 = 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
82
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
2
[ 1
−1
8
0
1
ⅈ𝑎
4 ⅈ𝑎
−5.333
0
1] [ⅈ𝑏 ] = [0] ⇒ [ⅈ𝑏 ] = [−1.333]
ⅈ1
4
0 ⅈ2
5.333
This implies that, ⅈ𝑎 = −5.333𝐴, ⅈ𝑏 = −1.333𝐴 and ⅈ1 = 5.333𝐴
It’s clear that, 𝑣2 = 8ⅈ𝑏 = −10.66𝑉
So that, 𝑣𝑥 = 𝑣1 + 𝑣2 = −16𝑉 − 10.66𝑉 = −𝟐𝟔. 𝟔𝟔𝑽
4.23
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
1
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, 45𝑉 source and 3𝛺 resistor are connected in series so that, they can be replaced with
15𝐴 source and 3𝛺 resistor connected in parallel!(Note that direction must be upward!)
1
1
1
It’s clear that, 6𝛺 and 3𝛺 are connected in parallel so 𝑅ⅇ𝑞 for them is 6 + 3 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 2𝛺
𝑒𝑞
Now, 15𝐴 source and 2𝛺 resistor are connected in parallel so that, they can be replaced with 30𝑉
source and 2𝛺 resistor connected in series!
It’s obvious that, 9𝐴 source and 10𝛺 resistor are connected in parallel so that, they can be replaced
with 90𝑉 source and 10𝛺 resistor connected in series!
So that the circuit becomes,
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −90 + 10ⅈ0 + 8ⅈ0 + 2ⅈ0 + 30 = 0 ⇒ ⅈ0 = 𝟑𝑨
This implies, 𝑣8𝛺 = 8ⅈ0 = 𝟐𝟒𝑽
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
=𝑣⋅ⅈ
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
This implies, 𝑝8𝛺 = 𝑣8𝛺 ⋅ ⅈ8𝛺 = 𝟕𝟐𝑾
83
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.25
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
1
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, 6𝐴 source and 5𝛺 resistor are connected in parallel so that, they can be replaced with
30𝑉 source and 5𝛺 resistor connected in series!(Note that direction of the voltage drop must be
upward!)
Now, 2𝐴 source and 9𝛺 resistor are connected in parallel so that, they can be replaced with 18𝑉
source and 9𝛺 resistor connected in series!(Note that direction of the voltage rise must be to the right!)
It’s clear that, 3𝐴 source and 4𝛺 resistor are connected in parallel so that, they can be replaced with
12𝑉 source and 4𝛺 resistor connected in series!(Note that direction of the voltage drop must be
upward!)
So that the circuit becomes,
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, 12 − 18 + 9𝐼 − 30 + 5𝐼 − 30 + 2𝐼 + 4𝐼 = 0 ⇒ 𝐼 = 𝟑. 𝟑𝑨
This implies, 𝑣0 = −2𝐼 = −𝟔. 𝟔𝑽
4.27
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
1
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, 40𝑉 source and 20𝛺 resistor are connected in series so that, they can be replaced
with 2𝐴 source and 20𝛺 resistor connected in parallel!(Note that direction must be upward!)
Now, in order to create single source current, just add two independent current sources 8𝐴 + 2𝐴 =
10𝐴
84
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s clear that, 10𝐴 source and 20𝛺 resistor are connected in parallel so that, they can be replaced with
200𝑉 source and 20𝛺 resistor connected in series!(Note that direction of the voltage rise must be
upward!)
It’s obvious that, 50𝑉 source and 10𝛺 resistor are connected in series so that, they can be replaced
with 5𝐴 source and 10𝛺 resistor connected in parallel!(Note that direction must be upward!)
1
1
1
Now, 10𝛺 and 40𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 10 + 40 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 8𝛺
𝑒𝑞
It’s clear that, 5𝐴 source and 8𝛺 resistor are connected in parallel so that, they can be replaced with
40𝑉 source and 8𝛺 resistor connected in series!(Note that direction of the voltage rise must be
upward!)
So that the circuit becomes,
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −40 + 8ⅈ0 + 12ⅈ0 + 20ⅈ0 + 200 = 0 ⇒ ⅈ0 = −𝟒𝑨
This implies, 𝑣𝑥 = 12ⅈ0 = −𝟒𝟖𝑽
4.29
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
1
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, dependent voltage source and 2𝑘𝛺 resistor are connected in series so that, they can be
3𝑣0
replaced with 2000
current source and 2𝑘𝛺 resistor connected in parallel!(Note that direction must be
to the right!)
1
1
1
Now, 4𝑘𝛺 and 2𝑘𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 4 + 2 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 1.333𝑘𝛺
3𝑣
𝑒𝑞
0
It’s clear that, 2000
source and 1.333𝑘𝛺 resistor are connected in parallel so that, they can be replaced
with 1.996𝑣0 source and 1.333𝑘𝛺 resistor connected in series!(Note that direction of the voltage rise
must be to the right!)
So that the circuit becomes,
CONT . . .
85
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s clear that, 𝑣0 = (3 × 10−3 𝐴) ⋅ (1 × 103 𝛺) = 𝟑𝑽
4.31
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
∞
1
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
𝑛=1 𝑅𝑛
resistors!
−1
) for parallel connections of
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, dependent voltage source and 6𝛺 resistor are connected in series so that, they can be
𝑣
replaced with 3𝑥 current source and 6𝛺 resistor connected in parallel!(Note that direction must be to
the upward!)
1
1
1
Now, 8𝛺 and 6𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 8 + 6 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 3.429𝛺
𝑣
𝑒𝑞
It’s clear that, 3𝑥 source and 3.429𝛺 resistor are connected in parallel so that, they can be replaced with
1.143𝑣𝑥 source and 3.429𝛺 resistor connected in series!(Note that direction of the voltage rise must be
to the upward!)
So that the circuit becomes,
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −12 + 3ⅈ𝑥 + 3.429ⅈ𝑥 + 1.143𝑣𝑥 = 0
It’s clear that, 𝑣𝑥 = 3ⅈ𝑥
Rearranging the equations gives,
CONT . . .
86
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
6.429ⅈ𝑥 + 1.143𝑣𝑥 = 12
−3ⅈ𝑥 + 𝑣𝑥 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
6.429 1.143] [ ⅈ𝑥 ] = [12] ⇒ [ ⅈ𝑥 ] = [ 1.217 ]
𝑣𝑥
𝑣𝑥
0
3.652
−3
1
[
This implies that, 𝑣𝑥 = 𝟑. 𝟔𝟓𝟐𝑽
4.33
(a) Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣12 = 𝑣4𝛺 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage 𝑣𝑇ℎ gives,
𝑣𝑇ℎ
𝑣𝑇ℎ −20
+ 40
= 0 ⇒ 𝑣𝑇ℎ = 16𝑉
10
Now, let create a short-circuit for the terminals 1 and 2!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 40𝛺 resistor becomes zero!
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −20 + 10ⅈ𝑠𝑐 = 0 ⇒ ⅈ𝑠𝑐 = 2𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 8𝛺
𝑠𝑐
(b) Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣12 = 𝑣30𝛺 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
𝑣
𝑣
−30
𝑇ℎ
+ 𝑇ℎ60
By KCL for node voltage 𝑣𝑇ℎ gives, −2 + 30
= 0 ⇒ 𝑣𝑇ℎ = 50𝑉
Now, let create a short-circuit for the terminals 1 and 2!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 30𝛺 resistor becomes zero!
30
By KCL for the short-circuit node gives, −2 + ⅈ𝑠𝑐 − 60 = 0 ⇒ ⅈ𝑠𝑐 = 2.5𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟐𝟎𝜴
𝑠𝑐
CONT . . .
87
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.37
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣12𝛺 directly!
Let redraw the circuit with node voltages and reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 − 180
𝑣1 − 𝑣𝑇ℎ
−3+
=0
20
40
𝑣𝑇ℎ − 𝑣1 𝑣𝑇ℎ
3+
+
=0
40
12
Rearranging the equations gives,
1
1
1
𝑣1 (20 + 40) + 𝑣𝑇ℎ (− 40) = 12
𝑣1 (−
1
1
1
) + 𝑣𝑇ℎ ( + ) = −3
40
40 12
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
1
−
40 ] [ 𝑣1 ] = [ 12 ] ⇒ [ 𝑣1 ] = [ 163.3 ]
[ 40
𝑣𝑇ℎ
1
13 𝑣𝑇ℎ
−3
10
−
40 120
This implies that, 𝑣𝑇ℎ = 10𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 12𝛺 resistor becomes zero!
88
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
CONT . . .
Let redraw the circuit with a node voltage and reference node!
By KCL for node 𝑣1 gives,
𝑣
𝑣1 −180
− 3 + 401 = 0 ⇒ 𝑣1 = 𝟏𝟔𝟎𝑽
20
𝑣
By KCL for node a gives, 3 − 401 + ⅈ𝑠 = 0 ⇒ ⅈ𝑠 = 𝟏𝑨
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏𝟎𝜴
𝑠𝑐
So that the Norton equivalent circuit is given below!
4.39
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 directly!
Let denote the bottom node as reference node, first middle node as 𝑣1 and 𝑣5𝛺 as 𝑣2 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
CONT . . .
89
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣1 − 8 𝑣1 − 𝑣2
+
=0
10
10
𝑣2 − 𝑣1 𝑣2 𝑣2 − 𝑣𝑇ℎ
+ +
=0
10
5
16
𝑣𝑇ℎ − 𝑣2
1+
=0
16
−1 +
Rearranging the equations gives,
𝑣1 (
1
1
1
9
+ ) + 𝑣2 (− ) + 0𝑣𝑇ℎ =
10 10
10
5
𝑣1 (−
1
1 1 1
1
) + 𝑣2 ( + + ) + 𝑣𝑇ℎ (− ) = 0
10
10 5 16
16
0𝑣1 + 𝑣2 (−
1
1
) + 𝑣𝑇ℎ ( ) = −1
16
16
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
5
1
−
10
[
0
1
− 10
29
80
1
− 16
0
9
𝑣1
𝑣1
8.8
1
5
𝑣2 ] = [ ] ⇒ [ 𝑣2 ] = [ −0.4 ]
−
[
0
16
𝑣𝑇ℎ
𝑣𝑇ℎ
−16.4
1
−1
]
16
This implies that, 𝑣𝑇ℎ = −16.4𝑉
Let denote short-circuit current as ⅈ𝑠𝑐 !
Let denote the bottom node as reference node, first middle node as 𝑣1 and 𝑣5𝛺 as 𝑣2 !
By KCL for first and second node voltages gives,
𝑣1 − 8 𝑣1 − 𝑣2
+
=0
10
10
𝑣2 − 𝑣1 𝑣2 𝑣2
+ +
=0
10
5 16
−1 +
Rearranging the equations gives,
𝑣1 (
1
1
1
9
+ ) + 𝑣2 (− ) =
10 10
10
5
𝑣1 (−
1
1 1 1
) + 𝑣2 ( + + ) = 0
10
10 5 16
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
1
9
−
𝑣1
𝑣1
10.44
5
10
]
[
] [𝑣 ] = [5] ⇒ [𝑣 ] = [
1
29
2.88
2
2
0
−
10
80
This implies that, 𝑣1 = 𝟏𝟎. 𝟒𝟒𝑽 and 𝑣2 = 𝟐. 𝟖𝟖𝑽
90
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
By KCL for node a gives, 1 − 162 + ⅈ𝑠𝑐 = 0 ⇒ ⅈ𝑠𝑐 = −𝟎. 𝟖𝟐𝑨
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟐𝟎𝜴
5
So that the Thevenin equivalent circuit is given below!
4.41
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣3𝛺 directly!
Let denote the bottom node as reference node and 𝑣6𝛺 as 𝑣1 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
−1 +
𝑣1 𝑣1 + 14 − 𝑣𝑇ℎ
+
=0
6
14
𝑣𝑇ℎ
𝑣𝑇ℎ − 14 − 𝑣1
+3+
=0
14
5
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣𝑇ℎ (− ) = 0
6 14
14
𝑣1 (−
1
1 1
) + 𝑣𝑇ℎ ( + ) = −2
14
14 5
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
CONT . . .
91
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5
21
[ 1
− 14
1
− 14 𝑣1
𝑣1
−2.4
0
] [ ] = [ ] ⇒ [𝑣 ] = [
]
19 𝑣2
−8
𝑇ℎ
−2
70
This implies that, 𝑣𝑇ℎ = −𝟖𝑽
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 5𝛺 resistor becomes zero!
Let denote the bottom node as reference node and 𝑣6𝛺 as 𝑣1 !
By KCL for node voltages gives,
−1 +
𝑣1 𝑣1 + 14
+
=0
14
6
−14 − 𝑣1
+ 3 + ⅈ𝑠𝑐 = 0
14
Rearranging the equations gives,
1 1
𝑣1 ( + ) + 0ⅈ𝑠𝑐 =
6 14
𝑣1 (−
1
) + ⅈ𝑠𝑐 = −2
14
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
5
21
[
1
−
14
0 𝑣
𝑣1
1
4.2
0
]
] [ⅈ ] = [ ] ⇒ [ⅈ ] = [
−2
−2
𝑠𝑐
𝑠𝑐
1
This implies that, ⅈ𝑠𝑐 = −𝟐𝑨
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = −𝟐𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟒𝜴
𝑠𝑐
So that the Thevenenin and Norton equaivalent circuits are given below!
92
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.43
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
It’s obvious that, 10𝛺 and 10𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝛺 + 10𝛺 = 20𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
10
This implies, 𝑣𝑎 = 20 20𝑉 = 10𝑉
𝑒𝑞
It’s clear that, 5𝛺 and 2𝐴 source are connected in same pair of terminals!
This implies that, 𝑣𝑏 = 2𝐴 ⋅ 5𝛺 = 10𝑉
So that, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣𝑎 − 𝑣𝑏 = 10 − 10 = 0
Now in order to find the 𝑅𝑇ℎ , let redraw the circuit with test a test source!
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
Let denote the bottom node as reference node and 𝑣10𝛺 as 𝑣1 and 𝑣5𝛺 as 𝑣2 !
It’s obvious that independent voltage source 𝑉𝑇 connects two non-reference nodes!
This implies that, 𝑉𝑇 source creates a supernode!
By KCL for supernode,
𝑣1 −20
𝑣
𝑣
+ 1 + 2−2 = 0
10
10
5
𝑣
𝑣 −20
It’s clear that, 𝑣𝑇 = 𝑣1 − 𝑣2 and 𝐼𝑇 = 101 + 110
By using substitution, 10𝐼𝑇 = 𝑉𝑇
𝑉
As it known, 𝑅𝑇ℎ = 𝐼 𝑇 = 𝟏𝟎𝜴
𝑇
93
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
So that the Thevenin equivalent circuit becomes,
4.45
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣4𝛺 directly!
Let denote the bottom node as reference node and 𝑣6𝛺 as 𝑣1 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 𝑣1 − 𝑣𝑇ℎ
+
=0
6
6
𝑣𝑇ℎ − 𝑣1 𝑣𝑇ℎ
+
=0
6
4
−6 +
Rearranging the equations gives,
1 1
1
𝑣1 ( + ) + 𝑣𝑇ℎ (− ) = 6
6 6
6
1
1 1
𝑣1 (− ) + 𝑣𝑇ℎ ( + ) = 0
6
6 4
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
3
1
[− 6
1
6 [ 𝑣1 ] = [6] ⇒ [ 𝑣1 ] = [22.5 ]
𝑣𝑇ℎ
5 𝑣𝑇ℎ
9
0
]
12
−
This implies that, 𝑣𝑇ℎ = 9𝑉
Now, let create a short-circuit for the terminals a and b!
CONT . . .
94
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 4𝛺 resistor becomes zero!
Let denote the bottom node as reference node and 𝑣6𝛺 as 𝑣1 !
𝑣
𝑣
By KCL for node voltage 𝑣1 gives, −6 + 61 + 61 = 0 ⇒ 𝑣1 = 18𝑉
𝑣
It’s clear that, ⅈ𝑠𝑐 = 61 = 3𝐴
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟑𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟑𝜴
So that the Norton equivalent circuit becomes,
𝑠𝑐
4.47
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑉𝑥 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage gives,
𝑣𝑇ℎ −50
𝑣𝑇ℎ
+ 60
+ 2𝑣𝑇ℎ = 0 ⇒ 𝑣𝑇ℎ = 1.9841𝑉
12
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 60𝛺 resistor becomes zero therefore dependent current
source becomes zero!
−50
By KCL for node voltage gives, 12 + ⅈ𝑠𝑐 = 0 ⇒ ⅈ𝑠𝑐 = 4.1667𝐴
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟒. 𝟏𝟔𝟔𝟕𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 0.47620𝛺 = 𝟒𝟕𝟔. 𝟐𝒎𝜴
𝑠𝑐
So that the Thevenin and Norton equivalent circuit are given below!
CONT . . .
95
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.51
(a) Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣4𝛺 directly!
Let denote the bottom node as reference node, 𝑣3𝛺 as 𝑣1 and 𝑣2𝛺 as 𝑣2 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣1 −120
𝑣
𝑣 −𝑣
+ 1+ 1 2=0
6
3
4
𝑣
𝑣2 −𝑣1
− 6 + 22 = 0
4
Rearranging the equations gives,
1
1
1
1
𝑣1 (6 + 3 + 4) + 𝑣2 (− 4) = 20
1
1
1
𝑣1 (− 4) + 𝑣2 (4 + 2) = 6
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
3
−
4] [𝑣1 ] = [20] ⇒ [𝑣1 ] = [ 33 ]
4
[
𝑣2
1
3 𝑣2
19
6
−
4
4
This implies that, 𝑣1 = 33𝑉 and 𝑣2 = 19𝑉
It’s clear that, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣1 − 𝑣2 = 14𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 4𝛺 resistor becomes zero!
Let denote the bottom node as reference node, 𝑣3𝛺 as 𝑣1 and 𝑣2𝛺 as 𝑣2 !
By KCL for node voltages gives,
CONT . . .
96
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
𝑣1 −120
+ 31 + ⅈ𝑠𝑐 = 0
6
−ⅈ𝑠 − 6 +
𝑣2
=0
2
It’s obvious that 𝑣4𝛺 = 𝑣1 − 𝑣2 = 0 ⇒ 𝑣1 = 𝑣2
Rearranging the equations gives,
1 1
𝑣1 ( + ) + ⅈ𝑠𝑐 = 20
6 3
1
𝑣1 ( ) − ⅈ𝑠𝑐 = 6
2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
[21
2
1 𝑣1
−1
] [ⅈ ] = [
𝑠𝑐
20] ⇒ [𝑣1 ] = [ 26 ]
6
ⅈ𝑠𝑐
7
This implies that, ⅈ𝑠𝑐 = 𝟕𝑨
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟕𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟐𝜴
𝑠𝑐
So that the Norton equivalent circuit is given below!
(b) Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑐ⅆ = 𝑣2𝛺 directly!
Let denote the bottom node as reference node, 𝑣3𝛺 as 𝑣1 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣
𝑣 −𝑣
𝑣1 −120
+ 31 + 1 4 𝑇ℎ = 0
6
𝑣
𝑣𝑇ℎ −𝑣1
− 6 + 2𝑇ℎ = 0
4
CONT . . .
97
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Rearranging the equations gives,
1
1
1
1
𝑣1 (6 + 3 + 4) + 𝑣𝑇ℎ (− 4) = 20
1
1
1
𝑣1 (− 4) + 𝑣𝑇ℎ (4 + 2) = 6
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
1
−
4
4] [ 𝑣1 ] = [20] ⇒ [ 𝑣1 ] = [ 33 ]
[
𝑣𝑇ℎ
1
3 𝑣𝑇ℎ
19
6
−
4
4
This implies that, 𝑣𝑇ℎ = 19𝑉
Now, let create a short-circuit for the terminals c and d!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 2𝛺 resistor becomes zero!
Let denote the bottom node as reference node, 𝑣3𝛺 as 𝑣1 !
By KCL for node voltage 𝑣1 gives,
𝑣
𝑣
𝑣
𝑣1 −120
+ 31 + 41 = 0
6
By KCL for node b gives, − 41 − 6 + ⅈ𝑠𝑐 = 0
Rearranging the equations gives,
1
6
1
3
1
4
𝑣1 ( + + ) + 0ⅈ𝑠𝑐 = 20
1
4
𝑣1 (− ) + ⅈ𝑠𝑐 = 6
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
[ 4
1
−
4
0 𝑣
1
1
] [ⅈ ] = [
𝑠𝑐
20] ⇒ [𝑣1 ] = [ 26.667 ]
6
ⅈ𝑠𝑐
12.667
This implies that, ⅈ𝑠𝑐 = 𝟏𝟐. 𝟔𝟔𝟕𝑨
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟏𝟐. 𝟔𝟔𝟕𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏. 𝟓𝜴
4.53
𝑠𝑐
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 directly!
Let denote the bottom node as reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
CONT . . .
98
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣
𝑣 −𝑣
𝑣0 −18
− 0.25𝑣0 + 30 + 0 2 𝑇ℎ = 0
6
𝑣
−𝑣
0.25𝑣0 + 𝑇ℎ2 0 = 0
Rearranging the equations gives,
1
1
1
4
1
2
1
1
1
𝑣0 (6 + 3 + 2 − 4) + 𝑣𝑇ℎ (− 2) = 3
1
2
𝑣0 ( − ) + 𝑣𝑇ℎ ( ) = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
[ 4
1
−
4
−
1
2] [ 𝑣0 ] = [3] ⇒ [ 𝑣0 ] = [ 6 ]
𝑣𝑇ℎ
1 𝑣𝑇ℎ
3
0
2
This implies that, 𝑣𝑇ℎ = 3𝑉
Now, let create a short-circuit for the terminals c and d!
Let denote short-circuit current as ⅈ𝑠𝑐 !
By KCL for node voltage 𝑣0 gives,
𝑣
𝑣
𝑣
𝑣0 −18
− 0.25𝑣0 + 30 + 20 = 0
6
It’s clear that, 0.25𝑣0 − 20 + ⅈ𝑠𝑐 = 0
Rearranging the equations gives,
1
1
1
1
1
1
𝑣0 (6 + 3 + 2 − 4) + 0ⅈ𝑠𝑐 = 3
𝑣0 (4 − 2) + ⅈ𝑠𝑐 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
[ 4
1
−
4
0 𝑣
𝑣0
0
4
3
] [ⅈ ] = [ ] ⇒ [ⅈ ] = [ ]
1
0
𝑠𝑐
𝑠𝑐
1
This implies that, ⅈ𝑠𝑐 = 𝟏𝑨
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟏𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟑𝜴
𝑠𝑐
CONT . . .
99
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
So that the Norton equivalent circuit is given below!
4.55
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑉𝑎𝑏 = 𝑣50𝑘𝛺 directly!
Let denote the bottom node as reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
By KVL for leftmost closed path gives,
−2 + 8000𝐼 + 0.001𝑉𝑎𝑏 = 0
It’s clear that, 𝑉𝑎𝑏 = −400000𝐼
Rearranging the equations gives,
8000𝐼 + 0.001𝑉𝑎𝑏 = 2
4000000𝐼 + 𝑉𝑎𝑏 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
8000
4000000
[
0.001] [ 𝐼 ] = [2] ⇒ [ 𝐼 ] = [ 5 × 10−4 ]
𝑉𝑎𝑏
𝑉𝑎𝑏
1
0
−2000
This implies, 𝑉𝑎𝑏 = 𝑣𝑇ℎ = −2000𝑉 = −2𝑘𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 50𝑘𝛺 resistor becomes zero therefore dependent voltage
source becomes zero!
By KVL for leftmost closed path gives, −2 + 8000𝐼 = 0
It’s clear that, 80𝐼 = −ⅈ𝑠
Rearranging the equations gives,
8000𝐼 + 0ⅈ𝑠 = 2
80𝐼 + ⅈ𝑠 = 0
100
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
0] [ 𝐼 ] = [2] ⇒ [ 𝐼 ] = [2.5 × 10−4 ]
ⅈ𝑠𝑐
1 ⅈ𝑠𝑐
0
−0.02
8000
80
[
This implies, ⅈ𝑠𝑐 = −0.02𝐴 = −20𝑚𝐴
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = −𝟐𝟎𝒎𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 100000𝛺 = 𝟏𝟎𝟎𝒌𝜴
So that the Norton equivalent circuit is given below!
𝑠𝑐
4.57
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣10𝛺 directly!
Let denote the bottom node as reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣
𝑣 −𝑣
𝑣𝑥 −50
+ 6𝑥 + 𝑥 2 𝑇ℎ = 0
3
𝑣𝑇ℎ
𝑣𝑇ℎ −𝑣𝑥
− 0.5𝑣𝑥 + 10
=0
2
Rearranging the equations gives,
1
3
1
6
1
2
1
2
𝑣𝑥 ( + + ) + 𝑣𝑇ℎ (− ) =
1
1
1
1
50
3
𝑣𝑥 (− 2 − 2) + 𝑣𝑇ℎ (2 + 10) = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
1
1 −2
−1
3
5
50
𝑣𝑥
𝑣𝑥
100
] [𝑣 ] = [ 3 ] ⇒ [𝑣 ] = [
]
166.67
𝑇ℎ
𝑇ℎ
0
This implies, 𝑣𝑇ℎ = 𝟏𝟔𝟔. 𝟔𝟕𝑽
101
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 10𝛺 resistor becomes zero!
By KCL for node voltages gives,
𝑣
𝑣
𝑣𝑥 −50
+ 6𝑥 + 2𝑥 = 0
3
−
𝑣𝑥
− 0.5𝑣𝑥 + ⅈ𝑠𝑐 = 0
2
Rearranging the equations gives,
1
1
1
50
𝑣𝑥 (3 + 6 + 2) + 0ⅈ𝑠𝑐 = 3
1
1
𝑣𝑥 (− 2 − 2) + ⅈ𝑠𝑐 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
−1
[
50
0] [𝑣𝑥 ] = [ 3 ] ⇒ [𝑣𝑥 ] = [16.667 ]
ⅈ𝑠𝑐
16.667
1 ⅈ𝑠𝑐
0
This implies, ⅈ𝑠𝑐 = 16.667𝐴
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟏𝟔. 𝟔𝟔𝟕𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏𝟎𝜴
𝑠𝑐
So that the Thevenin and Norton equivalent circuits are given below!
4.59
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections and 𝑅ⅇ𝑞 = (∑
of resistors!
∞
1
)
𝑅
𝑛=1 𝑛
−1
for parallel connections
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 directly!
CONT . . .
102
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑅
By Current Division equation, ⅈ𝑥 = 𝑅𝑒𝑞 ⅈ𝑠𝑜𝑢𝑟𝑐ⅇ
𝑥
It’s clear that, ⅈ10𝛺 = ⅈ50𝛺 and ⅈ20𝛺 = ⅈ40𝛺 !
It’s obvious that, 10𝛺 and 50𝛺 are connected in series so that 𝑅ⅇ𝑞 for them is 10𝛺 + 50𝛺 = 60𝛺
Likewise, 20𝛺 and 40𝛺 are connected in series so that 𝑅ⅇ𝑞 for them is 20𝛺 + 40𝛺 = 60𝛺
1
1
1
Now, 60𝛺 and 60𝛺 are connected in parallel so that 𝑅ⅇ𝑞 for them is 60 + 60 = 𝑅 ⇒ 𝑅ⅇ𝑞 = 30𝛺
30
30
By substitution, ⅈ10𝛺 = ⅈ50𝛺 = 60 8𝐴 = 4𝐴 and ⅈ20𝛺 = ⅈ40𝛺 = 60 8𝐴 = 4𝐴
𝑣
𝑒𝑞
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
This implies, 𝑣𝑎 = 𝑣50𝛺 = 50ⅈ50𝛺 = 200𝑉 and 𝑣𝑏 = 𝑣40𝛺 = 40ⅈ40𝛺 = 160𝑉
It’s clear that, 𝑣𝑇ℎ = 𝑣𝑎𝑏 = 𝑣𝑎 − 𝑣𝑏 = 40𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
Let redraw the circuit with mesh currents!
Assume that direction of voltage drops signed positive!
It’s clear that, ⅈ1 = 𝟖𝑨!
By KVL for other meshes gives,
20ⅈ2 + 10(ⅈ2 − 8) = 0
40ⅈ3 + 50(ⅈ3 − 8) = 0
Rearranging the equations gives,
30ⅈ2 + 0ⅈ3 = 80
0ⅈ2 + 90ⅈ3 = 400
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
CONT . . .
103
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
[
30
0
0 ] [ⅈ2 ] = [ 80 ] ⇒ [ⅈ2 ] = [2.6667 ]
ⅈ3
4.4445
90 ⅈ3
400
This implies, ⅈ2 = 2.6667𝐴 and ⅈ3 = 4.4445𝐴
It’s clear that, ⅈ𝑠𝑐 = ⅈ3 − ⅈ2 = −1.7778𝐴
Note that, negative sign indicates that the current flows in the direction opposite to the one assumed,
that is, from terminal b to a!
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟏. 𝟕𝟕𝟕𝟖𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟐𝟐. 𝟓𝜴
𝑠𝑐
So that the Thevenin and Norton equivalent circuits are given below!
4.61
Let redraw the circuit with mesh currents!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣𝑎𝑏 directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
2ⅈ1 − 12 + 6(ⅈ1 − ⅈ2 ) + 6(ⅈ1 − ⅈ3 ) − 12 = 0
2ⅈ2 + 6(ⅈ2 − ⅈ3 ) + 6(ⅈ2 − ⅈ1 ) = 0
12 + 2ⅈ3 + 12 + 6(ⅈ3 − ⅈ1 ) + 6(ⅈ3 − ⅈ2 ) = 0
Rearranging the equations gives
CONT . . .
104
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
14ⅈ1 − 6ⅈ2 − 6ⅈ3 = 24
−6ⅈ1 + 14ⅈ2 − 6ⅈ3 = 0
−6ⅈ1 − 6ⅈ2 + 14ⅈ3 = −24
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
14
[ −6
−6
ⅈ1
−6 ⅈ𝑎
1.2
24
−6] [ⅈ𝑏 ] = [ 0 ] ⇒ [ⅈ2 ] = [ 0 ]
ⅈ3
−1.2
−24
14 ⅈ2
−6
14
−6
This implies that, ⅈ1 = 1.2𝐴, ⅈ2 = 0𝐴 and ⅈ3 = −1.2
It’s clear that, 𝑣𝑇ℎ = 12 + 2ⅈ3 = 9.6𝑉
Let denote short-circuit current as ⅈ𝑠𝑐 !
Let redraw the circuit with mesh currents!
By KVL for each mesh gives,
2ⅈ1 − 12 + 6(ⅈ1 − ⅈ2 ) + 6(ⅈ1 − ⅈ3 ) − 12 = 0
2ⅈ2 + 6(ⅈ2 − ⅈ3 ) + 6(ⅈ2 − ⅈ1 ) = 0
12 + 2(ⅈ3 − ⅈ𝑠𝑐 ) + 12 + 6(ⅈ3 − ⅈ1 ) + 6(ⅈ3 − ⅈ2 ) = 0
2(ⅈ𝑠𝑐 − ⅈ3 ) − 12 = 0
Rearranging the equations gives,
14ⅈ1 − 6ⅈ2 − 6ⅈ3 + 0ⅈ𝑠𝑐 = 24
−6ⅈ1 + 14ⅈ2 − 6ⅈ3 + 0ⅈ𝑠𝑐 = 0
−6ⅈ1 − 6ⅈ2 + 14ⅈ3 − 2ⅈ𝑠𝑐 = −24
0ⅈ1 + 0ⅈ2 − 2ⅈ3 + 2ⅈ𝑠𝑐 = 12
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
14
−6
[
−6
0
−6
14
−6
0
−6
−6
14
−2
ⅈ1
0 ⅈ1
3.6
24
ⅈ2
0 ⅈ2
0
2.4
][ ] = [
]⇒[ ]=[ ]
ⅈ3
−2 ⅈ3
−24
2
ⅈ𝑠𝑐
8
2 ⅈ𝑠𝑐
12
This implies that, ⅈ𝑠𝑐 = 𝟖𝑨
105
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
As it known, 𝐼𝑁 = ⅈ𝑠𝑐 = 𝟖𝑨 and 𝑅𝑁 = 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏. 𝟐𝜴
𝑠𝑐
So that the Thevenin and Norton equivalent circuits are given below!
4.63
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ = 𝑣0.5𝑣0 directly!
Let denote the bottom node as reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage gives,
𝑣
𝑣
−𝑣
It’s clear that, 200 = 𝑇ℎ10 0
𝑣𝑇 −𝑣0
− 0.5𝑣0 = 0
10
Rearranging the equations gives,
1
1
1
𝑣0 (− 10 − 2) + 𝑣𝑇ℎ (10) = 0
1
1
1
𝑣0 (20 + 10) + 𝑣𝑇ℎ (− 10) = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
3
−5
3
20
1
− 10
𝑣0
𝑣0
0
0
] [𝑣 ] = [ ] ⇒ [𝑣 ] = [ ]
0
𝑇ℎ
𝑇ℎ
0
1
− 10
This implies, 𝑣𝑇ℎ = 𝟎𝑽
Let redraw the circuit with test voltage source and current in order to find Norton or Thevenin
resistance!
\
CONT . . .
106
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 𝑣𝑇 = 𝑣0.5𝑣0 because they are connected to the same pair of terminals!( if there is any
confusion you can verify it using Kirchhoff's Voltage Law!)
By KCL for node voltage gives,
𝑣
𝑣 −𝑣
It’s clear that, 200 = 𝑇10 0
𝑣𝑇 −𝑣0
− 0.5𝑣0 − 𝐼𝑇 = 0
10
By using substitution, −10𝐼𝑇 = 3𝑣𝑇
𝑣
As it known, 𝑅𝑇ℎ = 𝑅𝑁 = 𝐼 𝑇 = −𝟑. 𝟑𝟑𝟑𝜴
𝑇
So that the Norton equivalent circuit is given below!
4.65
The voltage 𝑣0 will be ignored in order to find the open-circuit voltage!
Open-circuit causes zero voltage drop across the 2𝛺 resistor!
This implies, 𝑣𝑇ℎ = 𝑣12𝛺 !
Let denote the bottom node as reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage gives,
107
𝑣𝑇ℎ
𝑣𝑇ℎ −64
+ 12
= 0 ⇒ 𝑣𝑇ℎ = 𝟒𝟖𝑽
4
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 and 𝑣12𝛺 as 𝑣1 !
By KCL for node voltage 𝑣1 gives,
𝑣
It’s clear that, 21 = ⅈ𝑠𝑐 !
𝑣
𝑣
𝑣1 −64
+ 121 + 21 = 0
4
Rearranging the equations gives,
1
1
1
𝑣1 (4 + 12 + 2) + 0ⅈ𝑠𝑐 = 16
1
2
𝑣1 ( ) − ⅈ𝑠𝑐 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
5
[61
2
0
−1
𝑣1
16] ⇒ [𝑣1 ] = [19.2 ]
ⅈ𝑠𝑐
9.6
0
] [ⅈ ] = [
𝑠𝑐
This implies, ⅈ𝑠𝑐 = 9.6𝐴
𝑣
As it known, 𝑅𝑇ℎ = 𝑅𝑁 = ⅈ𝑇ℎ = 𝟓𝜴
𝑠𝑐
Let redraw the Thevenin equivalent circuit with the circuit element now!
Assume that direction of voltage drops signed positive!
By KVL for closed path gives, −48 + 5𝐼0 + 𝑉0 = 0 ⇒ 𝑉0 = 48 − 5𝐼0
4.67
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
CONT . . .
108
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let redraw the circuit with mesh currents!
By KVL for each mesh gives,
90ⅈ1 + 10ⅈ1 + 40 = 0 ⇒ ⅈ1 = −0.4𝐴 = −400𝑚𝐴
20ⅈ2 − 40 + 80ⅈ2 = 0 ⇒ ⅈ2 = 0.4𝐴 = 400𝑚𝐴
Let denote the bottom node as reference node!
This implies, 𝑣𝑇ℎ = 𝑣20𝛺 + 𝑣90𝛺 = 20ⅈ1 + 90ⅈ2 = −𝟐𝟖𝑽
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
Let redraw the circuit with mesh currents!
By KVL for each mesh gives,
90(ⅈ1 − ⅈ𝑠𝑐 ) + 10ⅈ1 + 40 = 0
20(ⅈ2 − ⅈ𝑠𝑐 ) − 40 + 80ⅈ2 = 0
90(ⅈ𝑠𝑐 − ⅈ1 ) + 20(ⅈ𝑠𝑐 − ⅈ2 ) = 0
Rearranging the equations gives,
CONT . . .
109
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
100ⅈ1 + 0ⅈ2 − 90ⅈ𝑠𝑐 = −40
0ⅈ1 + 100ⅈ2 − 20ⅈ𝑠𝑐 = 40
−90ⅈ1 − 20ⅈ2 + 110ⅈ𝑠𝑐 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
[
100
0
−90
ⅈ1
−90 ⅈ𝑎
−40
−1.408
−20] [ⅈ𝑏 ] = [ 40 ] ⇒ [ ⅈ2 ] = [ 0.176]
ⅈ𝑠𝑐
110 ⅈ2
−1.12
0
0
100
−20
This implies that, ⅈ𝑠𝑐 = −1.12𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟐𝟓𝜴
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to 𝑅!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
=
⋅
= 𝑣 ⋅ ⅈ = ⅈ2 ⋅ 𝑅
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
) ⋅ 𝑅𝐿
Now, 𝑝 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑉𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 𝟎
] = 𝑉𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
𝑣
2
𝑣
2
𝑣2
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ = 𝟕. 𝟖𝟒𝑾
So that, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
110
𝐿
𝑇ℎ
𝑇ℎ
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.69
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage and voltage
drop across the 30𝑘𝛺 resistor directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltages gives,
𝑣0
𝑣0 −𝑣𝑇ℎ
𝑣0 −100
+ 40000
+ 22000
=0
10000
𝑣𝑇ℎ −𝑣0
𝑣
− 0.003𝑣0 + 𝑇ℎ = 0
22000
30000
Rearranging the equations gives,
1
1
1
1
𝑣0 (10 + 40 + 22) + 𝑣𝑇ℎ (− 22) = 10
𝑣0 (−
1
1
1
− 3) + 𝑣𝑇ℎ ( + ) = 0
22
22
30
15
[ 88
67
−
22
− 22
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
13
165
𝑣0
10] ⇒ [ 𝑣0 ] = [ 6.3030 ]
𝑣𝑇ℎ
−243.64
0
] [𝑣 ] = [
𝑇ℎ
This implies, 𝑣𝑇ℎ = −243.64𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 30𝑘𝛺 resistor becomes zero!
By KCL for node voltages gives,
𝑣0
𝑣0
𝑣0 −100
+ 40000
+ 22000
=0
10000
−𝑣0
− 0.003𝑣0 + ⅈ𝑠𝑐 = 0
22000
Rearranging the equations gives,
1
1
1
+ + ) + 0ⅈ𝑠𝑐 = 10
10
40
22
𝑣0 (
𝑣0 (−
1
− 3) + 1000ⅈ𝑠𝑐 = 0
22
CONT . . .
111
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
15
[ 88
67
−
22
0
10] ⇒ [𝑣0 ] = [ 58.667 ]
ⅈ𝑠𝑐
0.17867
0
𝑣0
] [ⅈ ] = [
𝑠𝑐
1000
This implies, ⅈ𝑠𝑐 = 0.17867𝐴 = 178.67𝑚𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = −1363.6𝛺 = −𝟏. 𝟑𝟔𝟑𝟔𝒌𝜴
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to 𝑅!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
) ⋅ 𝑅𝐿
Now, 𝑝 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑉𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 𝟎
] = 𝑉𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
However, we are facing a special situation right now!
If we let 𝑅𝐿 = −𝑅𝑇ℎ , the denominator will be zero!
So that,
𝑝𝑚𝑎𝑥 = (
2
2
𝑣𝑇ℎ
𝑣𝑇ℎ
−243.64 2
) ⋅ 𝑅𝐿 = 𝑙ⅈ𝑚 = (
) ⋅ 𝑅𝐿 = (
) ⋅ (1363.6𝛺) = ∞
𝑅𝑇 +𝑅𝐿 →0
0
𝑅𝑇ℎ + 𝑅𝐿
𝑅𝑇ℎ + 𝑅𝐿
So that the maximum power becomes infinity theoretically!
112
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
4.71
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage and voltage
drop across the 30𝑘𝛺 resistor directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage 𝑣𝑇ℎ gives,
𝑣𝑇ℎ + 120𝑣0
𝑣
+ 𝑇ℎ = 0
10000
40000
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 3𝑘𝛺 and 1𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 3𝑘𝛺 + 1𝑘𝛺 = 4𝑘𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
1
By substitution, 𝑣0 = 4 8𝑉 = 2𝑉
𝑒𝑞
Rearranging the equations gives,
1
1
𝑣𝑇ℎ (10 + 40) + 12𝑣0 = 0
0𝑣𝑇 + 𝑣0 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
1
[8
0
12] [𝑣𝑇ℎ ] = [0] ⇒ [𝑣𝑇ℎ ] = [−192 ]
𝑣0
𝑣0
2
2
1
This implies, 𝑣𝑇ℎ = −192𝑉
Let denote short-circuit current as ⅈ𝑠𝑐 !
This implies that, voltage drop across the 40𝑘𝛺 resistor becomes zero!
120𝑣
By KCL for rightmost node gives, 100000 + ⅈ𝑠𝑐 = 0
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
1
By substitution, 𝑣0 = 4 8𝑉 = 2𝑉
𝑒𝑞
Rearranging the equations gives,
CONT . . .
113
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
12
𝑣0 (1000) + ⅈ𝑠𝑐 = 0
𝑣0 + 0ⅈ𝑠𝑐 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
3
[250
1
2
1 ] [𝑣0 ] = [0] ⇒ [𝑣0 ] = [
]
ⅈ𝑠𝑐
ⅈ𝑠𝑐
−0.024
2
0
This implies, ⅈ𝑠𝑐 = −0.024𝐴 = −24𝑚𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 8000𝛺 = 8𝑘𝛺
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to load resistor!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
Now, 𝑝 = (𝑅 𝑇ℎ
) ⋅ 𝑅𝐿
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑉𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 0
] = 𝑉𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
𝑣
2
𝑣
2
𝑣2
So that, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ = 𝟏. 𝟏𝟓𝟐𝑾
+𝑅
4.73
𝑇ℎ
𝐿
𝑇ℎ
𝑇ℎ
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
CONT . . .
114
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Let redraw the circuit as open-circuit with a reference node!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 10𝛺 and 20𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝛺 + 20𝛺 = 30𝛺
Likewise, 25𝛺 and 5𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 25𝛺 + 5𝛺 = 30𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
It’s obvious that, voltage drop across the 30𝛺 resistors are same because they are both connected to
the same pair of terminals!( if there is any confusion you can verify it using Kirchhoff's Voltage Law!)
20
5
This implies, 𝑣20𝛺 = 𝑣𝑎 = 30 60𝑉 = 40𝑉 and 𝑣5𝛺 = 𝑣𝑏 = 30 60𝑉 = 10𝑉
It’s obvious that, 𝑣𝑇ℎ = 𝑣𝑎 − 𝑣𝑏 = 30𝑉
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
Let redraw the circuit as short-circuit with mesh currents!
Assume that direction of voltage drops signed positive!
By KVL for each mesh gives,
CONT . . .
115
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
−60 + 10(ⅈ1 − ⅈ2 ) + 20(ⅈ1 − ⅈ3 ) = 0
25ⅈ2 + 10(ⅈ2 − ⅈ1 ) = 0
5ⅈ3 + 20(ⅈ3 − ⅈ1 ) = 0
Rearranging the equations gives,
30ⅈ1 − 10ⅈ2 − 20ⅈ3 = 60
−10ⅈ1 + 35ⅈ2 + 0ⅈ3 = 0
−20ⅈ1 + 0ⅈ2 + 25ⅈ3 = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
30
[−10
−20
ⅈ1
−20 ⅈ1
60
5.3846
0 ] [ⅈ2 ] = [ 0 ] ⇒ [ⅈ2 ] = [ 1.5385]
ⅈ3
25 ⅈ3
0
4.3077
−10
35
0
This implies that, ⅈ1 = 5.3846𝐴, ⅈ2 = 1.5385𝐴 and ⅈ3 = 4.3077𝐴
It’s clear that, ⅈ𝑠 = ⅈ3 − ⅈ2 = 2.7692𝐴
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏𝟎. 𝟖𝟑𝟑𝜴
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to load resistor!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
Now, 𝑝 = (𝑅 𝑇ℎ
) ⋅ 𝑅𝐿
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
CONT . . .
116
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑉𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 𝟎
] = 𝑉𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
𝑣
2
𝑣
2
𝑣2
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ = 𝟐𝟎. 𝟕𝟕𝑾
So that, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
4.75
𝐿
𝑇ℎ
𝑇ℎ
The load resistor will be ignored in order to find the open-circuit voltage!
Because of we are trying to find Thevenin equivalent, 𝑣𝑇ℎ is equal to open-circuit voltage directly!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Assume that direction of voltage drops signed positive!
It’s clear that, −10 − 20𝐼 + 𝑣𝑇ℎ = 0 ⇒ 𝑣𝑇ℎ = 20𝐼 + 10
Now, let create a short-circuit for the terminals a and b!
Let denote short-circuit current as ⅈ𝑠𝑐 !
By KVL for closed path gives, −10 + 10ⅈ𝑠𝑐 − 20𝐼 = 0 ⇒ ⅈ𝑠𝑐 =
𝑣
As it known, 𝑅𝑇ℎ = ⅈ𝑇ℎ = 𝟏𝟎𝜴
20𝐼+10
10
𝑠𝑐
So that the Thevenin equivalent is given below!
Here 𝑅𝐿 corresponds to load resistor!
By the definition of power, 𝑝 =
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
= ⅆ𝑞 ⋅ ⅆ𝑡 = 𝑣 ⋅ ⅈ = ⅈ 2 ⋅ 𝑅
ⅆ𝑡
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections!
This implies that, 𝑅ⅇ𝑞 for 𝑅𝑇ℎ and 𝑅𝐿 becomes 𝑅ⅇ𝑞 = 𝑅𝑇ℎ + 𝑅𝐿
𝑣
2
) ⋅ 𝑅𝐿
Now, 𝑝 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
𝐿
In order to find the maximum power that transferred to load, we must differentiate the equation given
above!
CONT . . .
117
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
ⅆ𝑝
2
So that, ⅆ𝑅 = 𝑣𝑇ℎ
[
𝐿
(𝑅𝑇ℎ +𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑇ℎ +𝑅𝐿 )
𝑅 −𝑅
2
[(𝑅 𝑇ℎ+𝑅 𝐿)3 ] = 0
] = 𝑣𝑇ℎ
(𝑅𝑇ℎ +𝑅𝐿 )4
𝑇ℎ
2
This implies that, the maximum power is transferred to the load when the load resistance equals to the
Thevenin resistance!( 𝑅𝐿 = 𝑅𝑇ℎ !)
𝑣
2
𝑣
2
𝑣2
) ⋅ 𝑅𝐿 = (2𝑅𝑇ℎ ) ⋅ 𝑅𝑇ℎ = 4𝑅𝑇ℎ =
So that, 𝑝𝑚𝑎𝑥 = (𝑅 𝑇ℎ
+𝑅
𝑇ℎ
𝐿
𝑇ℎ
𝑇ℎ
(20𝐼+10)2
40
This implies, maximum power is transferred as 𝑰approaches to infinity!
So that, 𝑙ⅈ𝑚
𝐼→∞
(20𝐼+10)2
40
=∞
END!
Chapter 5-Practice Problems
5.2
Assume that the op amp is operating in its linear region!
Let redraw the op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 −𝑣
𝑛
0
𝑛
𝑠
+ 20000
+ ⅈ𝑛 = 0
By KCL for node voltage 𝑣𝑛 gives, 10000
Rearranging the equation gives, −2𝑣𝑠 − 𝑣0 = 0!
𝑣
So that the open loop gain becomes, 𝑣0 = −2
118
𝑠
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣 −𝑣
𝑣
𝑛
0
0
= − 20000
It’s clear that, ⅈ = 20000
When 𝑣𝑠 equals 2𝑉, 𝑣0 becomes −4𝑉!
𝑣
0
This implies, ⅈ = − 20000
= 2 × 10−4 𝐴 = 𝟎. 𝟐𝒎𝑨
5.3
Assume that the op amp is operating in its linear region!
Let denote, 30𝑚𝑉 as 𝑣𝑠 , 3𝑘𝛺 as 𝑅1 and 3𝑘𝛺 as 𝑅𝑓 !
Now let redraw the op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node voltage 𝑣𝑛 gives,
𝑣 −𝑣
𝑣𝑛 −𝑣𝑠
+ 𝑛𝑅 0 + ⅈ𝑛 = 0
𝑅1
𝑓
𝑅
Rearranging the equation gives, 𝑣0 = − 𝑅𝑓 𝑣𝑠 !
120
1
This implies, 𝑣0 = − 3 (30𝑚𝑉) = −120𝑚𝑉 = −1.2𝑉
Let denote the current passing through the feedback resistor as ⅈ𝑓 !
𝑣 −𝑣
𝑣
𝑛
0
0
It’s obvious that, ⅈ𝑓 = 120000
= − 120000
= 1 × 10−5 𝐴 = 𝟏𝟎𝝁𝑨
5.4
(a) Assume that the op amp is operating in its linear region!
Now let redraw the op amp circuit with some important voltages and currents!
CONT . . .
119
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
By KCL for node voltage 𝑣𝑛 gives, −ⅈ𝑠 + 𝑛𝑅 0 + ⅈ𝑛 = 𝟎
𝑣
𝑣
𝑣
Rearranging the equation gives, ⅈ𝑠 = − 𝑅0 ⇒ − ⅈ 0 = 𝑅 ⇒ ⅈ 0 = −𝑹
𝑠
𝑠
(b) Assume that the op amp is operating in its linear region!
Now let redraw the op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
120
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
By KCL for node voltage 𝑣𝑛 gives, −ⅈ𝑠 + 𝑛𝑅 1 + ⅈ𝑛 = 0
1
𝑣 −𝑣
𝑣 −𝑣
𝑣
By KCL for node 𝑣1 gives, 1𝑅 𝑛 + 1𝑅 0 + 𝑅1 = 0
1
3
2
Rearranging the equations gives,
𝑣
ⅈ𝑠 = − 𝑅1
𝑣0 = 𝑣1 (
1
1
1
+
+ ) ⋅ 𝑅3
𝑅1 𝑅2 𝑅3
𝒗
𝑹
𝑹
This implies, 𝒊 𝟎 = −𝑹𝟏 (𝑹𝟑 + 𝑹𝟑 + 𝟏)
5.5
𝒔
𝟏
𝟐
Assume that the op amp is operating in its linear region!
Now let redraw the op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 4𝛺 and 6𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 4𝑘𝛺 + 8𝑘𝛺 = 12𝑘𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
121
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
8
By substitution, 𝑣𝑝 = 12 3𝑉 = 2𝑉
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 −𝑣
𝑛
By KCL for node voltage 𝑣2𝑘𝛺 gives, 2000
+ ⅈ𝑛 + 𝑛 0 = 0
5000
As it known, 𝑣𝑛 = 𝑣𝑝 = 2𝑉
Rearranging the equations gives,
1
1
1
𝑣𝑛 (2 + 5) + 𝑣0 (− 5) = 0
𝑣𝑛 + 0𝑣0 = 2
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
[10
1
1
− 5] [𝑣𝑛] = [0] ⇒ [𝑣𝑛] = [2]
𝑣0
𝑣0
7
2
0
This implies, 𝑣0 = 𝟕𝑽
5.6
Assume that the op amp is operating in its linear region!
Now let redraw the op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
CONT . . .
122
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣 −1.5
𝑣 −2
𝑣 −1.2
𝑣 −𝑣
𝑛
𝑛
𝑛
0
𝑛
+ 10000
+ 6000
+ 8000
+ ⅈ𝑛 = 𝟎
By KCL for node 𝑣𝑛 gives, 20000
Rearranging the equation gives,
1.5
2
1.2
𝑣
− 20 − 10 − 6 − 80 = 0 ⇒ 𝑣0 = −𝟑. 𝟖𝑽
𝑣 −𝑣
𝑣
3𝑣
0
0
0
1
By KCL for node 𝑣0 gives, 8000
− ⅈ0 + 4000
⇒ ⅈ0 = 8000
= −1.425 × 10−3 𝐴 = −𝟏. 𝟒𝟏𝟓𝒎𝑨
5.7
Assume that the op amp is operating in its linear region!
Now draw a basic difference op amp circuit with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
𝑣 −𝑣
𝑣𝑛 −𝑣1
+ 𝑛𝑅 0 + ⅈ𝑛 = 0
𝑅1
2
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 𝑅3 and 𝑅4 are connected in series so 𝑅ⅇ𝑞 for them is 𝑅3 + 𝑅4 !
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑅
4
𝑣2
By substitution, 𝑣𝑝 = 𝑅 +𝑅
3
4
𝑒𝑞
𝑅 (1+𝑅 ∕𝑅 )
𝑅
Substituting the equations gives, 𝑣0 = 𝑅2(1+𝑅1∕𝑅2) 𝑣2 − 𝑅2 𝑣1
1
123
3
4
1
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the
property that 𝑣0 = 0 when 𝑣1 = 𝑣2 !
𝑅
𝑅
This implies, 𝑅1 = 𝑅3!
2
4
Now by using APENDIX resistors can be, 𝑅1 = 𝑅3 = 𝟏𝟎𝒌𝜴 and 𝑅2 = 𝑅4 = 𝟓𝟎𝒌𝜴
5.8
Let redraw the instrumentation amplifier with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
It’s clear that, 𝑣1 = 8.00𝑉 and 𝑣2 = 8.01𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 −𝑣
𝑜𝑢𝑡𝑝𝑢𝑡
𝑛
1
By KCL for node 𝑣𝑛 gives, 20000
+ 𝑛 40000
+ ⅈ𝑛 = 0
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
𝑣 −𝑣
𝑣
𝑝
2
𝑝
By KCL for node 𝑣𝑝 gives, 20000
+ 40000
+ ⅈ𝑝 = 0
By substitution, 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 = 2(𝑣2 − 𝑣1 ) = 0.02𝑉
𝑣
This implies that, ⅈ0 = 𝑜𝑢𝑡𝑝𝑢𝑡
= 2 × 10−5 𝐴 = 20µ𝐴
1000
5.9
Let redraw the amplifier with some important voltages and currents!
CONT . . .
124
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
It’s clear that, 𝑣1 = 8𝑉
This implies, 𝑣𝑝 = 𝑣𝑛 = 8𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
8
8−𝑣
By KCL for node 𝑣𝑛 gives, 4000 + 80000 + ⅈ𝑛 = 0
Rearranging the equation gives, 𝑣0 = 𝟐𝟒𝑽
𝑣
𝑛
It’s obvious that, ⅈ0 = 4000
= 2 × 10−3 𝐴 = 𝟐𝒎𝑨
5.10
Let redraw the amplifier with some important voltages and currents!
125
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1!
This implies, 𝑣𝑝1 = 𝑣𝑛1 = 𝑣1
Likewise, 𝑣𝑝2 − 𝑣𝑛2 = 0 ⇒ 𝑣𝑝2 = 𝑣𝑛2!
This implies, 𝑣𝑝2 = 𝑣𝑛2 = 0
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
−𝑣
𝑣
−𝑣
𝑛2
𝑎
𝑛2
2
+ 50000
=0
By KCL for node 𝑣𝑛2 gives, 10000
Rearranging the equation gives, 𝑣𝑎 = −5𝑣2 !
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑛
𝑎
𝑛
0
𝑛
𝑛1
By KCL for node 𝑣𝑛 gives, 20000
+ 30000
+ 60000
+ ⅈ𝑛 = 0
Rearranging the equation gives, 𝑣0 = −3𝑣𝑛1 − 2𝑣𝑎 = −3𝑣1 + 10𝑣2
By substitution, 𝑣0 = 𝟏𝟖𝑽
5.9
(a) Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
It’s clear that, 𝑣𝑝 = 6𝑉
This implies, 𝑣𝑝 = 𝑣𝑛 = 6𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
126
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑛
0
+ ⅈ𝑛 = 𝟎
By KCL for node 𝑣𝑛 gives, −0.002 + 2000
Rearranging the equation gives, 𝑣0 = −4 + 𝑣𝑛 = 𝟐𝑽
(b) Let redraw the op amp with some important voltages and currents!
It’s clear that, 𝑣𝑛 = 𝟓𝑽
This implies, 𝑣𝑝 = 𝑣𝑛 = 𝟓𝑽
It’s clear that, 𝑣𝑝 − 𝑣0 = 2 ⇒ 𝑣0 = 𝟑𝑽
5.13
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
127
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 10𝑘𝛺 and 90𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 10𝑘𝛺 + 90𝑘𝛺 = 100𝑘𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
90
By substitution, 𝑣90𝑘𝛺 = 𝑣𝑛 = 𝑣𝑝 = 100 1𝑉 = 0.9𝑉
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣
𝑝
𝑝
0
+ 100000
+ ⅈ𝑝 = 0
By KCL for node 𝑣𝑝 gives, 50000
This implies, 𝑣0 = 3𝑣𝑝 = 𝟐. 𝟕𝑽
𝑣 −𝑣
𝑣
0
𝑝
0
By KCL for node ⅈ0 gives, −ⅈ0 + 100000
+ 10000
=𝟎
11𝑣 −𝑣
0
𝑃
= 288 × 10−4 𝐴 = 𝟐𝟖𝟖µ𝑨
This implies, ⅈ0 = 100000
5.15
(a) Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
By KCL for node 𝑣𝑛 gives, −ⅈ𝑠 + 𝑛𝑅 1 + ⅈ𝑛 = 0
By KCL for node 𝑣1 gives,
128
1
𝑣
𝑣 −𝑣
𝑣1 −𝑣𝑛
+ 𝑅1 + 1𝑅 0 = 0
𝑅1
2
3
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Rearranging the equations gives,
𝑣
ⅈ𝑠 = − 𝑅1
1
𝑅
𝑅
𝑣0 = 𝑣1 (𝑅3 + 𝑅3 + 1)
1
𝒗
2
𝑹
𝑹
𝑹 𝑹
This implies, 𝒊 𝟎 = −𝑹𝟏 (𝑹𝟑 + 𝑹𝟑 + 𝟏) = −𝑹𝟐 − 𝑹𝟑 𝟏 − 𝑹𝟏
𝒔
𝑣
𝟏
𝟐
𝟐
(b) By substitution, ⅈ 0 = −𝟗𝟐𝒌𝜴
𝑠
5.17
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 −𝑣
𝑛
𝑖
(a) By KCL for node 𝑣𝑛 gives, 10000
+ 𝑛 0 + ⅈ𝑛 = 0
12000
Rearranging the equation gives, 5𝑣0 = −6𝑣ⅈ
𝑣
This implies, 𝑣0 = −1.2
𝑖
𝑣 −𝑣
𝑣 −𝑣
𝑛
𝑖
𝑛
0
(b) By KCL for node 𝑣𝑛 gives, 10000
+ 80000
+ ⅈ𝑛 = 0
Rearranging the equation gives, 𝑣0 = −8𝑣ⅈ
129
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
This implies, 𝑣0 = −8
𝑖
𝑣 −𝑣
𝑣 −𝑣
𝑛
𝑖
𝑛
0
(c) By KCL for node 𝑣𝑛 gives, 10000
+ 2000000
+ ⅈ𝑛 = 0
Rearranging the equation gives, 𝑣0 = −𝟐𝟎𝟎𝒗𝒊
𝑣
This implies, 𝑣0 = −𝟐𝟎𝟎
5.19
𝑖
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −1
𝑣
𝑣 −𝑣
1
1
𝑛
1
+ 4000
+ 4000
= 0 ⇒ 𝑣1 = 0.5𝑉
By KCL for node 𝑣1 gives, 2000
𝑣 −𝑣
𝑣 −𝑣
𝑛
1
By KCL for node 𝑣𝑛 gives, 4000
+ 𝑛 0 + ⅈ𝑛 = 0 ⇒ 𝑣0 = −1.25𝑉
10000
𝑣 −𝑣
𝑣
3𝑣
0
𝑛
0
− ⅈ0 + 0 = 0 ⇒ ⅈ0 =
= −3.75 × 10−4 𝐴 = −3.75µ𝐴
By KCL for node 𝑣0 gives, 10000
5000
10000
5.21
Let redraw the op amp with some important voltages and currents!
CONT . . .
130
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 1𝑉!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −3
𝑣 −𝑣
𝑛
0
𝑛
+ 10000
+ ⅈ𝑛 = 𝟎
By KCL for node 𝑣𝑛 gives, 4000
Rearranging the equation gives, 𝑣0 =
5.25
7𝑣𝑛 −15
= −𝟒𝑽
2
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 2𝑉!
CONT . . .
131
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 12𝑘𝛺 and 20𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 12𝑘𝛺 + 20𝑘𝛺 = 32𝑘𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
20
By substitution, 𝑣0 = 32 2𝑉 = 𝟏. 𝟐𝟓𝑽
5.27
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 16𝛺 and 24𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 16𝛺 + 24𝛺 = 40𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
24
By substitution, 𝑣1 = 40 5𝑉 = 3𝑉
𝑒𝑞
This implies, 𝑣1 = 𝑣𝑝 = 𝑣𝑛 = 𝑣2 = 3𝑉!
Likewise, 8𝛺 and 12𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 8𝛺 + 12𝛺 = 20𝛺
12
By substitution, 𝑣0 = 20 3𝑉 = 𝟏. 𝟖𝑽
132
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.29
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 𝑅1 and 𝑅2 are connected in series so 𝑅ⅇ𝑞 for them is 𝑅1 + 𝑅2 !
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
𝑅
2
𝑣ⅈ
By substitution, 𝑣𝑅2 = 𝑣𝑝 = 𝑣𝑛 = 𝑅 +𝑅
1
2
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 −𝑣
By KCL for node 𝑣𝑛 gives, 𝑅𝑛 + 𝑛𝑅 0 + ⅈ𝑛 = 0
1
2
𝑅
Rearranging the equations gives, 𝑣0 = 𝑣𝑛 (𝑅2 + 1)!
𝑅
𝑅
1
) ⋅ ( 2 + 1)
By substitution, 𝑣0 = 𝑣ⅈ ⋅ (𝑅 +𝑅
𝑅
1
2
𝒗
1
𝑹
This implies, the voltage gain is 𝒗𝟎 = 𝑹𝟐!
133
𝒊
𝟏
1
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.31
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 𝑣0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 −𝑣
𝑣 −𝑣
1
𝑝
1
1
0
+ 6000
+ 12000
=0
By KCL for node 𝑣1 gives, −0.004 + 3000
𝑣 −𝑣
𝑣
𝑝
1
𝑝
+ 6000
+ ⅈ𝑝 = 0
By KCL for node 𝑣𝑝 gives, 6000
Rearranging the equations gives,
1
1
1
1
1
𝑣1 (3 + 6 + 12) + 𝑣0 (− 6 − 12) = 4
1
1
1
𝑣1 (− 6) + 𝑣0 (6 + 6) = 0
As it known, 𝑨𝒙 = 𝒃 ⇒ 𝑨−1 𝑨𝒙 = 𝑨−1 𝒃 ⇒ 𝒙 = 𝑨−1 𝒃
7
12
[ 1
−6
1
− 4 𝑣1
𝑣1
8.7273
4
]
] [ ] = [ ] ⇒ [𝑣 ] = [
1 𝑣0
4.3636
0
0
3
This implies, 𝑣1 = 𝑣𝑝 = 𝑣𝑛 = 8.7273𝑉 and 𝑣0 = 𝑣𝑝 = 𝑣𝑛 = 4.3636𝑉
𝑣
𝑝
It’s clear that, ⅈ𝑥 = 6000
= 7.2727 × 10−3 𝐴 = 𝟕. 𝟐𝟕𝟐𝟕𝒎𝑨
134
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.33
Let redraw the op amp with some important voltages and currents!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑝
+ ⅈ𝑝 = 0
By KCL for node 𝑣𝑝 gives, −0.003 + 4000
This implies, 𝑣𝑝 = 𝑣𝑛 = 12𝑉!
𝑣
𝑣 −𝑣
𝑛
0
𝑛
+ 1000
+ ⅈ𝑛 = 0
By KCL for node 𝑣𝑛 gives, 2000
This implies, 𝑣0 = 1.5𝑣𝑛 = 1.5𝑣𝑝 = 18𝑉
𝑣 −𝑣
𝑛
0
It’s clear that, ⅈ𝑥 = 1000
= −6 × 10−3 𝐴 = −𝟔𝒎𝑨
By the definition of power, 𝑝 =
𝑣2
ⅆ𝑤(𝑞)
ⅆ𝑤(𝑞) ⅆ𝑞(𝑡)
𝑣2
=
⋅
=
𝑣
⋅
ⅈ
=
𝑅
ⅆ𝑡
ⅆ𝑞
ⅆ𝑡
0
This implies that, 𝑝𝑣0 = 3000
= 0.108𝑊 = 𝟏𝟎𝟖𝒎𝑾
135
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.35
Let draw a basic non-inverting op amp circuit with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 𝑣ⅈ !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣 −𝑣
By KCL for node 𝑣𝑛 gives, 𝑅𝑛 + 𝑛𝑅 0 + ⅈ𝑛 = 0
1
𝑅
𝑓
This implies that, 𝑣0 = (1 + 𝑅𝑓) 𝑣ⅈ
𝑅
𝑅
1
So that, 1 + 𝑅𝑓 = 10 ⇒ 𝑅𝑓 = 𝟗
1
1
By using APENDIX, if 𝑅𝑓 = 90𝑘𝛺 then, 𝑅1 = 𝟏𝟎𝒌𝜴!
5.37
Let redraw the op amp with some important voltages and currents!
136
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −2
𝑣 −1
𝑣 +3
𝑣 −𝑣
𝑛
𝑛
𝑛
0
𝑛
+ 20000
+ 30000
+ 30000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = −𝟑𝑽
By KCL for node 𝑣𝑛 gives, 10000
5.39
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −2
𝑣 −𝑣
𝑣 +1
𝑣 −𝑣
𝑛
2
𝑛
𝑛
0
𝑛
By KCL for node 𝑣𝑛 gives, 10000
+ 20000
+ 50000
+ 50000
+ ⅈ𝑛 = 𝟎 ⇒ 𝑣2 =
137
−2𝑣0 −18
⇒ 𝑣2 = 𝟑𝒗
5
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.41
Let draw the basic summing op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
𝑅
𝑅
𝑣𝑛 −𝑣1
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
+ 𝑛 2 + 𝑛 3 + 𝑛 4 + 𝑛 0 + ⅈ𝑛 = 0
𝑅1
𝑅2
𝑅3
𝑅4
𝑅𝑓
𝑅𝑓
𝑅𝑓
⇒ 𝑣0 = − (𝑅𝑓 𝑣1 + 𝑅𝑓 𝑣2 + 𝑅 𝑣3 + 𝑅 𝑣4 )
𝑅
1
𝑅
𝑅
2
𝑅
3
1
So that, 𝑅𝑓 = 𝑅𝑓 = 𝑅𝑓 = 𝑅𝑓 = 4
1
2
3
4
4
By using APENDIX, if 𝑅𝑓 = 10𝑘𝛺 then, 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 = 𝟒𝟎𝒌𝜴
5.43
Let draw the basic summing op amp with some important voltages and currents!
138
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
𝑅
𝑅
𝑣𝑛 −𝑣1
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
+ 𝑛 2 + 𝑛 3 + 𝑛 4 + 𝑛 0 + ⅈ𝑛 = 0
𝑅1
𝑅2
𝑅3
𝑅4
𝑅𝑓
𝑅𝑓
𝑅𝑓
⇒ 𝑣0 = − (𝑅𝑓 𝑣1 + 𝑅𝑓 𝑣2 + 𝑅 𝑣3 + 𝑅 𝑣4 )
1
2
3
4
𝑹
𝑹
𝑹
𝑹
𝟏
To create an averaging amplifier, the following equation must be satisfied 𝑹𝒇 = 𝑹𝒇 = 𝑹𝒇 = 𝑹𝒇 = 𝟒!
As it known, 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 = 𝟒𝒌𝜴 ⇒ 𝑅𝑓 = 𝟑𝒌𝜴
𝟏
𝟐
𝟑
𝟒
5.47
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 2𝑘𝛺 and 20𝑘𝛺 are connected in series so 𝑅ⅇ𝑞 for them is 2𝑘𝛺 + 20𝑘𝛺 = 22𝑘𝛺
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
CONT . . .
139
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
20
By substitution, 𝑣𝑝 = 𝑣𝑛 = 22 𝑣2
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 −𝑣
𝑛
1
𝑛
0
+ 30000
+ ⅈ𝑛 = 0
By KCL for node 𝑣𝑛 gives, 2000
20
This implies, 𝑣0 = 16𝑣𝑛 − 15𝑣1 = 16 22 𝑣2 − 15𝑣1 = 𝟏𝟒. 𝟎𝟗𝟏𝑽
5.49
Let draw the basic difference op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
𝑣 −𝑣
𝑣𝑛 −𝑣1
+ 𝑛𝑅 0 + ⅈ𝑛 = 0
𝑅1
2
Let denote equivalent resistance as 𝑅ⅇ𝑞 !
As it known, 𝑅ⅇ𝑞 = ∑∞
𝑛=1 𝑅𝑛 for the series connections of resistors!
It’s obvious that, 𝑅3 and 𝑅4 are connected in series so 𝑅ⅇ𝑞 for them is 𝑅3 + 𝑅4 !
𝑅
By Voltage Division equation, 𝑣𝑥 = 𝑅 𝑥 𝑣𝑠𝑜𝑢𝑟𝑐ⅇ
𝑒𝑞
140
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
𝑅
4
𝑣2
By substitution, 𝑣𝑝 = 𝑅 +𝑅
3
4
𝑅 (1+𝑅 ∕𝑅 )
𝑅
Substituting the equations gives, 𝑣0 = 𝑅2(1+𝑅1∕𝑅2) 𝑣2 − 𝑅2 𝑣1
1
3
4
1
Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the
property that 𝑣0 = 0 when 𝑣1 = 𝑣2 !
𝑹
𝑹
𝑹
This implies, 𝑹𝟏 = 𝑹𝟑 ⇒ 𝒗𝟎 = 𝑹𝟐 (𝒗𝟐 − 𝒗𝟏 )
𝟐
𝟒
𝟏
So that, if 𝑅1 = 𝑅3 = 10𝑘𝛺 then 𝑅2 = 𝑅4 = 𝟐𝟎𝒌𝜴!
5.53
(a) Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
By KCL for node 𝑣𝑝 gives,
𝑅
𝑣 −𝑣
𝑣𝑛 −𝑣1
+ 𝑛𝑅 0 + ⅈ𝑛 = 0
𝑅1
2
𝑣𝑝 −𝑣2
𝑅1
2
𝑣2 =
By setting 𝑣𝑝 = 𝑣𝑛 ⇒ 𝑅 +𝑅
1
2
It’s clear that, 𝑣ⅈ = 𝑣2 − 𝑣1 !
𝒗
𝑹
This implies, 𝒗𝟎 = 𝑹𝟐!
141
𝒊
𝟏
𝑣
+ 𝑝 + ⅈ𝑝 = 0
𝑅2
𝑅2 𝑣1 +𝑅1 𝑣0
𝑅1 +𝑅2
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
(b) Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
By KCL for node 𝑣𝑎 gives, 𝑅𝑎 ∕21 + 𝑎𝑅 𝑏 + 𝑅𝑎 ∕2𝑛 = 0
1
𝐺
1
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
By KCL for node 𝑣𝑏 gives, 𝑅𝑏 ∕22 + 𝑏𝑅 𝑎 + 𝑅𝑏 ∕2𝑝 = 0
1
𝐺
1
It’s clear that, 𝑣𝑝 = 𝑣𝑛 !
This implies,
𝑅
𝑣
𝑣2 −𝑣1
= (1 + 2𝑅1 ) (𝑣𝑏 − 𝑣𝑎 ) = 2𝑖!
2
6
𝑅
2
(𝑣𝑏 − 𝑣𝑎 )
Equation for the difference amplifier gives, 𝑣0 = 𝑅 ∕2
𝒗
𝑹
By using substitution, 𝒗𝟎 = 𝑹𝟐 ⋅
142
𝒊
𝟏
𝟏
𝟏+
𝟏
𝟐𝑹𝑮
1
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
(c) Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛 gives,
By KCL for node 𝑣𝑝 gives,
𝑣 −𝑣
𝑣𝑛 −𝑣1
+ 𝑅𝑛 ∕2𝑎 + ⅈ𝑛 = 0 (1)
𝑅1
2
𝑣𝑝 −𝑣2
𝑅1
𝑣 −𝑣
𝑣 −𝑣
+ 𝑅𝑝 ∕2𝑏 + ⅈ𝑝 = 0 (2)
2
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑣
By KCL for node 𝑣𝑎 gives, 𝑅𝑎 ∕2𝑛 + 𝑎𝑅 𝑏 + 𝑅𝐴 ∕20 = 0 (3)
2
𝑣 −𝑣
6
2
By KCL for node 𝑣𝑏 gives, 𝑅𝑏 ∕2𝑝 + 𝑏𝑅 𝑎 + 𝑅 𝑏∕2 = 0 (4)
It’s clear that,
2
𝐺
2
𝑅
𝑣
𝑣2 −𝑣1
= (1 + 2𝑅1 ) (𝑣𝑏 − 𝑣𝑎 ) = 2𝑖 (5)
2
6
𝑅
By subtracting the equation 4 from 3 gives, −2(𝑣𝑏 − 𝑣𝑎 ) ⋅ (1 + 2𝑅2 ) = 𝒗𝟎
𝑹
𝑹
𝒗
𝑹
𝑹
By substitution, 𝑹𝟐 𝒗𝒊 (𝟏 + 𝟐𝑹𝟐 ) = 𝒗𝟎 ⇒ 𝒗𝟎 = 𝑹𝟐 (𝟏 + 𝟐𝑹𝟐 )
𝟏
143
𝟔
𝒊
𝟏
𝑮
𝐺
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.57
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0!
Likewise, 𝑣𝑝2 − 𝑣𝑛2 = 0 ⇒ 𝑣𝑝2 = 𝑣𝑛2 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛1 gives,
𝑣𝑛1 −𝑣𝑠1
𝑣 −𝑣
+ 𝑛1 1 = 0 ⇒ 𝑣1 = −2𝑣𝑠1
25000
50000
𝑣
−𝑣
𝑣
−𝑣
𝑣
−𝑣
𝑛2
1
+ 𝑛2 𝑠2 + 𝑛2 𝑝 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = −(𝑣1 + 2𝑣𝑠2 )
By KCL for node 𝑣𝑛2 gives, 100000
100000
50000
𝑣
𝑣 −𝑣
𝑛
0
𝑛
+ 100000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = 3𝑣𝑛
By KCL for node 𝑣𝑛 gives, 50000
By substitution, 𝒗𝟎 = 𝟑𝒗𝒏 = −𝟑(𝒗𝟏 + 𝟐𝒗𝒔𝟐 ) = 𝟔𝒗𝒔𝟏 − 𝟔𝒗𝒔𝟐!
5.59
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 𝑣𝑠 !
144
CONT . . .
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Likewise, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛1 gives,
By KCL for node 𝒗𝒏 gives,
𝑣
𝑣 −𝑣
𝑣𝑛1
+ 𝑛13𝑅 1 = 0 ⇒ 𝑣1 = 4𝑣𝑛1
𝑅
𝒗 −𝒗
𝒗𝒏 −𝒗𝟏
+ 𝒏𝟒𝑹 𝟎 + 𝒊𝒏 = 𝟎 ⇒ 𝒗𝟎 = −𝟒𝒗𝟏 = −𝟏𝟔𝒗𝒏𝟏 = −𝟏𝟔𝒗𝒔
𝑹
This implies that, 𝑣0 = −𝟏𝟔
5.61
𝑠
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0!
Likewise, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
+0.8
𝑣
−𝑣
𝑛1
1
𝑛1
+ 20000
= 0 ⇒ 𝑣1 = 𝟏. 𝟔𝑽
By KCL for node 𝑣𝑛1 gives, 10000
𝑣 −0.4
𝑣 −𝑣
𝑣 −𝑣
𝑛
1
𝑛
0
𝑛
By KCL for node 𝑣𝑛 gives, 10000
+ 20000
+ 40000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = −𝟒. 𝟖𝑽
145
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.63
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0!
Likewise, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
By KCL for node 𝑣𝑛1 gives,
By KCL for node 𝑣𝑝 gives,
𝑣 −𝑣
𝑣 −𝑣
𝑅
𝑅
𝑣𝑛1 −𝑣𝑖
+ 𝑛1𝑅 1 + 𝑛1𝑅 0 = 0 ⇒ 𝑣1 = − 𝑅2 𝑣ⅈ − 𝑅2 𝑣0
𝑅1
2
3
1
3
𝑣𝑝 −𝑣1
𝑹 𝑹
𝑅5
+
𝑣𝑝 −𝑣𝑖
𝑅6
+
𝑹 𝑹
𝑣𝑝 −𝑣0
𝑅4
𝑹
𝑅
𝒗
By substitution, 𝒗𝟎 (𝟏 − 𝑹𝟐𝑹𝟒 ) = 𝒗𝒊 (𝑹𝟐𝑹𝟒 − 𝑹𝟒) ⇒ 𝒗𝟎 =
𝟑 𝟓
𝟏 𝟓
𝟔
𝑅
+ ⅈ𝑝 = 0 ⇒ 𝑣0 = − 𝑅4 𝑣1 − 𝑅4 𝑣ⅈ
𝒊
𝑹𝟐 𝑹𝟒 𝑹𝟒
−
𝑹𝟏 𝑹𝟓 𝑹𝟔
𝑹𝟐 𝑹𝟒
𝟏−
𝑹𝟑 𝑹𝟓
5
6
5.65
Let redraw the op amp with some important voltages and currents!
CONT . . .
146
FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 6𝑚𝑉!
Likewise, 𝑣𝑝2 − 𝑣𝑛2 = 0 ⇒ 𝑣𝑝2 = 𝑣𝑛2 = 0!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
−0.006
𝑣
−𝑣
𝑛2
1
By KCL for node 𝑣𝑛2 gives, 𝑛210000 + 30000
= 0 ⇒ 𝑣1 = −18𝑚𝑉
𝑣
𝑣 −𝑣
𝑛
0
𝑛
By KCL for node 𝑣𝑛 gives, 40000
+ 8000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = 1.2𝑣𝑛 = 1.2𝑣𝑝
As it known, ⅈ𝑛 = ⅈ𝑝 = 0!
This implies that, the voltage drop across the 20𝑘𝛺 resistor becomes zero!
So that, 𝑣𝑝 = 𝑣1 ⇒ 𝑣0 = 1.2𝑣𝑝 = 1.2𝑣1 = −𝟐𝟏. 𝟔𝒎𝑽
5.67
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0.4𝑉 !
Likewise, 𝑣𝑝2 − 𝑣𝑛2 = 0 ⇒ 𝑣𝑝2 = 𝑣𝑛2 = 0.2𝑉!
It’s clear that, 𝑣𝑝3 − 𝑣𝑛3 = 0 ⇒ 𝑣𝑝3 = 𝑣𝑛3 = 0!
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FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
−𝑣
𝑣
−𝑣
𝑛3
1
𝑛3
𝑛1
By KCL for node 𝑣𝑛3 gives, 20000
+ 80000
= 0 ⇒ 𝑣1 = −4𝑣𝑛1 = −𝟏. 𝟔𝑽
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑛
𝑛2
𝑛
0
𝑛
1
+ 20000
+ 80000
= 0 ⇒ 𝑣0 = −2𝑣1 − 4𝑣𝑛2 = 𝟐. 𝟒𝑽
By KCL for node 𝑣𝑛 gives, 40000
5.69
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0!
Likewise, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 !
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣𝑛1 −𝑣𝑝
0.030+𝑣𝑝
𝑣 −𝑣
𝑣𝑛1 −0.010
+ 15000 + 𝑛1𝑅 0 = 0 ⇒ 𝑣0 = −𝑅𝑓 ( 15000 )
5000
𝑓
By KCL for node 𝑣𝑛1 gives,
𝑣 −𝑣
𝑣
𝑛
0
𝑛
By KCL for node 𝑣𝑛 gives, 2000
+ 6000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = 4𝑣𝑛 = 4𝑣𝑝
0.30+10𝑣𝑝
While 𝑅𝑓 = 10𝑘𝛺, 𝑣0 = − (
15
0.30+10𝑣𝑝
By substitution, 4𝑣𝑝 = − (
15
)!
) ⇒ 𝑣𝑝 = −𝟒. 𝟐𝟖𝟓𝟕𝒎𝑽
This implies, 𝑣0 = 4𝑣𝑝 = −𝟏𝟕. 𝟏𝟒𝟑𝒎𝑽
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5.71
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 0!
Likewise, 𝑣𝑝2 − 𝑣𝑛2 = 0 ⇒ 𝑣𝑝2 = 𝑣𝑛2 = 3𝑉!
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
−2
𝑣
−𝑣
𝑛1
1
𝑛1
By KCL for node 𝑣𝑛1 gives, 5000
+ 20000
= 0 ⇒ 𝑣1 = −8𝑉
The voltage drop across the 10𝑘𝛺 resistor becomes zero!
This implies, 𝑣𝑝2 = 𝑣22 = 3𝑉!
3
3−𝑣
2
By KCL for node 𝑣𝑝2 gives, 30000 + 50000
= 0 ⇒ 𝑣2 = 𝟖𝑽
𝑣 −𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑛
2
𝑛
0
𝑛
1
By KCL for node 𝑣𝑛 gives, 40000
+ 80000
+ 100000
+ ⅈ𝑛 = 0 ⇒ 𝑣0 = −𝟐. 𝟓𝒗𝟏 − 𝟏. 𝟐𝟓𝒗𝟐 = 𝟏𝟎𝑽
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FUNDAMENTALS of ELECTRIC CIRCUITS-SOLUTION MANUAL
5.73
Let redraw the op amp with some important voltages and currents!
Assume that the op amp is operating in its linear region!
By the definition of ideal op amp, ⅈ𝑛 = ⅈ𝑝 = 0!
It’s obvious that, 𝑣𝑝1 − 𝑣𝑛1 = 0 ⇒ 𝑣𝑝1 = 𝑣𝑛1 = 3𝑉 !
It’s obvious that, 𝑣𝑝 − 𝑣𝑛 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 𝑣0!
𝑣
By Ohm’s Law, 𝑣 = ⅈ ⋅ 𝑅 ⇒ ⅈ = 𝑅
As it known, current flows from a higher potential to lower!
𝑣
−𝑣
This implies that, ⅈ = ℎ𝑖𝑔ℎ𝑒𝑟𝑅 𝑙𝑜𝑤𝑒𝑟
Let denote currents leaving a node positive!
𝑣
𝑣
−𝑣
𝑛1
𝑝
𝑛1
+ 50000 = 0 ⇒ 𝑣𝑝 = 𝑣𝑛 = 𝑣0 = 6𝑣𝑛1 = 𝟏𝟖𝑽
By KCL for node 𝑣𝑛1 gives, 10000
END!
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