ELECTRICAL TECHNOLOGY
EET 103/4
 Explain and analyze series and
parallel circuits
 Explain, derive and analyze Ohm’s
Law, Kirchhoff Current Law,
Kirchhoff Voltage Law, Source
Transformation, Thevenin theorem.
1
SERIES DC CIRCUIT
(CHAPTER 5)
2
5.1 - Introduction
 Two types of current are readily available,
direct current (dc) and sinusoidal alternating
current (ac)
 We will first consider direct current (dc)
3
5.1 - Introduction
 If a wire is an ideal conductor, the
potential difference (V) across the
resistor will equal the applied voltage of
the battery.
 V (volts) = E (volts)
 Current is limited only by the resistor
(R). The higher the resistance, the less
the current.
4
5.1 - Introduction
Defining the direction of conventional flow
for single-source dc circuits.
5
5.1 - Introduction
Defining the polarity resulting from a
conventional current I through a resistive
element.
6
5.2 – Series Resistors
RT  R1  R2  R3  R4  ...  RN
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5.2 – Series Resistors
The total resistance of a series
configuration is the sum of the
resistance levels.
RT  R1  R2  R3  R4  ...  RN
 The more resistors we add in series,
the greater the resistance (no matter
what their value).
8
5.2 – Series Resistors
 When series resistors have the same value,
RT  NR
Where N = the number of resistors in the
string.
 The total series resistance is not affected by
the order in which the components are
connected.
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5.2 – Series Resistors
Example 5.1
Determine the total resistance of the series
connection in the following figure:
10
5.2 – Series Resistors
Solution:
RT  R1  R2  R3  R4
 20   220  1.2 k  5.6 k
 7040   7.04 k
11
5.2 – Series Resistors
Example 5.2
Find the total
resistance of the
series resistors in
the following figure:
12
5.2 – Series Resistors
Solution:
RT  NR  43.3 k  13.2 k
13
5.2 – Series Resistors
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5.3 – Series Resistors
Schematic representation for a dc series
circuit:
Only one current flows through all the resistors
15
5.3 – Series Resistors
 Total resistance (RT) is all the source “sees.”
 Once RT is known, the current drawn from
the source can be determined using Ohm’s
law:
E
Is 
RT
 Since E is fixed, the magnitude of the source
current will be totally dependent on the
magnitude of RT .
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5.3 – Series Resistors
8.4 V
Is 
 0.06 A  60 mA
140 
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5.3 – Series Resistors
 The polarity of the voltage across a resistor
is determined by the direction of the current.
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5.3 – Series Resistors
 The magnitude of the voltage drop across
each resistor can then be found by applying
Ohm’s law using only resistance of each
resistor;
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5.3 – Series Resistors
V1  I s R1;
E
Is 
RT
V2  I s R2 ;
V3  I s R3 ;
RT  R1  R2  R3
20
5.3 – Series Resistors
Example 5.4
(a) Find RT
(b) Calculate Is
(c) Determine the voltage across each resistor
21
5.3 – Series Resistors
Solution
(a)
RT  R1  R2  R3
 2 1 5
8
(b)
E 20 V
Is 

 2.5 A
RT
8
22
5.3 – Series Resistors
Solution (cont’d)
(c)
V1  I s R1
 2.5 A  2 
5V
V2  I s R2
 2.5 A 1 
 2.5 V
V3  I s R3
 2.5 A  5 
 12.5 V
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5.3 – Series Resistors
Example 5.6
Calculate R1 and E in the following circuit;
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5.3 – Series Resistors
Solution
RT  R1  R2  R3
R1  RT  R2  R3 
 12  4  6  2 k
E  I 3 RT  6 mA 12 k  72 V
25
5.3 – Series Resistors
Voltage measurement
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5.3 – Series Resistors
Current measurement
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5.4 – Power distribution in
Series Circuits
The power applied by the dc supply must
equal that dissipated by the resistive
elements.
PE  PR1  PR2  ...  PRN
I RT  I R1  I R2  ...  I RN
2
2
2
2
2
2
2
2
VN
E
V1 V2


 ... 
RT
R1 R2
RN
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5.4 – Power distribution in
Series Circuits
Example 5.7
Determine;
(a) RT
(b) Is
(c) voltage across
each resistor
(d) PE
(e) power dissipated
by each resistor
29
5.4 – Power distribution in
Series Circuits
Solution
(a)
RT  R1  R2  R3
 1  3  2  6 k
(b)
E 36 V
Is 

 6 mA
RT 6 k
30
5.4 – Power distribution in
Series Circuits
Solution (cont’d)
(c)
V1  I s R1
 6 mA 1 k
6V
V2  I s R2
V3  I s R3
 6 mA  3 k
 6 mA  2 k
 18 V
 12 V
31
5.4 – Power distribution in
Series Circuits
Solution (cont’d)
(d)
PE  EI s
 36 V  6 mA
 216 mW
32
5.4 – Power distribution in
Series Circuits
Solution (cont’d)
(e)
2
V1
P1 
R1
62

 36 mW
1 k
Or;

P1  I s R1  6 10
2
 10
3 2
3
 0.036 W
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5.4 – Power distribution in
Series Circuits
Solution
(e) – cont’d
2
V2
P2 
R2
182

 108 mW
3 k
Or;

P2  I s R2  6 10
2
  3 10
3 2
3
 0.108 W
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5.4 – Power distribution in
Series Circuits
Solution
(e) – cont’d
2
V3
P3 
R3
12 2

 72 mW
2 k
Or;

P3  I s R3  6 10
2
  2 10
3 2
3
 0.072 W
35
5.5 – Voltage Source in Series
36
5.5 – Voltage Source in Series
37
5.5 – Voltage Source in Series
38
5.5 – Kirchhoff’s Voltage Law
• Kirchhoff’s voltage law (KVL) states that
the algebraic sum of the potential rises
and drops around a closed loop (or path)
is zero.
39
5.5 – Kirchhoff’s Voltage Law
 The applied voltage of a series circuit equals
the sum of the voltage drops across the series
elements:
V
rises
 Vdrops
The sum of the rises around a closed loop must
equal the sum of the drops.
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5.5 – Kirchhoff’s Voltage Law
The application of Kirchhoff’s voltage law need
not follow a path that includes current-carrying
elements.
 When applying Kirchhoff’s voltage law, be sure to
concentrate on the polarities of the voltage rise or
drop rather than on the type of element.
 Do not treat a voltage drop across a resistive
element differently from a voltage drop across a
source.
41
Gustav Robert Kirchhoff
(1824 – 27)
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5.5 – Kirchhoff’s Voltage Law
Example 5.8
Use Kirchhoff’s voltage law to determine the
unknown voltage in the following figure
43
5.5 – Kirchhoff’s Voltage Law
Example 5.8 – solution
 E1  V1  V2  E2  0
V1  E1  V2  E2
 16  4.2  9
 2.8 V
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5.5 – Kirchhoff’s Voltage Law
Example 5.10
Determine V1 and V2:
Solution
Start at a and back to
a through loop 1;
 25  V1  15  0
V1  40 V
45
5.5 – Kirchhoff’s Voltage Law
Solution (cont’d)
Start at a and back to
a through loop 2;
V2  20  0
V2  20 V
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5.5 – Kirchhoff’s Voltage Law
Example 5.13
(a) Determine V2.
(b) Determine I2.
(c) Find R1 and R2.
47
5.5 – Kirchhoff’s Voltage Law
Example 5.13 – solution
(a)
 V1  V2  V3  E  0
18  V2 15  54  0
V2  21 V
(b)
I2 
V2 21

3A
R2 7
(Ohm’s law)
48
5.5 – Kirchhoff’s Voltage Law
Example 5.13 – solution (cont’d)
(c) (Ohm’s law)
V1 18
R1  
6
I2 3
V3 15
R3 

5
I2 3
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5.9 – Notation
Ground symbol
Voltage source
symbol
50
5.9 – Notation
Replacing the notation for a negative dc
supply with the standard notation.
51
5.9 – Notation
The expected voltage level at a particular point in a
network if the system is functioning properly.
52
5.9 – Notation
Double-subscript notation
 Because voltage is an “across” variable and exists
between two points, the double-subscript notation
defines differences in potential.
 The double-subscript notation Vab specifies point a
as the higher potential. If this is not the case, a
negative sign must be associated with the
magnitude of Vab .
 The voltage Vab is the voltage at point (a) with
respect to point (b).
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5.9 – Notation
Defining the sign for double-subscript notation.
Vab  Va  Vb
54
5.9 – Notation
Example 5.21
Find Vab:
Solution
Vab  Va  Vb  16  20  4 V
Vba  Vb  Va  20  16  4 V
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5.9 – Notation
Example 5.22
Find Va:
Solution
Vab  Va  Vb
Va  Vab  Vb  5  4  9 V
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5.9 – Notation
Example 5.23
Find Vab:
Solution
Vab  Va  Vb
  20  15  35 V
57
5.9 – Notation
Example 5.25
Determine Vab, Vcb
and Vc:
58
5.9 – Notation
Example 5.25 – solution
Va  E2  35 V
Vc  E1  19 V
Vac  Va  Vc  35   19  54 V
Vac
54 V
I

 1.2 A
R1  R2 20  25 
Vab  VR 2  IR2
 1.2 A  25 
 30 V
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5.9 – Notation
Example 5.25 – solution (cont’d)
Vab  Va  Vb
Vb  Va  Vab
 35  30  5 V
Vcb  Vc  Vb
 19  5  24 V
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