Fundamentals of Electric Circuit - Chapter 2: Basic laws

tion.
II. Ohm’s law.
III. Nodes, branches and loops.
IV. Kirchhoff’s laws.
V. Series resistors and voltage division.
VI. Parallel resistors and current division.
VII. Wye – Delta transformations
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
I. Introduction
 In order to determine the values of current, voltage, and power in an electric
circuit, we should understand some fundamental laws.
 This chapter presents:
 Ohm’s law, Kirchhoff’s laws
 Some techniques commonly applied in circuit design and analysis:
 Combining resistors in series and parallel
 Voltage division
 Current division
 Delta – Wye and Wye – Delta transformations
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Chapter 2: Basic laws
II. Ohm’s law
 In general, material have a characteristic behavior of resisting the flow of
electric charge.
 Resistance (R) is known as the ability to resist current.
l
R
A

ρ: resistivity of the material [Ωm]
l : length of material [m]
A: cross sectional area [m2]
R+
-
v
i
 Ohm’s law: The voltage v across a resistor is directly proportional
to the current i flowing through the resistor.
v Ri
 The resistance R of an element denotes its ability to resist
the flow of electric current; it is measured in Ohms [Ω]
v V
R
i A
1 1   
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Chapter 2: Basic laws
II. Ohm’s law
 There are two extreme possible values of R
 Short circuit: is a circuit element with resistance approaching zero (current
could be anything).
R v iR0 0   
 Open circuit: is a circuit element with resistance approaching infinity
(voltage could be anything).
R
v
i
R
lim 0

 
v = 0 R = 0
+
-
i
v R = ∞
+
-
i = 0
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Chapter 2: Basic laws
II. Ohm’s law
 Resister classification:
 Fixed resistors:
 Variable resistors:
Wire wound 
(small resistance
Composition (large 
resistance)
R
R
Symbol for 
fixed resistor
 Linear resistor:
 Nonlinear resistor: Do not consider
v
i
Slope = R
v
i
Slope = R
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
II. Ohm’s law
 Conductance is the ability of an element to conduct electric current, it is
measured in Siemens [S]
 a resistor always absorbs power from the circuit (passive element)
i A
G 1S=1
R v V
1
 
 Power dissipated by a resistor (conductance):
v
p vi i R
R
2
2  
i
p vi v G
G
2
2  
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Chapter 2: Basic laws
III. Nodes, branches, and loops
 Since the elements of an electric circuit can be interconnected in several ways
 we need to understand some basic concepts of network topology.
 We regard a network as an interconnection of elements or devices, whereas a
circuit is a network providing one or more closed paths
 We regard a network as an interconnection of elements or devices, whereas a
circuit is a network providing one or more closed paths, and we study the
properties relating to the placement of element in the network and the
geometric configuration of the network.
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Chapter 2: Basic laws
III. Nodes, branches, and loops
 A branch (b) represents a single element
such as a voltage source or a resistor
(represent any two terminal element).
5 V
R 1
R 2 R 3
1A
2 Ω3 Ω
10 Ω
c
b
a
 A node (n) is the point of connection between two or more branches.
Ex: The circuit has three nodes: a, b, and c.
 A loop is any closed path in a circuit, formed by starting at a node, passing
through a set of nodes, and returning to the starting node without passing
through any node more than once.
Ex: abca is a loop, containing the R1, R2 and voltage source.
Ex: The circuit has five branches: the 5-V voltage source, the 1-A current source,
and three resistors.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
III. Nodes, branches, and loops
 A loop is said to be independent if it contains
a branch which is not in any other loop.
5 V
R 1
R 2 R 3
1A
2 Ω3 Ω
10 Ω
c
b
a
 A network with b branches, n nodes, and l independent loops, has an equation:
 Two or more elements are in series if they are cascaded or connected
sequentially and consequently carry the same current.
Ex: The circuit has totally six loops, but only
three of them are independent.
b l n 1  
 Two or more elements are in parallel if they are connected to the same two
nodes and sequentially have the same voltage across them.
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Chapter 2: Basic laws
IV. Kirchhoff’s laws
 Kirchhoff’s laws include: Kirchhoff’s current law (KCL) and Kirchhoff’s voltage
law (KVL).
 Kirchhoff’s laws, coupled with Ohm’s law, make a sufficient and powerful set of
tools for analyzing a large variety of electric circuits.
 Kirchhoff’s current law (KCL) states that the algebraic sum of currents
entering a node (or a closed boundary) is zero.
N
n
n
i
1
0


Convention:
 Current entering a node may be regarded as positive.
 Current leaving the node may be taken as negative.
i1
i2
i3
i4i5
i i i i i
1 2 3 4 5
0    
e
I I I I
1 2 3
   
a
b
I3I2I1
Ie
a
b
Ie
Ie
Equivalent
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Chapter 2: Basic laws
IV. Kirchhoff’s laws
 Kirchhoff’s voltage law (KVL) states that the algebraic sum of all
voltages around a closed path (or loop) is zero.
M
m
m
v
1
0


v v v v v
v v v v v
1 2 3 4 5
2 3 5 1 4
0     
   
-
v3v2
v1
+v5
v4
- +
-
+
Ex1: Write KVL for this circuit.
 Start with any branch and go around the loop
either clockwise or counterclockwise
Ex2: When voltage sources are connected in series,
KVL can be applied to obtain the total voltage.
ab
V V V V
1 2 3
  
-
-
a bVab
+
V3V2V1
Vab ba
+
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
IV. Kirchhoff’s laws
v i v i
1 2
4 ; 2  
Ex3: Find v1 and v2 in the circuit.
 Assume that current I flows through the loop as
indicating in the Figure.
v v
1 2
10 8 18   
-v1
4Ω
+
10 V 8 V
2Ω
-v2+
 From Ohm’s law:
 Applying KVL around the loop gives:
i i A6 18 3  
v i V
1
4 12 
 Substituting i in Ohm’s law to KVL:
 Finally we have:
v i V
2
2 6   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
IV. Kirchhoff’s laws
v i v i v i
1 1 2 2 3 3
2 ; 8 ; 4  
Ex4: Find the currents and voltages in the circuit.
i i i
1 2 3
0  
 From Ohm’s law:
 At node a, applying KCL gives:
v v i i
v v i i
1 2 1 2
2 3 2 3
5 2 8 5
3 8 4 3
    
 
      
 Applying KVL to loop 1 and loop 2:
 Finally we have:
-
v1
2Ω
+
8Ω
5 V
3 V
4Ω
-+-+
v2
v3i2
i3i1 a
Loop 1
Loop 2
i i i
i i
i i
1 2 3
1 2
2 3
0
2 8 5
8 4 3
  

 
  
i A
i A
i A
1
2
3
1,5
0,25
1,25


 
 
v V
i V
i V
1
2
3
3
2
5


 
 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
V. Series resistors and voltage division
 The equivalent resistance of any number of resistors connected in series is the
sum of the individual resistances.
N
eq N n
n
R R R R R
1 2
1
...

    
 Voltage divider: The voltage v is divided among the resistors in direct
proportion to their resistances, the larger the resistance, the larger the voltage
drop.
n
n
N
R
v v
R R R
1 2
...

  
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
VI. Parallel resistors and current division
 The equivalent resistance of two parallel resistors is equal to the product of
their resistances divided by their sum
eq
R R
R
R R
1 2
1 2

 eq NR R R R1 2
1 1 1 1
...  
 The equivalent conductance of resistors connected in parallel is the sum of
their individual conductances.
N
eq N n
n
G G G G G
1 2
1
...

    
 Current divider:
n
n
N
G
i i
G G G
1 2
...

  
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
VII. Wye – Delta transformations
 How do we combine resistors when they are neither
in series nor in parallel ?
R 1
R 2 R 3
v
R 5 R 6
R 4
Bridge circuit
4
3
2
1
R c
R b R a
Delta (Δ) or Π network
R 1 R 2
R 3
4
3
2
1
Wye (Y) or T network
 Delta to Wye conversion:
b c c a a b
a b C a b C a b C
R R R R R R
R R R
R R R R R R R R R
1 2 3
; ;  
     
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
VII. Wye – Delta transformations
4
3
2
1
R c
R b R a
Delta (Δ) or Π network
R 1 R 2
R 3
4
3
2
1
Wye (Y) or T network
 Wye to Delta conversion:
a b
c
R R R R
R R R R R R
R R
R R
 R R R
R
2 3 1 3
2 3 1 3
1 2
1 2
1 2
3
;     
  
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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Chapter 2: Basic laws
VII. Wye – Delta transformations
Ex: For the bridge circuit, find Req and i
R 2
R 6
R 3
R 5R 4
R 1
1 0 0 V
50Ω
20Ω
30Ω
24Ω
13Ω
10Ω
i
b
a
 Applying the Y to Δ transformation:
 In this circuit, there are two Y networks: (R2,
R4, R6) and (R3, R5, R6)  transforming just
one of them will simplify the circuit
a
R R
R R R =85
R
3 5
3 5
6
   
c
R R
R R R
R
3 6
3 6
5
34    
b
R R
R R R =170
R
5 6
5 6
3
   
R 2
R a
R c
R bR 4
R 1
 Combining all resistors, we obtain:
    eq C b a
eq
R R R R R R R
R
1 2 4
/ / / / / /
40
    
 
ab
eq
u
i A
R
100
2,5
40
   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
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v
v
R
R
i
i
v
v
+
-
v
+
-
v
i
i