Basic Laws (a)
1
Ohm’s Law
• Resistance (R)
– The ability of an element to resist the flow of electric
current, in ohms ().

R
A

(2.1)
• Resistivity ()
– A general property of
materials: the ability to
resist current measured
in ohm-meters (-m).

A
2
Resistivity
3
Ohm’s Law
• Resistor
– The circuit element used to model the current-resisting
behavior of a material.
• Ohm’s law
– The voltage v across a resistor is directly proportional to
the current i flowing through the resistor.
vi
v  iR
(2.2)
(2.3)
v
R
(2.4)
i
1  = 1 V/A
4
Ohm’s Law
• Short circuit
– A circuit element with resistance approaching zero.
R0
v  iR  0
(2.5)
• Open circuit
– A circuit element with resistance approaching infinity.
R
v
i  lim  0
(2.6)
R  R
5
Ohm’s Law
6
Resisters
7
Resisters
8
Resisters
9
Resisters
10
Resistors
11
Conductance
• Conductance (G)
– The ability of an element to conduct electric current, in
mhos ( ) or siemens (S).
1 i
G 
R v
(2.7)
1 S  1   1 A/V
i  Gv
(2.8)
(2.9)
– For a resistor,
10   0.1 S
12
Power Dissipated
• Using Eqs. (1.7) and (2.3):
2
v
p  vi  i 2 R 
R
2
i
p  vi  v 2G 
G
(2.10)
(2.11)
13
Example 2.1
14
Practice Problem 2.1
• The essential component of a toaster is an electrical element (a
resistor) that converts electrical energy to heat energy. How
much current is drawn by a toaster with resistance 10 at 110
V?
15
Example 2.2
16
Example 2.2 (cont.)
17
Practice Problem 2.2
• For the circuit shown in Fig. 2.9, calculate the voltage v, the
conductance G, and the power p.
18
Example 2.3
19
Practice Problem 2.3
• A resistor absorbs an instantaneous power of 20cos2t mW
when connected to a voltage source v = 10 cost V. Find i and R.
20
Nodes, Branches, and Loops
• Branch
– A branch represents a single element such as a voltage
source or a resistor…
• Node
– A node is the point of connection between two or more
branches
• Loop
– A loop is any closed path in a circuit
21
Nodes, Branches, and Loops
22
Nodes, Branches, and Loops
• The fundamental theorem of network topology
– A network with b branches, n nodes, and l independent
loops will satisfy:
b  l  n 1
(2.12)
• Series
– 2 or more elements exclusively share a single node and
consequently carry the same current.
• Parallel
– 2 or more elements are connected to the same 2 nodes and
consequently have the same voltage across them.
23
Example 2.4
24
Example 2.4 (cont.)
25
Practice Problem 2.4
• How many branches and nodes does the circuit in Fig. 2.14
have? Identify the elements that are in series and in parallel.
26
Kirchhoff’s Laws
• Kirchhoff’s current law (KCL)
– The algebraic sum of currents entering a node (or a closed
boundary) is zero.
N
i
n 1
n
0
(2.13)
Law of conservation of
electric charge
i1  (i2 )  i3  i4  (i5 )  0
(2.16)
27
Kirchhoff’s Laws
i1  i3  i4  i2  i5
(2.17)
• KCL
– The sum of the currents
entering a node is equal
to the sum of the
currents leaving the
node.
28
Kirchhoff’s Laws
(2.18)
• A current can not contain 2 different currents, I1 and I2, in
series, unless I1 = I2; otherwise KCL will be violated.
29
Kirchhoff’s Laws
• Kirchhoff’s voltage law (KVL)
– The algebraic sum of all voltages around a closed path (or
loop) is zero.
M
v
m 1
m
0
(2.19)
v1  v2  v3  v4  v5  0
(2.20)
30
Kirchhoff’s Laws
v2  v3  v5  v1  v4
(2.21)
• KVL: Sum of voltage drops = Sum of voltage rises (2.22)
Vab  V1  V2  V3
(2.23)
31
Example 2.5
32
Example 2.5 (cont.)
33
Practice Problem 2.5
• Find v1 and v2 in the circuit of Fig. 2.22.
34
Example 2.6
35
Example 2.6 (cont.)
36
Practice Problem 2.6
• Find vx and vo in the circuit of Fig. 2.24.
37
Example 2.7
38
Example 2.7 (cont.)
39
Practice Problem 2.7
• Find vo and io in the circuit of Fig. 2.26.
40
Example 2.8
41
Example 2.8 (cont.)
42
Example 2.8 (cont.)
43
Example 2.8 (cont.)
44
Practice Problem 2.8
• Find the currents and voltages in the circuit shown in Fig. 2.28.
45
Series Resistors and Voltage Division
• Applying Ohm’s law
v1  iR1 , v2  iR2
(2.24)
• Applying KVL
v  v1  v2  0
(2.25)
v
i
( R1  R2 )
(2.27)
v  v1  v2  i ( R1  R2 )
(2.26)
46
Series Resistors and Voltage Division
v  iReq
(2.28)
Req  R1  R2
• Equivalent resistance of series resistors
– Sum of individual resistances
– For N resistors in series,
(2.29)
N
Req  R1  R2    RN   Rn
(2.30)
n 1
• Principle of voltage division
R1
R2
v1 
v, v2 
v
(2.31)
R1  R2
R1  R2
– For N resistors in series, the nth resistor have a voltage drop:
Rn
vn 
v
(2.32)
R1  R2    RN
47
Parallel Resistors and Current Division
• From Ohm’s law:
v  i1 R1  i2 R2
• Applying KCL: i  i1  i2
v
v
i1  , i2 
R1
R2
(2.33)
(2.34)
48
Parallel Resistors and Current Division
1 1  v
v
v
 v   
i 
R1 R2
 R1 R2  Req
1
1 1
 
Req R1 R2
(2.36)
(2.35)
R1 R2
Req 
R1  R2
(2.37)
• Equivalent resistance of 2 parallel resistors
– Product of their resistance divided by their sum
• Equivalent resistance of N parallel resistors
1
1 1
1
   
(2.38)
Req R1 R2
RN
R
(2.39)
– For N equal resistors in parallel: Req 
N
49
Parallel Resistors and Current Division
• Equivalent conductance of N parallel resistors
– Sum of individual conductances
Geq  G1  G2  G3    GN
(2.40)
• Principle of current division
iR1 R2
v  iReq 
(2.42)
R1  R2
R2i
R1i
i1 
, i2 
(2.43)
R1  R2
R1  R2
G1i
G2i
i1 
, i2 
(2.44)
G1  G2
G1  G2
Gn
in 
i
(2.45)
G1  G2    GN
50
Parallel Resistors and Current Division
• 2 extreme cases:
51