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ELEC1103 44
ELEC1103 – Semester 1 Chapter 2. Introduction
Dr. Craig Jin
12 March 2012
ELEC1103 45
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Five Ideal Basic Circuit Elements + Ideal Wire: - Ideal Wire
- Ideal Resistor
- Ideal Current Source
- Ideal Voltage Source
- Ideal Capacitor
- Ideal Inductor
› We will first study resistive networks, leaving capacitors and inductors for the second half of the semester because their operation is governed by differential equations.
We will study these first.
ELEC1103 46
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Ideal Wire: - In an ideal wire, electrons can move freely without any change in energy. In a
real wire there will be some resistance, but this resistance is usually negligible when designing or analysing the circuit.
- Voltage difference is a measure of energy difference between two points in a circuit. Because there can be no change in energy along an ideal wire, the entire wire must be at the same voltage.
ELEC1103 47
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Ideal Independent Voltage Source: - An ideal independent voltage source maintains a specified voltage across its
terminals regardless of the current flowing through it and other elements connected to it.
- The current flowing through an ideal independent voltage source is determined by the rest of the circuit.
dc voltage source ac voltage source
12 V +_ 5 cos(2πt) +_
Batteries can be modeled as a dc voltage source
ELEC1103 48
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Ideal Independent Current Source: - An ideal independent current source maintains a specified current flow through
its terminals regardless of the voltage across it and other elements connected to it.
- The voltage across an ideal independent current source is determined by the rest of the circuit.
dc current source ac current source
2 A 3 sin(100πt)
Keithley current source
ELEC1103 49
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Ideal Dependent Sources or Controlled Sources: - Drawn with a diamond shape around it.
- Come in four varieties, discussed in following slides.
Independent Sources
Dependent Sources
ELEC1103 50
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
vx
+
_2 vx
+_
› Voltage Controlled Voltage Source (VCVS) › A VCVS maintains a voltage across its terminals that is proportional to the
voltage across a pair of terminals elsewhere in the circuit.
› In general the voltage across a VCVS is where is a gain with units (V/V) so it is really unitless.
xvµ µ
VCVS
ELEC1103 51
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Current Controlled Voltage Source (CCVS) › A CCVS maintains a voltage across its terminals that is proportional to the
current through a circuit element elsewhere in the circuit.
› In general the voltage across a VCVS is where is a gain with units (V/A).
xiρ ρ
CCVS
3 ix+_
ix
ELEC1103 52
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Voltage Controlled Current Source (VCCS) › A VCCS maintains a current through its terminals that is proportional to the
voltage across a pair of terminals elsewhere in the circuit.
› In general the current through a VCCS is where is a gain with units (A/V).
xvα α
VCCS
vx
+
_3 vx
ELEC1103 53
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Current Controlled Current Source (CCCS) › A CCCS maintains a current through its terminals that is proportional to the
current through an element elsewhere in the circuit.
› In general the current through a CCCS is where is a gain with units (A/A) so it is unitless.
xiβ β
CCCS
iy
2 iy
ELEC1103 54
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Valid or Not Valid ?
Yes No
ELEC1103 55
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Valid or Not Valid ?
Yes No
ELEC1103 56
BASIC CIRCUIT ELEMENTS
Ch2. I. Ideal Basic Circuit Elements
› Valid or Not Valid ?
Yes No
ELEC1103 57
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Ohm’s Law: The voltage across an ideal resistor is proportional to the current through it.
› The circuit symbol for a resistor is: i(t)
+
_v(t) R
( ) ( )v t i t R=
i-v characteristic of an ideal resistor
i
vv = i R
( ) ( ) / ( )i t v t R G v t= =
1GR
=
The constant of proportionality is the resistance R. The units of resistance are V/A, which are called ohms (Ω).
G is called the conductance G. The units of conductance are A/V, which are called siemens (S).
Ohms’ Law
ELEC1103 58
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Passive sign convention and Ohm’s Law: (Current flows into the positive terminal of the device, then use + sign to right Ohm’s law.)
v = iR v = - iR
ELEC1103 59
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Real Resistors:
Area
AreaLR ρ
=
Resistivity Resistance increases with length, but decreases with cross-sectional area.
The units of resistivity are Ωm.
ELEC1103 60
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Resistivity of Materials at 300 K:
Conductors Aluminium 2.73 ×10–8
Carbon (amorphous) 3.5 ×10–5
Copper 1.72 ×10–8
Gold 2.27 ×10–8
Nichrome 1.12 ×10–6
Silver 1.63 ×10–8
Tungsten 5.44 ×10–8
Semiconductors Silicon (device grade) 10–5 to 1 depends on impurity concentration
Insulators Fused quartz > 1021
Glass (typical) 1 ×1012
Teflon 1 ×1019
Smaller the resistivity, the better the conductor. Resistivity depends on the temperature. A short circuit (i.e. an ideal conductor) is a resistor of 0 Ω resistance (no such material exists,
but a thick copper wire can be a good approximation). An open circuit is a resistor of 0 S conductance (or equivalently ∞ Ω resistance). Again no
such material exists, but some insulators can be a good approximation.
ELEC1103 61
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Resistance Calculation:
Calculate the resistance of a copper wire having a diameter of 2.05 mm and a length of 10 m.
2 3 26 2(2.05 10 ) 3.3 10 m
4 4dA π π −
−×= = = ×
8
6
(1.72 10 )(10) 0.052(3.3 10 )
LRAρ −
−
×⇒ = = = Ω
×
ELEC1103 62
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Real Resistors:
http://www.electronics-tutorials.ws/resistor/res_1.html
ELEC1103 63
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Real Resistors:
http://www.electronics-tutorials.ws/resistor/res_1.html
ELEC1103 64
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Chip Resistors:
ELEC1103 65
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Chip Resistors:
ELEC1103 66
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Power in a Resistor:
› We can calculate the power in a resistor in terms of voltage using Ohm’s Law (v=iR) and the expression for power (p=vi).
2v vP vi vR R
= = =
( ) 2P vi iR i i R= = =
› We can calculate the power in a resistor in terms of current using Ohm’s Law (v=iR) and the expression for power (p=vi).
ELEC1103 67
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Using Ohm’s Law:
Given current and resistanceFind the voltage
AI 2=Ω= 5R
Given current and resistanceFind the voltage
AI 2=Ω= 5R
−=
+][10 VV
v iR=
ELEC1103 68
Given Current and VoltageFind Resistance
−
+][20 V
][4 AI =
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Using Ohm’s Law:
vRi
=
Given Current and VoltageFind Resistance
−
+][20 V
][4 AI =
Ω= 5R
ELEC1103 69
Given Voltage and ResistanceCompute Current
−
+][12 V Ω=3R
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Using Ohm’s Law:
viR
=
Given Voltage and ResistanceCompute Current
−
+][12 V Ω=3R
][4 AI =
Note: we determine the direction of current (blue arrow) using the passive sign convention.
ELEC1103 70
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› Using Ohm’s Law and Keeping Units Straight:
Table 1 Keeping Units Straight
Voltage Current Resistance
Volts Amps Ohms
Volts mA kΩ
mV A mΩ
mV mA Ω
ELEC1103 71
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› More Ohm’s Law:
› Find the current and power absorbed by the resistor ?
[ ][ ] [ ]12 V
6 mA2 k
I = =Ω
[ ]( ) [ ]( ) [ ]12 V 6 mA 72 mWP VI= = =
+
-
Use passive sign convention.
ELEC1103 72
RESISTOR AND OHM’S LAW
Ch2. II. Resistor and Ohm’s Law
› More Ohm’s Law:
› Find the voltage of the voltage source and power absorbed by the resistor ?
3
S 6
0.5mA 0.5 10 A 10V50 S 50 10 S
IV IRG µ
−
−
⋅= = = = =
⋅
GIRIP
22 ==
( )232
6
0.5 10 [ ]0.5 10 [ ] 5[ ]
50 10 [ ]
AP W mW
S
−−
−
×= = × =
×
+
-
Use passive sign convention.
ELEC1103 73
NODES, LOOPS, AND BRANCHES
Ch2. III. Nodes, Loops, and Branches
› A branch represents a single element such as a voltage source or a resistor.
› A node is the point of connection between two or more branches.
› A loop is any closed path in a circuit.
› A network with b branches, n nodes, and L independent loops will satisfy the fundamental theorem of network topology:
› b = L + n - 1
How many branches, nodes, loops ?
3 Nodes 3 Loops 5 Branches 5 = 3 + 3 - 1
ELEC1103 74
NODES, LOOPS, AND BRANCHES
Ch2. III. Nodes, Loops, and Branches
› How many branches, nodes and loops are there ?
3 Nodes 3 Loops 5 Branches 5 = 3 + 3 - 1
ELEC1103 75
NODES, LOOPS, AND BRANCHES
Ch2. III. Nodes, Loops, and Branches
› An essential node is a node where three or more circuit elements join.
› An essential branch is a branch path connecting two essential nodes.
› A network with bE branches, nE nodes, and L independent loops will satisfy the fundamental theorem of network topology:
› bE = L + nE - 1
How many essential branches, essential nodes, loops ?
2 Essential Nodes 3 Loops 4 Essential Branches 4 = 3 + 2 - 1
Node 3 is not an essential node.
ELEC1103 76
KIRCHHOFF’S CURRENT LAW
Ch2. IV. Kirchhoff’s Current Law (KCL)
› Charge is not created nor destroyed in electric circuits. For this physical reason, we have Kirchoff’s Current Law.
› Kirchhoff’s Current Law states that the algebraic sum of currents entering (or leaving) a node (or a closed boundary) is zero.
1 2 3 4 5 0i i i i i− + + − =
i1
i2
i3
i4 i5
1 2 3 4 5 0i i i i i− + + + − =
Sum currents entering node: Sum currents
leaving boundary:
ELEC1103 77
KIRCHHOFF’S CURRENT LAW
Ch2. IV. Kirchhoff’s Current Law
B
C
i2
i1 Supernode
F
A
E
D
G
i3
id
ieic
1 2 3i.e. KCL at supernode: i i i= +
1
2
3
c e
c d
d e
i i ii i ii i i
= += ++ =
To verify the claim note that:
(now just add these equations)
KCL can be applied to supernodes. A supernode is obtained by defining a boundary that enclosed more than one node.
ELEC1103 78
KIRCHHOFF’S CURRENT LAW
Ch2. IV. Kirchhoff’s Current Law mAI 501 −=
mAmAmAIT 204010 ++=
0410 1 =−− ImAmA 01241 =−+ mAmAI03 12 =−+ ImAI
1I Find TI Find
1I Find21 I and I Find
ELEC1103 79
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
› Electric circuits obey the principle of conservation of energy.
› Similar to a gravitational field, the electric field is conservative. This means that if a charge moves around a loop and comes back to the same point, the net energy gain or loss must be zero.
› Recall that ΔW = qV
+
AV
BBV
q
CV
−
+ABV −
+BC
V
−+ CAV
ABqVW =∆
BCqVW =∆
CAqVW =∆
Total Energy Gain Around Loop ( ) 0AB BC CDq V V V= + + =
- VCA +
ELEC1103 80
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
› Electric circuits obey the principle of conservation of energy, for this physical reason we have Kirchoff’s Voltage Law.
› Kirchhoff’s Voltage Law states that the algebraic sum of the voltages around a closed path, or loop, is zero.
01
=∑=
M
mnv
ELEC1103 81
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
Loop 3
A
B
C E
D
va
+
_
+_ vd+ _vb
vc
+
_ve
+
_
0e c dv v v− − =
Note that in obtaining loop equations, we sum the voltage drops around the loop (could also sum voltage rise). In the circuit analysis you are free to choose the direction of the KVL loop. KVL for loop 3 can be obtained by taking the difference of KVL for loop 2 and 1, i.e. it does not contain any new information.
Loop 1Loop 2
KVL for loop 1:
KVL for loop 2:
KVL for loop 3:
0a c bv v v− − =
0a e d bv v v v− + − =
ELEC1103 82
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
› Find Vae and Vec?
ELEC1103 83
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
› Find Vae and Vec?
10 24 014[V]
ae
ae
VV
+ − ==
GIVEN THE CHOICE USE THE SIMPLEST LOOP
ELEC1103 84
KIRCHHOFF’S VOLTAGE LAW
Ch2. V. Kirchhoff’s Voltage Law (KVL)
4 6 010[V]
ec
ec
VV
+ + == −
› Find Vae and Vec?
ELEC1103 85
KIRCHOFF’S LAWS & OHM’S LAW
Ch2. VI. Kirchhoff’s Laws and Ohm’s Law
Find io ?
KCL at node b: 1 6 0oi i− + =
Ohm’s Law for 10Ω Resistor: 120 10b oV i− =
Ohm’s Law for 50Ω Resistor: 150bV i=
1 6 0oi i− + =
1120 50 10oi i− =
( )120 50 6 10o oi i− + =
180 60oi− =
[ ]3 Aoi = −
ELEC1103 86
KIRCHOFF’S LAWS & OHM’S LAW
Ch2. VI. Kirchhoff’s Laws, Ohm’s Law, and Dependent Sources
Find vo ? (Note vo = Vb)
KCL at node b: 0 5 0i i i∆ ∆− + =
Ohm’s Law for 5Ω Resistor: 500 5bV i∆− =
Ohm’s Law for 20Ω Resistor: 0 20bV i=
( )6 20 120bV i i∆ ∆= =
500 5120
bb
VV− =
12000 24 b bV V− =
[ ]480 VbV =
ELEC1103 87
KIRCHOFF’S LAWS & OHM’S LAW
Ch2. VI. Kirchhoff’s Laws, Ohm’s Law, and Dependent Sources
Find vo ?
KCL at Boundary means no current here!
ELEC1103 88
KIRCHOFF’S LAWS & OHM’S LAW
Ch2. VI. Kirchhoff’s Laws, Ohm’s Law, and Dependent Sources
Find vo ?
Ohm’s Law for 6Ω Resistor: [ ]5 A3si =
Ohm’s Law for 2Ω and 3Ω Resistor: ( )053 2 33
i = +
[ ]0 1 Ai =
[ ]0 3 Vv =
ELEC1103 89
› How to electrically model the human body:
› Rskin (dry)=15 kΩ, Rskin (wet)=150 Ω
› Rlimb=100 Ω
› Rtrunk=200 Ω
ELECTRICAL SAFETY
Ch2. VII. Electrical Safety
ELEC1103 90
ELECTRICAL SAFETY
Ch2. VII. Electrical Safety
› First, consider electrostatic shock on a dry day:
› Around 20,000 to 40,000 V and 40 A – but only
for a few microseconds.
› Current flow is mainly over the body surface
› Annoying to you, deadly for electrical components
› Consider 50 Hz, 230 V electrical outlets, danger is shown to the right. 0.1 to 0.2 A is fatal causing ventricular fibrillation. 60 Hz current penetrates the body more deeply and electrical outlet sustains the current.
ELEC1103 91
ELECTRICAL SAFETY
Ch2. VII. Electrical Safety
Man working in damp basement using a drill. Insulation on wire nicked and touches casing of drill. Damp concrete relatively good conductor. If the case is not grounded, man could receive fatal shock. If case is grounded, the circuit breaker shorts and lights go out. Which do you prefer ?
ELEC1103 92
ELECTRICAL SAFETY
Ch2. VII. Electrical Safety
Boys in and near a pool. Ground fault in pool lighting. Vinyl lining of pool means water is electrically insulated. One boy in pool and one outside. If they touch and outside boy is grounded, who is more likely to die?
Outside boy, because current more likely to go through the heart.
ELEC1103 93
ELECTRICAL SAFETY
Ch2. VII. Electrical Safety
A crane operator touches high-voltage line at 7200 V. The crane is about 10 m from the pole which is earthed. He jumps out of the crane runs toward the pole and dies – why?
There is a 7200 V/10 m or 720 V/m potential gradient between the crane and the pole. If the man’s running stride is 1 m, his body experiences 720 V. He dies, but a man standing still survives.