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2
8
BASIC COMPONENTS AND ELECTRIC CIRCUITS
2
Basic Components and Electric Circuits
Base Quantity
length
mass
time
electric current
thermodynamic temperature
amount of substance
luminous intensity
Name
meter
kilogram
second
ampere
kelvin
mole
candela
Symbol
m
kg
s
A
K
mol
cd
Table 1: SI (International System of Units) Base Units
2.0.1
Examples
• Force unit: newton (N): force required to accelerate 1kg mass by 1m/s/s.
• Work/ Energy:
– Joule=1 Newton.Meter=kg.m.m/s/s
– Cal=4.187 J
– kwh=3.6e6 J
– BTU=1055.1 J
• Power=rate of change of work (w.r. to time)
– watt=J/s
– hp=745.7 watts
– Note: J=watt.s ⇒ KJ=Kw.s ⇒ 3600KJ=kw.h ⇒ 3.6 Meg.J=kwh
• Find the cost of heating 50 liters of water from 10 to 75 degrees kelvin if the cost of 1 KWH=0.08 JD. If the power
consumed by the heating element is 2 KW, find the time needed for heating the water. (The small calorie or gram calorie
approximates the energy needed to increase the temperature of 1 gram of water by 1 C. This is about 4.184 joules=4.184
Watt.second.)
Answer: Energy required = 50*1000*(75-10)*1 Calories = 50*1000*(75-10)*1*4.187 Joules (or Watt.second) = 13607750
Watt.second = 13607750/1000/60/60 KWH = 3.7799 KHW.
Cost of heating = 3.7799*0.08 = 0.3024 JD.
If 2 KW heater is used, then time needed to heat the water = 3.7799/2 = 1.89 Hours.
2.1
Introduction
An electric circuit is an interconnection of electrical elements.
2.1.1
SI System of Units
Meter (m), Kilogram (kg), second (s), Ampere (A), Kelvin (K), Luminous intensity Candela (cd), amount of Substance Mole
(mol)
2.1.2
SI Prefixes
E=exa= 1018 , P=peta= 1015 , T=tera= 1012
G=giga= 109 , M=mega= 106 , k=kilo= 103
h=hecto= 102 , da=deca= 10, d=deci= 10−1
c=centi= 10−2 , m=milli= 10−3 , µ=micro= 10−6
n=nano= 10−9 , p=pico= 10−12 , f=femto= 10−15
a=atto= 10−18
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BASIC COMPONENTS AND ELECTRIC CIRCUITS
2.2
9
Charge and Current
Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C).
1. In a coulomb there are
1
1.602×10−19
= 6.24 × 1018 electrons.
2. charges occur in multiples of electron charge, i.e. multiples of 1.602 × 10−19 C.
3. law of conservation of charge: charge can neither be created nor be destroyed, only transferred.
⇒ algebraic sum of electric charges in a system does not change
Universally accepted convention is that current is the net flow of positive charges.
Electric current is the time rate of change of charge, measured in ampere (A)- flow of +ve charges
A point can not hold or accumulate charge and hence the algebraic sum of currents entering (+ve) and leaving
(-ve) any node (point) equals zero.
∆ dq
dt
i=
, q(t) instantaneous value of charge.
∆
and q(t) =
t
t0
idt + q(t0 )
A direct current (dc) is a current that remains constant with time
An alternating current (ac) is a current that varies sinusoidally with time
current: value and direction
2.3
Voltage
Electromotive force (emf)=voltage=potential difference.
∆ dw
=work
dq
vab =
needed to move a unit of charge from a to b
w in joules, q in coulombs, v in volts.
1 volt=1 joule/coulomb=1 newton meter/coulomb
Voltage (or potential difference) is the energy required to move a unit charge through an element, measured in volts
vab = −vba
Voltage: value plus polarity (+ and - signs)
2.4
Power and Energy
Power is the energy (+ve) if supplied (delivered) or (-ve) if absorbed (consumed) per unit of time. It is also the product of
voltage and current
dw
= vi watt or J/s
p=
dt
According to the passive sign convention power assumes a positive sign when the current enters the positive polarity of the
voltage across an element.
Absorbed is +ve, delivered is -ve.
2.4.1
Law of conservation of Energy
The algebraic sum of power in a circuit, at any instant of time, must be zero (power (+ve) absorbed + power (-ve) delivered=0)
:
p=0
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BASIC COMPONENTS AND ELECTRIC CIRCUITS
2.5
Circuit Elements
• Circuit element = mathematical model.
• Simple circuit element = mathematical model of two terminal electrical device. Completely characterized by its v − i
relationship.
1. An ideal voltage source produces a specific potential difference across it terminals regardless of what is connected to it.
2. An ideal current source produces a specific current through its terminals regardless of what is connected to it.
3. Voltage and current sources can be independent or dependent on either current or voltage.
1.2ix
+
−
H
9.2iy
F
2.7v6
+
−
E
2.1v2
G
• VS (V), Independent Voltage Source
• IS (I), Independent Current Source
• VCVS (E), (Voltage Controlled Voltage Source)
• CCCS (F), (Current Controlled Current Source)
• VCCS (G), (Voltage Controlled Current Source)
• CCVS (H), (Current Controlled Voltage Source)
Voltage sources’ symbols are HEV
Current sources’ symbols are FIG
As an example, (voltage) transformer can be viewed as a VCVS,
vp
N
= Nps
vs
As an example, (current) transformer can be viewed as a CCCS,
ip Np = is Ns =Ampere.turns
12V
+
− V
1.5A
I
sources.m4
Symbols used for Electric Sources
2.5.1
G=
i
v
Conductance
=
1
R
measured in siemens (S). Notes:
1. power dissipated in a resistor is always positive. (can not be negative)
2. short circuit means R = 0
3. open circuit means R = ∞
2.5.2
Networks and Circuits
Circuits must be closed while networks need not be. Any circuit is a network.
2.5.3
Ohm’s Law
v = iR. R=resistance (bidirectional, constant value, linear element).
2.5.4
Power Absorption
p = vi = i2 R = v 2 /R
2.5.5
PSPICE Syntax
First line is comment line. Next comes the description lines, one line per element, and then comes the (dot) command lines,
and finally (.) END line.
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BASIC COMPONENTS AND ELECTRIC CIRCUITS
2.6
Tutorial # 1
1. Practical Application: Wire Gauge R = ρ Al
Conductor size (AWG)
28
24
18
14
12
6
4
2
2.6.1
Cross-sectional Area (mm2 )
0.0804
0.205
0.823
2.08
3.31
13.3
21.1
33.6
Ohms per 1000 ft at 20o C
65.3
25.7
6.39
2.52
1.59
0.3952
0.2485
0.1563
Example 2.3, page 21, 6th ed.
A wire of length 4000-ft, 4 AWG, has resistance R = (4000f t)(0.2485Ω)/(1000f t) = 0.992Ω.
If i = 100A, then power dissipated in the wires=i2 R = 1002 × 0.994 = 9940W .
2. Practice Problem:
2V
+−
8A
p2
+
5V p1
−
I = 5A
3A
p3 + 0.6I
−
p4
+
3V
−
ch1pp7.m4
p1 = −40W, p2 = 16W, p3 = 9W, p4 = 15W
Figure 3: HKD7eC1PP7
p1 = (+5)(−8) = −40W, p2 = (+2)(+8) = +16W, p3 = (0.6I)(+3) = (0.6 ∗ 5)(3) = +9W, p4 = (+3)(+5) = +15W
3. Find the cost of heating 50 liters of water from 10 to 75 degrees kelvin if the cost of 1 KWH=0.08 JD. If the power
consumed by the heating element is 2 KW, find the time needed for heating the water. (The small calorie or gram calorie
approximates the energy needed to increase the temperature of 1 gram of water by 1 C. This is about 4.184 joules=4.184
Watt.second.)
Answer: Energy required = 50*1000*(75-10)*1 Calories = 50*1000*(75-10)*1*4.187 Joules (or Watt.second) = 13607750
Watt.second = 13607750/1000/60/60 KWH = 3.7799 KHW.
Cost of heating = 3.7799*0.08 = 0.3024 JD.
If 2 KW heater is used, then time needed to heat the water = 3.7799/2 = 1.89 Hours.
4. How do you provide good earth for a building? Practically, what is considered as an acceptable resistance in Ω’s for a
proper earth?
Answer:
Dig a hole for approximately half a meter and insert a copper rod of
approximately 1 meter height inside the hole. Connect a reasonably
thick wire to the rod and this point will be your EARTH for the building.
For a normal house, as recommended by IEEE, good earth should be less
than 5 Ω.
Picture to the right from Wikipedia - The free encyclopedia.