The learning goals of Chapter 1 are:
reviewing the SI system of units and standard prefixes
knowing the definitions of basic electrical quantities:
Charge, Current, Voltage, Power, and Energy
1
Quantity
Basic Unit
Symbol
Length
meter
m
Mass
kilogram
kg
Time
second
s
Electric current
ampere
A
Thermodynamic temperature
degree kelvin
K
Amount of substance
mole
mol
Luminous intensity
candela
cd
National Institute of Standards and Technology Special Publication 330, 2008 Edition, Natl. Inst. Stand.
Technol. Spec. Pub. 330, 2008 Ed., 96 pages (March 2008)
2
Quantity
Frequency
Force
Energy or work
Power
Electric charge
Electric potential
Electric resistance
Electric conductance
Electric capacitance
Magnetic flux
Inductance
Unit Name (Symbol)
hertz (Hz)
newton (N)
joule (J)
watt (W)
coulomb (C)
volt (V)
ohm (Ω)
Siemens (S)
farad (F)
weber (Wb)
henry (H)
Formula
s−1
kg • m/s2
N•m
J/s
A•s
J/C
V/A
A/V
C/V
V•s
Wb/A
3
Quantity
Length
Mass
Time
Current
Voltage
Power
Energy
Charge
Frequency
Resistance
Conductance
Capacitance
Inductance
Symbol
x
M
t
I or i
V or v
P or p
W (E) or w
Q or q
f
R
G
C
L
Units
Meter
Kilogram
Second
Ampere
Volt
Watt
Joule (watt-sec)
Coulomb
Hertz
Ohm
Siemens
Farad
Henry
m
kg
s
A
V
W
J
C
Hz
Ω
S
F
H
4
Prefix
atto
femto
pico
nano
micro
milli
centi
deci
deka
hecto
kilo
mega
giga
tera
peta
Symbol
a
f
p
n
µ
m
c
d
da
h
k
M
G
T
P
Power
10−18
10−15
10−12
10−9
10−6
10−3
10−2
10−1
10
102
103
106
109
1012
1015
5
Electric circuits or electric networks are interconnections of
electrical devices or elements with at least one closed path in
which electric current may flow.
6
Charge (q) is the quantity of electricity responsible for electric phenomena.
The unit of charge is coulomb (C).
Current (i) is the time rate of flow of electric charge, q, past a given time
and its unit is ampere (A). The current can be expressed as,
𝑑𝑞
𝑖(𝑡) =
𝑑𝑡
Voltage (v) is the difference in potential between two terminals (or nodes).
Voltage represents the energy, w , required to move a unit positive electric
charge from the negative (–) terminal to the positive (+) terminal,
𝑑𝑤
𝑣(𝑡) = 𝑑𝑞
The unit of voltage is volt (V).
7
Current is represented by a value and a direction (arrow or
subscript).
▪ 𝑖𝑎𝑏 or 𝐼𝑎𝑏 means current goes through element from terminal
a to terminal b.
▪ 𝑖𝑏𝑎 or 𝐼𝑏𝑎 means current goes through element from terminal
b to terminal a.
𝐼𝑎𝑏 = 𝐼1 = 5 A
𝐼𝑏𝑎 = −𝐼𝑎𝑏 = 𝐼2 = −5 A
8
Direct current has a constant value that is independent of time.
𝑑𝑞
𝑑𝑡
= constant w. r. t. time
Alternating current is a function of time.
𝑑𝑞
𝑑𝑡
= varying with time
9
Voltage is represented by a value and polarity marks + and –
The subscripts denote the positive and negative terminals:
▪ 𝑣𝑎𝑏 or 𝑉𝑎𝑏 means terminal a is positive (+) and terminal b is
negative ( –).
▪ 𝑣𝑏𝑎 or 𝑉𝑏𝑎 means terminal b is positive (+) and terminal a is
negative ( –).
𝑉𝑎𝑏 = 10 V
𝑉𝑏𝑎 = −𝑉𝑎𝑏 = −10 V
10
Power is the rate at which energy is developed (supplied
or delivered) or dissipated (consumed or absorbed):
𝑑𝑤
𝑑𝑤
𝑝(𝑡) =
=
𝑑𝑡
𝑑𝑞
𝑑𝑞
= 𝑣(𝑡) ∗ 𝑖(𝑡)
𝑑𝑡
Unit of power is watt (W = joules/second (J/s)).
Energy (J or Watt-second) is:
𝑡
𝑤(𝑡) = 𝑤 𝑡0 + 𝜏𝑑 𝜏 𝑝 𝑡
0
11
pab = vab* iab = positive quantity
→ the element absorbs (or consumes) power.
pab = vab*iab = negative quantity
→ the element delivers (or supplies) power.
12
a.
Given q(t) = 10t C for t ≥ 0, find i(t).
𝑖 𝑡 =
b.
:
𝑑𝑞(𝑡)
𝑑𝑡
=
𝑑(10𝑡)
𝑑𝑡
= 10 A for t > 0
Given i(t) = 10e–2t A for t ≥ 0, and q(0) = 0. Find q(t).
𝑡
𝑞 𝑡 = 𝑞 0 + 0 𝑖 𝜏 𝑑𝜏
𝑡
= 𝑞 0 + 0 10𝑒 −2𝜏 𝑑𝜏 = 5 − 5𝑒 −2𝑡 C for 𝑡 ≥ 0
c.
Given q(t) = 0.1sin(377t) mC, find i(t).
𝑖 𝑡 =
𝑑𝑞(𝑡)
𝑑𝑡
= 37.7 cos 377𝑡 mA
13
a
𝑝𝑎𝑏 = 𝑣𝑎𝑏 ∗ 𝑖𝑎𝑏 = −20 3 = −60 W 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
𝑝𝑏𝑎 = 𝑣𝑏𝑎 ∗ 𝑖𝑏𝑎 = 20 −3 = −60 W 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
b
a
𝑝𝑎𝑏 = 𝑣𝑎𝑏 ∗ 𝑖𝑎𝑏 = −20 3 = −60 W 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
𝑝𝑏𝑎 = 𝑣𝑏𝑎 ∗ 𝑖𝑏𝑎 = 20 −3 = −60 W 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
b
14
𝑝𝑎𝑏 = 𝑣𝑎𝑏 𝑖𝑎𝑏 = 10 V ∗ 5 A = 50 W
𝑝𝑏𝑎 = 𝑣𝑏𝑎𝑖𝑏𝑎 = −10 V ∗ −5 A = 50 W
⇒ consuming power.
𝑝𝑎𝑏 = 𝑣𝑎𝑏 𝑖𝑎𝑏 = 5 V ∗ −2 A = −10 W
𝑝𝑏𝑎 = 𝑣𝑏𝑎𝑖𝑏𝑎 = −5 V ∗ 2 A = −10 W
⇒ supplying power.
𝑝𝑎𝑏 = 𝑣𝑎𝑏 𝑖𝑎𝑏 = −2 V ∗ −3 A = 6 W
𝑝𝑏𝑎 = 𝑣𝑏𝑎𝑖𝑏𝑎 = 2 V ∗ 3 A = 6 W
⇒ consuming power.
15
Calculate power (dissipated or delivered) by each
element in the circuit below.
•
•
•
•
𝑝1 = 𝑝𝑎𝑐 = 𝑣𝑎𝑐 𝑖𝑎𝑐 = 5 V −8 A = −40 W
𝑝2 = 𝑣𝑎𝑏 𝑖𝑎𝑏 = 2 V 8 A = 16 W
𝑝3 = 𝑣𝑏𝑐 𝑖𝑏𝑐 = 3 V 3 A = 9 W
𝑝4 = 𝑣𝑏𝑐 𝑖𝑏𝑐 = 3 V 5 A = 15 W
16
𝑝𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 (𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑜𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑)
= |sum of all negative powers|
𝑝𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 (𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑜𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑)
= sum of all positive powers
𝑝𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 = 𝑝𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑
17
Calculating power (absorbed or supplied) by each
element in the circuit below.
•
•
•
•
𝑝1 = 𝑝𝑎𝑐 = 𝑣𝑎𝑐 𝑖𝑎𝑐 = 5 V −8 A = −40 W
𝑝2 = 𝑣𝑎𝑏 𝑖𝑎𝑏 = 2 V 8 A = 16 W
𝑝3 = 𝑣𝑏𝑐 𝑖𝑏𝑐 = 3 V 3 A = 9 W
𝑝4 = 𝑣𝑏𝑐 𝑖𝑏𝑐 = 3 V 5 A = 15 W
• 𝑝𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 = 𝑝2 + 𝑝3 + 𝑝4 = 40 W
• 𝑝𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 = |𝑝1 | = 40 W
18
The current 𝑖 𝑡 through a given point in the circuit is given below.
Find the total charge that has passed through this point from 0 to 3s.
3
• 𝑞 3 − 𝑞 0 = 0 𝑖 𝑡 𝑑𝑡
3
1
2
3
• 0 𝑖 𝑡 𝑑𝑡 = 1 + 2 +
2∗2
2
=5C
19
The current through and the voltage across an element vary with
time as shown in the figure below.
Derive the expression of power delivered to the element for 𝑡 > 0.
What is the total energy delivered to the element between 𝑡 = 0 and 𝑡 = 25 s.
20
30 V
0 ≤ 𝑡 ≤ 10
𝑣 𝑡 = ቐ −5𝑡 + 80 V 10 ≤ 𝑡 ≤ 15
5V
15 ≤ 𝑡 ≤ 25
2𝑡 A
0 ≤ 𝑡 ≤ 10
𝑖 𝑡 = ቐ 2𝑡 A
10 ≤ 𝑡 ≤ 15
−3𝑡 + 75 A 15 ≤ 𝑡 ≤ 25
21
The expressions for power at different intervals are:
60𝑡 W
0 ≤ 𝑡 ≤ 10
2
𝑝 𝑡 = 𝑣 𝑡 𝑖(𝑡) = ቐ−10𝑡 + 160𝑡 W 10 ≤ 𝑡 ≤ 15
−15𝑡 + 375 W 15 ≤ 𝑡 ≤ 25
The total energy is:
10
15
25
𝑤 25 − 𝑤(0) = න 𝑝 𝜏 𝑑𝜏 + න 𝑝 𝜏 𝑑𝜏 + න 𝑝 𝜏 𝑑𝜏
10
0
15
𝑤 25 − 𝑤(0) = න 60𝜏𝑑𝜏 + න
0
10
15
25
−10𝜏 2 + 160𝜏 𝑑𝜏 + න
10
−15𝜏 + 375 𝑑𝜏
15
𝑤 25 − 𝑤(0) = 6750 J
22
0
