ELEC 3202
Electric Circuits I
Dr. Hazem N. Nounou
Department of Electrical Engineering
United Arab Emirates University
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Chapter One
Basic Concepts
• This course basically deals with the analysis of electric circuits.
• The most basic quantity used in the analysis of electrical circuits
is the electric charge (electron).
Basic Quantities
(1) Electron :electron is a mobile charge carrier.
•The electron is measured in coulumb [ C ]
• e = 1.6*10-19 C
• Multiple of electrons constitute charge (q).
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
•The movement of charge (q) over time causes current.
(2) Current :
the time rate of change of charge produces an electrical current
t
i(t)
q(t) =
Or
∫ i( τ ) d τ
τ = −∞
• the electric current is measured in Amper [A]
dq(t)
=
dt
1 A = 1 C / 1 sec
•.current convention.
- -e
e- e
i
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
There are 2 types of currents
1. Direct current (DC)
i(t)
time
2. Alternating current (AC)
i (t)
time
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
(3) Voltage :
The voltage is defined as the work or energy (in Joules) required
per unit charge to move a test charge though an element
W
V =
q
And
1J
1V =
1C
• Since we are dealing with a changing charge and energy, we have
v =
dw
dq
(4) Power :
Power is the time rate of change of energy.
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
P(t)
P (t ) =
=
dw(t)
dt
dw(t) dw(t) dq
=
⋅
dt
dq dt
P ( t ) = V(t) i(t)
•The unit of power is Watt [W].
• 1 W = 1 V * 1A
(5) Energy: energy can be expressed as
t2
t2
w(t) = ∫ p(t) dt = ∫ v(t) i(t) dt
t = t1
t = t1
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Passive sign convention
Current flow from the positive to the negative terminal.
(+)
i(t)
R
(-)
• Power can be absorbed or supplied by an element.
• Power is absorbed (or dissipated) by an element if the sign of
power is (+)
• Power is supplied (delivered or generated) by an element if
the sign of power is (-)
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Circuit Active Elements:
+-
There are 4 types of active elements (sources):
1. Independent voltage source:
It is a 2-terminal sources that maintains a specific voltage
across its terminals regardless of the current through it
2. Independent current source:
It is a 2-terminal sources that maintains a specific current
through it regardless of the voltage across it terminals.
3. Dependent voltage source:
It is a 2-terminal sources that generates a voltage that is
determined by a voltage or current at a specified location
in the circuit.
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
4. Dependent current source:
It is a 2-terminal sources that generates a current that is
determined by voltage or current at a specified location in the
circuit.
Example :
Compute the power that is absorbed or supplied by each of
the elements in the following circuit
1 Ix
Ix = 4 A
R1
+ 12 V -
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+-
Vs = 36 V
I R2
+
24 V
R2
-
Department of Electrical Engineering
I R3=2 A
+
28 V
-
R3
Dr.Hazem N.Nounou
P vs = V s I x = (36)( − 4) = − 144W (supplies )
P R1 = V R1 I x = (12)(4) = 48W
(absorbs )
PR2 = VR2 I R2 = VR2 (I x - I R3 ) =
(24)(4 - 2) = 48W (absorbs)
PDs = VDs I R3 = (1 I x )(I R3 ) = (4)(-2) = - 8W (supplies)
PR3 = V R3 I R3 = (28)(2) = 56 W (absorbs)
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Important Units
QUANTITY
SYMBOL
Length
l
Current
Temperature
Mass
Time
I, i
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T
m
t
UNIT
meter
ampere
kelvin
kilogram
second
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ABBREV.
m
A
K
kg
s
Dr.Hazem N.Nounou
Important Units
Voltage
Charge
Resistance
Power
Capacitance
Inductance
Frequency
Magnetic Flux
Mag. Flux Density
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V, v,
Q, q
R
P, p
C
L
f
Φ
B
volt
coulomb
ohm
watt
farad
henry
hertz
weber
tesla
Department of Electrical Engineering
V
C
Ω
W
F
H
Hz
Wb
T
Dr.Hazem N.Nounou
Unit Conversions
UNIT
MULTIPLY BY
in
ft
mi
lb
hp
kWh
ft-lb
0.0254
0.3048
1.609
4.448
746
3.6 x 106
1.356
UAE University
Department of Electrical Engineering
TO GET
m
m
km
N
W
J
J
Dr.Hazem N.Nounou
Prefixes For Engineering Notation
POWER OF 10
1012
109
106
103
10-3
10-6
10-9
10-12
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PREFIX
tera
giga
mega
kilo
milli
micro
nano
pico
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SYMBOL
T
G
M
k
m
µ
n
p
Dr.Hazem N.Nounou
Chapter 2
Ohm’s law :
Resistive Circuits
The voltage across a resistor is directly proportional to the current
flowing through it.
v (t)
V (t) = R i(t)
R≥0
The symbol of ohm is (Ω )
R
1
i (t)
1V
1Ω =
1A
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
The instantaneous power P (t):
P(t) = v(t) i(t) = R i(t) i(t)
v(t) v 2 (t)
= R i (t) = v(t)
=
R
R
2
2
v
(t)
2
∴ P(t) = v(t) i(t) = R i (t) =
R
Note: Last equation says that the power at a resistor is always
positive
Resistors always absorb power.
The inverse of resistance is conductance
1
G=
R
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
The unit of conductance is Siemens (S)
1A
1S =
1V
The current can be also expressed as
i(t)= G V(t)
And the instantaneous power is
i(t)
i 2 (t)
P(t) = v(t)i(t) = i(t) =
G
G
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
P(t) = v(t) i(t) = v(t) G v(t) = G v 2 (t)
i 2 (t)
⇒ = v(t) i(t) =
= G v 2 (t)
G
Open and short Circuits
Open circuit ( R = ∞ )
circuit
R= ∞
G=0
circuit
Open circuit
v(t) v(t)
i(t) =
=
=0
R
∞
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Short circuit ( R = 0)
circuit
G=∞
R= 0
circuit
Short circuit
v(t) = Ri (t ) = 0 * i (t ) = 0
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
I
Consider the circuit:
R= 2kΩ
+-
vs=12 v
Find the current and power absorbed by the resistor
v s 12 v
I= =
=6mA
R 2kΩ
P = v R I = (12) (6 m) = 72 m w
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example:
The power absorbed by a 10 k Ω resistor in the circuit is 3.6 mW.
I
Find voltage and current in the resistor.
P
I=
=
R
vs
+ -
P = Vs I = I 2 R
P
I2 =
R
R=10kΩ
(3.6 *10 ) (10 *10 )
−3
3
I = 3.6 *10 −7 = 0.6 mA
V = I R = (0.6 m A) (10 kΩ)
V=6V
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
Find the value of the voltage source and the power absorbed
by the resistance
I=0.5 m A
G = 50 µ S
+-
Vs
G = 50 µ S
R=1/G =2*104
Vs = I R = (0.5 m A) (20 * 10 4 Ω) = 10 V
PR = IV = (10 V) (0.5 mA) = 5 m W
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
Find R and the voltage across
The resistor?
Is=4mA
P = V Is
P=80mW
R
P 80 *10 −3 W
V= =
Is
4 *10 −3 A
V = 20V .
V = I R = (4 *10 −3 A) R
V
20 V.
R= =
= 5kΩ
−3
I 4 *10 A
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Kirchoff’s Laws:
(1) kirchoff’s current law (KCL) :
the sum of all currents entering any node is zero.
N
∑ i (t) = 0
k =1
k
Where N= number of currents.
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
R1
i1
Example:
Write the KCL equation
R2
i2
(1)
+
Here we have (4) nodes:
1
2
i3
Vs
50 i2
(3)
i4
R3
-
At node (1) : i (t) + i (t) = i (t)
(2)
i5
5
(4)
At node (2) : i 2 (t) + 50 i 2 (t) = i 3 (t)
At node (3) : 50 i 2 (t) + i 4 (t) = i1 (t)
At node (4) : i 3 (t) + i 4 (t) = i 5 (t)
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
R4
(2) Kirchoff’s voltage Law (KVL):
The sum of the voltage around any loop is zero.
N
∑v (t)= 0
using KVL
-
Find VR3 ?
Vs2= 5 V
R1
VR1=18 V
+
Vs1=30 V
VR2= 12 V
-
--
VR3
+
Vs3= 15 V
+
-30+18-5+12-15+ VR3 = 0
VR3 = 20 V
UAE University
+
Example:
-
k=1
N = # of voltage
k
R3
Department of Electrical Engineering
Dr.Hazem N.Nounou
R2
Example :
Find the KVL equation for the two paths abda and bcdb
Path abda:
v R1 + v R2 − v s = 0
Path bcdb:
+
a
20 v R1 + v R3 − v R2 = 0
R1
b
-
20 VR1
c
VR1
+
+-
Vs
VR2
-
R2
R3
+
VR3
-
d
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Single Loop circuits
We will discuss (2) issues :
1. Voltage divider rule:
Voltage is divided between resistor in direct proportion to
their resistance
v1 (t)
R1
v(t)
R1 + R 2
How?
v1 = R 1 i = R 1 (
R2
v 2 (t) =
v(t)
R1 + R 2
UAE University
+-
V(t)
+
V1
-
R1
+
V2
-
R2
Department of Electrical Engineering
v1 =
v
)
R1 + R 2
R1
v
R1 + R 2
Dr.Hazem N.Nounou
Multi Sources / resistors :
•Source can be added v=v1+v2+……
•Resistors can be added R= R1+R2+…..
v2
R1
+-
R3
V
+-
+-
v1
R2
R
Where:
v = v1 + v2
R = R1 + R2 + R3
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Single Node-Pair circuits :
We will discuss (2) issues:
i1(t)
1. Current-divider Rule .
i (t)
R2
i1 (t) =
i (t)
R1 + R 2
i 2 (t) =
Why ??
R1
i2(t)
R2
R1
i (t)
R1 + R 2
v = i1 R 1 = i 2 R 2
R2
∴i1 =
i2
R1
i = i1 + i 2
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⇒
i 2 = i − i1
Department of Electrical Engineering
Dr.Hazem N.Nounou
R2
(i − i 1 )
i1 =
R1
R2
R2
i1 (1 +
)=
i
R1
R1
R1 + R2
R2
i1 (
)=
i
R1
R1
R2
i1 =
i
R1 + R2
2. Multiple sources/resistors :
•Current source can be added.
•Resistors can added as reciprocals
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
i1(t)
R1
R2
i2(t)
R3
i(t) = i1 (t) + i 2 (t)
1 1
1
1
=
+
+
R R1 R 2 R 3
UAE University
i (t)
Department of Electrical Engineering
R
Dr.Hazem N.Nounou
Series and parallel resistors :
Series :
Parallel
R = R1 + R 2 +K+ R N
N
⇒ Rs =∑Rk
k =1
1
1
1
1
=
+
+K+
R P R1 R 2
RN
N
1
1
=∑
R P k =1 R k
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
Find equivalent resistance
2kΩ
2kΩ
4kΩ
6kΩ
10 k Ω
6kΩ
1k Ω
6k Ω
9kΩ
2kΩ
R 1 = {[(1 k Ω ) + (2 k Ω ) ] || 6k Ω} + 10 k Ω
2kΩ
2kΩ
4kΩ
6kΩ
6kΩ
R 1 = 12 k Ω
9kΩ
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
R2 = [12 k // 6 k ] + 2 k = 6k
2k Ω
4k Ω
6k Ω
R2= 6 k
Ω
9kΩ
R 3 = (6 k // 6 k ) = 3k Ω
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
2k Ω
4k Ω
R3 = 13
2 kkΩ
9kΩ
R eq = (12 k || 4 k ) + 2 k = 5 k
R eq = 5 k Ω
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
Find all currents and voltages
I3 3 k Ω
I1 9 k Ω
9 k Ω I5
I4
I2
+
+-
12 V
Va
6k Ω
+
4 k Ω Vb
-
-
The equivalent circuit is :
I1
Vc
-
3k Ω
9kΩ
+-
12 V
+
+
Va
R eq
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
R eq = [[(3 k + 9 k ) // 4 k ] + 3 k ] // 6 k
= 3 kΩ
12 V
I1 =
=1 m A
⇒ Va = R eq I i = 3 V
9 k + 3k
Va
3 1
∴ I2 =
=
= mA
6kΩ 6k 2
I 3 = I1 − I 2
1
1
⇒ I 3 =1 m A − m A = m A
2
2
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
1
V3 k Ω = I 3 (3 k Ω ) = ( m A) (3 k Ω ) = 1.5 V
2
∴ Vb − Va + V3 k Ω = 0
Vb − 3 + 1.5 = 0
⇒ Vb = 1.5 V
Vb
1.5 1
I5 =
=
= mA
9 k + 3 k 12 k 8
1
3
Vc = I 5 (3kΩ ) = m A (3kΩ ) = V
8
8
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example :
Find the source voltage Vo if I4=1/2 m A ?
1
Vb = I 4 (6k) = ( m A) (6 k Ω) = 3 V
2
Vb 3
=
=1 m A
∴ I3 =
3k 3k
I1
Vo
I2
+
Va
2k
1k
-
+-
1
I 2 = I 3 + I 4 = 1 m A + m A = 1.5 m A
2
3 k Ω I5
6k
I3
3k
+
Vb
-
I4
6k
4k
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
V0 = (6 k Ω ) I1 + Va + Vb + 4 k (I1 ) =10 k(3 m) + 3 + 3
V0 = 30 + 6 = 36
V0 = 36 V
∴ Va = 2 k Ω I 2 = (2 k) (1.5 m) = 3 V
Va + Vb 3 + 3
I5 =
=
= 1.5 m A
3k + 1k
4k
∴ I1 = I 2 + I 5 = 1.5 m A + 1.5 m A = 3 m A
I1
3 k Ω I5
6k
I2
+
Va
1k
-
+-
Vo
2k
I3
3k
+
Vb
-
I4
6k
4k
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
R1
3k ohm
I1
Vs1
+
12 V
-
+
2000 I1
-
Example :
Find V0 ? Using KVL
+
Vo
R2
5k ohm
-− 12 + 3 k (I1 ) − 2000 I1 + 5 k (I1 ) = 0
6kI1 = 12
⇒ I1 = 2 m A
V0 = (5 k Ω) (I1 ) (5 k) (2 m)
V0 =10 V
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example:
Find V0 using KCL:
10 m A
+
2k
3k
4k
+
V0
I0
4 I0
VS
-
-
10 m +
Vs
V
V
+ s −4( s )=0
6k 3k
3k
1
1
4
Vs
+
−
= − 10 m A
6 k 3k 3k
V s = 12 V
V0 =
4k
2
V s = (12) = 8 V
2k+4k
3
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example:
Find V0 using KVL
I
3k
2 V0
+
12 V
+-
1k
V0
-
− 12 + (3 k) I − 2 V0 + V0 = 0
− V0 + (3k) I − 12 = 0
V0
I=
1k
V0
− V0 + (3k) = 12
1k
− V0 + 3 V0 = 12
V0 = 6 V
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
Example:
Find V0 in the network
i1
6kΩ
i2
R1
1 k Ω R2
V0
2000
2kΩ
R3
+
2mA
V0
-
V0
− i1 − i 2 + 2 m A = 0
2000
R 1 i1 = (R 2 + R 3 )i 2
Also
(6 k Ω ) i1 = (3 k Ω ) i 2
1
∴ i1 = i 2
2
UAE University
Department of Electrical Engineering
Dr.Hazem N.Nounou
V0
1
∴
− i2 − i2 + 2 m A = 0
2000 2
V0
3
− i2 + 2 m A = 0
2000 2
V0
Q V0 = R 3 i 2 ⇒ i 2 =
2k
V0 3 V0
+ 2 m A = 0
−
∴
2k 2 2k
1 V0
= 2 m A
2 2k
UAE University
⇒ V0 = 8 V
Department of Electrical Engineering
Dr.Hazem N.Nounou
