LECTURE # 3 & 4
Circuit Laws
Network
Analysis
 Kirchhoff’s Current Law
 Kirchhoff’s Voltage Law
 Superposition Theorem
 Thevenin’s Theorem
 Norton’s Theorem
𝑺𝒆𝒍𝒇 𝑺𝒕𝒖𝒅𝒚
Common Electrical Circuit and Network Terms
n/n
Term
1
Electric Circuit
2
Circuit Param eters
3
Node
4
Branch
5
Lum ped Param eters
6
Distributed Param eters
7
Loop
8
Linear Circuit
9
Non-Linear Circuit
10
Bilateral Circuit
11
Unilateral Circuit
12
Short Circuit
13
Electric Network
14
Passive Network
15
Acive network
16
M esh
17
Open Circuit
18
Constant Current Source
19
Equivalent Resistance
20
Constant Voltage Source
Description
KIRCHHOFF’S LAWS
Kirchhoff’s Current Law – [KCL]
The sum of the currents entering 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 = 𝐼𝑎 + 𝐼𝑐 + 𝐼𝑥
a node is always equal to the sum 𝐼 + 𝐼 + 𝐼 + 𝐼 + (−𝐼 − 𝐼 − 𝐼 ) = 0
𝑎
𝑐
𝑥
of the currents leaving that node. 1 2 3 𝑛4
𝐼𝑘 = 0
𝑰𝟐
𝑘=1
𝑰𝒙
𝑰𝒂
Example #:
𝑰𝟒
𝑰𝟏
𝑰𝟑
𝑰𝒄
The algebraic sum of the currents at a
node is always equal to zero. Those
currents entering considered positive
while leaving are negative
𝐼1
𝐼2
𝐼5
𝐼6
𝐼4
𝐼3
𝑰𝟏 = 𝟐𝑨
𝑰𝟓 =? ?
𝑰𝟓 = 𝟒𝑨
𝑰𝟐 = 𝟐𝑨
𝑰𝟒 = 𝟐𝑨
𝑰𝟑 = 𝟐𝑨
𝑰𝟔 =? ?
𝑰𝟔 = 𝟎𝟎𝟎𝟎𝟎𝑨
Kirchhoff’s Voltage Law – [KVL]
According to Kirchhoff’s Voltage Law the sum of
the emf (voltage) sources acting upon a loop is
equal to the sum of the voltage drops experienced
along the loop.
𝑛
𝑛
𝐸𝑘 =
𝑘=1
𝑹𝟐
𝐼𝑅𝑘
𝑘=1
𝑹𝟑
𝑹𝟏
𝑬𝟐
𝐸1 + (−𝐸2 ) = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + 𝐼𝑅4
𝑬𝟏
𝑹𝟒
𝐸1 + −𝐸2 − 𝐼 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 = 0
 The direction of current is initially assumed,
since the resultant emf is not known
 Resultant magnitude and polarity of sources will
determine the larger source
𝑰
Example #:
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑎𝑙𝑙 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 ℎ𝑒𝑙𝑝 𝑜𝑓 𝐾𝐶𝐿 𝑎𝑛𝑑 𝐾𝑉𝐿
b
c
𝑰𝒂
1. Identify currents in the
4V
1Ω
branches
0.5Ω
2. Choose any closed loops
3. Assume the direction of
d
a
3Ω
current flow
2V
𝑰𝒃
4. Equate the sum of voltage
drops to emfs
3V
𝑰
𝒄
5. Equate the algebraic sums of
e
1Ω
f
emfs and voltage drops to
Consider loop abcda
zero
1 ∗ 𝐼𝑎 + 0.5𝐼𝑎 + 3𝐼𝑏 = 2 + 4
6. Find currents at a node
1.5𝐼𝑎 + 3𝐼𝑏 = 6
𝑨𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝑲𝑪𝑳:
7. Solve simultaneous equations
𝑰𝒂 + 𝟐𝑰𝒃 = 𝟒
For loop aefda
1 ∗ 𝐼𝑐 − 3𝐼𝑏 = 3 − 2
𝐼𝑐 − 3𝐼𝑏 = 1 −𝒊
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒊 𝒃𝒚 𝟑
𝟑𝑰𝒄 − 𝟗𝑰𝒃 = 𝟑 −𝒊𝒊𝒊
𝑰𝒂 = 𝑰𝒃 + 𝑰𝒄 −𝒊𝒊
𝑰𝒄 + 𝟑𝑰𝒃 = 𝟒 −𝒊𝒗
𝑨𝒅𝒅 𝒊𝒊𝒊 𝒂𝒏𝒅 𝒗
𝑰𝒂 = 𝟐. 𝟓𝑨𝑰𝒃 + 𝑰𝒄
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒊𝒗 𝒃𝒚 𝟑
𝟑𝑰𝒄 + 𝟗𝑰𝒃 = 𝟏𝟐 −𝒗
Open Circuit
An open circuit is a circuit where the path has been interrupted or
"opened" at some point so that current will not flow
Active Circuit
[with sources and circuit
elements]
Open-circuit voltage (OCV or 𝑉𝑜𝑐 )
𝒂 is the difference of electrical
potential between two terminals of a
𝑉𝑜𝑐 device when disconnected from any
𝑉𝑎𝑏
circuit
𝒃 The open circuit condition is the same
as a circuit with infinite resistance
No current flows through an open circuit
P.D.
measured
between
terminals of an open circuit is
the same as source voltage for
a simple d.c. circuit
Short Circuit
𝑺𝒆𝒍𝒇 𝑺𝒕𝒖𝒅𝒚
NETWORK ANNALYSIS
R1
LECTURE # 6
E1 – Voltage Source
R2
Superposition
Theorem
E2 – Voltage Source
R3
A-?
R4
IA – Current Source
Superposition Theorem
E1
R3
In any linear network with more than
one source (voltage source/current
source) the resultant current flowing
I4
through any branch is the sum of the
individual currents due to the different
sources considered separately,
whereby (voltage sources are R1
R4
R
2
replaced by short circuits or their
internal resistances and current
sources by an open circuit or
E2
infinite resistance)
The direction of the resultant currents
are determined by the size of the and
direction of values of currents flowing
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼4 𝑖𝑠 𝑎 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒
along the branch.
𝑒𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝑠𝑜𝑟𝑐𝑒𝑠 𝑬𝟏 . 𝑬𝟐 𝑎𝑛𝑑 𝑰𝑨
IA
Current due to Voltage Source E1
𝑰𝑬𝟏
−
𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏
𝑬𝟏
𝐈𝐄𝟏 =
𝑹𝑬𝟏
E1
R3
𝐼𝐸1
R1
E1
R3
𝐼𝐸1
R1
𝐼4𝐸1
R2
R4
𝐼4𝐸1
R4
R2
𝑂𝑝𝑒𝑛 𝐶𝑖𝑟𝑐𝑢𝑖𝑡
𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡
𝑹𝑬𝟏 − 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
𝑹𝑬𝟏 = 𝑹𝟏 +
𝑹𝟐 ∗ (𝑹𝟑 + 𝑹𝟒)
𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒
𝑰𝟒𝑬𝟏
−
𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒
𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏
𝐈𝐄𝟏 ∗ 𝑹𝟐
𝑰𝟒𝑬𝟏 =
𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒
Current due to Voltage Source E2
𝑰𝑬𝟐
−
𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐
𝑬𝟐
𝐈𝐄𝟐 =
𝑹𝑬𝟐
𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡
R3
𝐼4𝐸2
𝐼𝐸2
R1
R2
R4
E2
𝑹𝑬𝟏 − 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
𝑹𝟏 ∗ (𝑹𝟑 + 𝑹𝟒)
𝑹𝑬𝟐 = 𝑹𝟐 +
𝑹𝟏 + 𝑹𝟑 + 𝑹𝟒
𝑰𝟒𝑬𝟐
−
𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒
𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐
𝑰𝟒𝑬𝟐 =
𝐈𝐄𝟐 ∗ 𝑹𝟐
𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒
Current due to Voltage Source IA
𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡
R3
𝑰𝟒𝑨
R4
R1
𝑰𝑨
R2
𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡
𝑰𝟒𝑨 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒 𝒅𝒖𝒆 𝒕𝒐 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝑰𝑨
𝑶𝒏𝒍𝒚 𝒂𝒑𝒑𝒍𝒚 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒆𝒗𝒊𝒅𝒆𝒓 𝒓𝒖𝒍𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒃𝒓𝒂𝒏𝒄𝒉 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝑰𝟒𝑨
𝑹 𝑹𝟐
[ 𝑹 𝟏+ 𝑹
+ 𝑹𝟑]
𝟏
𝟐
𝑰𝟒𝑨 = 𝑰𝑨 ∗
𝑹 𝑹𝟐
[ 𝑹 𝟏+ 𝑹
+ 𝑹 𝟑] + 𝑹 𝟒
𝟏
𝟐
Resultant Current Due to All Sources
𝑰𝟒𝑬𝟏
𝑰𝟒
𝑰𝟒𝑨
𝑰𝟒𝑬𝟐
𝑹𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 𝑰𝟒 𝒊𝒔 𝒕𝒉𝒆
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝒔
𝑰𝟒 = 𝑰𝟒𝑬𝟏 + 𝑰𝟒𝑬𝟐 + 𝑰𝟒𝑨
Example#:
𝐸1 = 12𝑉 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒
𝐸2 = 24𝑉
𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝐼𝐴 = 𝐴4
𝑡ℎ𝑟𝑜𝑢𝑔ℎ
𝑅1 = 4Ω
𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟
𝑅2 = 2Ω
𝑅2 =? ?
𝑅3 = 6Ω
𝑅4 = 4Ω
R3
E1
I2
R1
R2
R4
E2
IA
EEB 211/231 & GEC 255 CONTACT TIME TABLE
EEB 211/LAB/01/2 Monday
08:00 AM
10:00 AM
BLDG 251 ROOM 201
Lab Work Group 1 &2
EEB 211/231/Lec
Monday
04:00 PM
05:30 PM
BLDG 252 ROOM R4
Lecture [26 August]
EEB 211/231/Lec
Tuesday
04:00 PM
05:30 PM
BLDG 252 ROOM R4
Lecture [27 august]
EEB 211/LAB/3/4
Wednesday
08:00 AM
10:00 AM
BLDG 251 ROOM 201
Lab Work Group 2/3
EEB 211/LAB/5/6
Thursday
08:00 AM
10:00 AM
BLDG 251 ROOM 201
Lab Work Group 5/6
EEB 211/LAB/7/8
Monday
02:00 PM
04:00 PM
BLDG 251 ROOM 201
Lab Work Group 7/8
EEB 211/LAB/9/10 Wednesday
03:00 PM
05:00 PM
BLDG 251 ROOM 201
Lab Work Group 9/10
GEC 255
Thursday
12:00 AM
13:00 PM
BLDG 248 ROOM 011
Tutorials
EEB 211/231/Lec
Thursday
02:00 PM
04:00 PM
BLDG 252 ROOM G1
Lecture
EEB 211/TUT/01
Monday
11:00 AM
12:00 PM
BLDG 248 ROOM 011
Tutorial Group 1
EEB 211/TUT/02
Tuesday
11:00 AM
12:00 PM
BLDG 248 ROOM 009
Tutorial Group 2
EEB 211/TUT/03
Thursday
11:00 AM
12:00 PM
BLDG 248 ROOM 009
Tutorial Group 3
GEC 255
Friday
11:00 AM
13:00 PM
BLDG 248 ROOM 009
Lecture
EEB 211/TUT/04
Friday
11:00 AM
12:00 PM
BLDG 248 ROOM 009
Tutorial Group 4
Example #
R3 𝑷𝟑 =?
E2
𝑰𝟐 =?
R1
R4
R2
IA
E1
𝑹𝟏 = 𝑹𝟒 = 𝟐Ω
𝑰𝑨 = 𝟖𝑨
𝑹𝟐 = 𝟒Ω,
𝑹𝟑 = 𝟑Ω
𝑬𝟏 = 𝟏𝟎𝑽
𝑬𝟐 = 𝟔𝑽
Solution
𝑷𝟑 =?
𝑰𝟐 = 𝑰𝟐𝑬𝟏 + 𝑰𝟐𝑬𝟐 + 𝑰𝟐𝑨
𝑰𝟐 =?
𝑷𝟑 = 𝑰𝟐 𝟑 ∗ 𝑹𝟑
𝑰𝟑 = 𝑰𝟑𝑬𝟏 + 𝑰𝟑𝑬𝟐 + 𝑰𝟑𝑨
𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏 =?
𝐼𝐸1 =
R3
R1
R2
R4
E1
𝑰𝟐𝑬𝟏
R3
R2
E1
R04
𝑹𝟎𝟒 =
𝟎 ∗ 𝑹𝟒
𝟎 + 𝑹𝟒
𝑹𝑬𝟏 = 𝑹𝟐 +
𝑰𝟐𝑬𝟏 = 𝑰𝑬𝟏
R0
𝐼3𝐸1 = 𝐼𝐸1 ∗
𝑰𝟑𝑬𝟏
R1
𝐸1
𝑅𝐸1
𝑅1
𝑅1 + 𝑅3
𝑹𝟏 ∗ 𝑹𝟐
𝑹𝟏 + 𝑹𝟐
Solution
𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑰𝑨 =?
𝑅21 ∗
R3
R1
𝑅31 ∗
R2
R4
IA
R0
R3
𝑰𝟐𝑨
R2
R0
IA
R04
𝑹𝟎𝟒 =
𝑅3 ∗ 𝑅1
𝑅3 + 𝑅1
R0
𝐼3𝐴
𝑅21
= 𝐼𝐴 ∗
𝑅21 + 𝑅3
𝐼2𝐴
𝑅31
= 𝐼𝐴 ∗
𝑅31 + 𝑅2
𝑰𝟑𝑨
R1
𝑅2 ∗ 𝑅1
𝑅2 + 𝑅1
𝟎 ∗ 𝑹𝟒
= 𝟎Ω
𝟎 + 𝑹𝟒
Solution
𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐 =? 𝑰
𝟑𝑬𝟐
R3
𝑰𝟐𝑬𝟐
R1
𝑰𝑬𝟐
𝑬𝟐
=
𝑹𝑬𝟐
𝑰𝑬𝟐
E2
R4
R2 𝑹 = ∞
𝑨
𝑹𝟑𝟐𝟏 = 𝑹𝟑 +
𝑰𝟑𝑬𝟐 R3
R0
𝑹𝟐 ∗ 𝑹𝟏
𝑹𝟐 + 𝑹𝟏
𝑰𝟑𝑬𝟐
𝑰𝟐𝑬𝟐
𝑹𝟏
= 𝑰𝟑𝑬𝟐 ∗
𝑹𝟏 + 𝑹𝟐
𝑰𝟐 = 𝑰𝟐𝑬𝟏 + −𝑰𝟐𝑬𝟐 + (−𝑰𝟐𝑨 )
𝑰𝑬𝟐
𝑰𝟐𝑬𝟐
R1
𝑹𝟒 ∗ 𝑹𝟑𝟐𝟏
𝑹𝟒 + 𝑹𝟑𝟐𝟏
𝑹𝟒
= 𝑰𝑬𝟐 ∗
𝑹𝟒 + 𝑹𝟑𝟐𝟏
R0
R2
𝑹𝑬𝟐 =
R4
E2
𝑰𝟐 = 𝑰𝟐𝑬𝟏 − 𝑰𝟐𝑬𝟐 − 𝑰𝟐𝑨
𝑰𝟑 = 𝑰𝟑𝑬𝟏 + (−𝑰𝟑𝑬𝟐 ) + 𝑰𝟑𝑨
𝑰𝟑 = 𝑰𝟑𝑬𝟏 − 𝑰𝟑𝑬𝟐 + 𝑰𝟑𝑨
𝑷𝟑 = 𝑰𝟐 𝟑 ∗ 𝑹𝟑