NETWORK ANNALYSIS

Mandatory Reading
 Electrical circuit
 Electrical Network
 Active Network
 Passive Network
 Open Circuit
 Short Circuit
 Node, Branch, Mesh
 Linear Circuit
 CONSTANT CURRENT SOURCE
 COMSTANT VOLTAGE SOURCE
 Unilateral Network
 Bilateral Network
 Non-linear Network
 Linear Parameters/non-linear parameters of a circuit
 𝑰𝑨
LECTURE # 8

𝑹𝟏

𝑹𝟐 𝑹𝟑

Norton’s
𝑹𝟒
Theorem 𝑹𝟓

𝑰𝑳
𝑹𝟔
𝑬
𝑹𝟏=∞ Ω 𝑹𝟐=𝑹𝟒=𝑹𝟔=𝟒Ω 𝑹𝟑=𝟔Ω 𝑹𝟓=𝟐Ω 𝑬=𝟒𝟎𝑽 𝑰 𝑨=𝟖 𝑨
A current flowing in any portion (between any two terminals) of an active
network of linear sources and resistances can be determined through CDR
as the current from a constant current source in parallel with a resistance.
This implies that any active network of linear sources and resistances can
be represented by a constant current source in parallel with a resistance.
Norton’s Equivalent Source
𝑨
𝑨 𝑰𝑳
Active Network
𝑰𝑳
with linear 𝑹𝑳 𝑹𝑵 𝑹𝑳
Sources and 𝑰 𝑺𝑪
Resistances
𝑩
𝑩
The constant current source is the short circuit current measured between the terminals
– referred to as Norton’s equivalent resistance is the resistance measured between the
terminals A & B while all voltage sources are replaced by internal resistances or short circuits
and current sources by open circuit. 𝑰 𝑨𝑩= 𝑰𝑳=
𝑰 𝒔𝒄 𝑹 𝑵
𝑹 𝑵 + 𝑹𝑳
 How to Find RN and ISC

Active 𝑨 The constant current source is the short circuit current

Network with that will flow between terminals A and B of the original
linear 𝑰 𝑺𝑪 network. If forms part of the Norton’s equivalent source
Sources and
Resistances 𝑩
Norton’s equivalent resistance – is the resistance measured
𝑨 between the terminals A & B when all network sources are
Passive
deemed passive. It is determined facing the network from
Network RN terminals A and B of the passive network
with linear
Resistances The Procedure
𝑩 Identify terminals between which to determine current
Draw the Norton equivalent circuit
C
C
I and RN in the Norton equivalent source
Incorporate the load Resistance RL (branch) in the Norton equivalent source
Compute the load current between terminals A&B in Norton’s Equivalent Circuit
…..and a little bit of some thinking
 Example # 2
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝐿 𝑡h𝑟𝑜𝑢𝑔h 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅 2 𝑢𝑠𝑖𝑛𝑔 𝑁𝑜𝑟𝑡𝑜𝑛′ 𝑠 𝑇h𝑒𝑜𝑟𝑒𝑚

𝑅 1=4 Ω 𝑅 3=6 Ω
𝐸
𝑅1 𝑹𝟐 𝑹𝟑 𝑅 2=8 Ω 𝑟 𝑖=1 Ω
𝐸=24 𝑉
 Thevenin’s Theorem is a network analysis
procedure meant to simplify the computations of a
complex network (with several sources and
LECTURE # 7 resistances).
It does so by reducing a complex network to simple
equivalent circuit (Thevenin’s Equivalent
Circuit) containing a Single Voltage Source in
series with a two Resistances.
Whereby:
The voltage source is the open circuit voltage
Thevenin’s measured between terminals where its
required to compute the current
Theorem One of the resistors is the Thevenin’s
equivalent resistance, measured between
terminals (where its required to compute
current) in a passive circuit
While the other resistance represents the load
resistance (connected between terminals where
its required to find current).
 resistance Rth
𝑨

𝑹𝒕𝒉 𝑰𝑳
𝐴
Active Network
with linear 𝑹𝑳
Sources and 𝑹𝑳
Resistances 𝑬 𝒕𝒉
𝐵
′ 𝑩
𝐸 𝑡h − 𝑇h𝑒𝑣𝑒𝑛𝑖 𝑛 𝑠 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
𝑇h𝑒𝑣𝑒𝑛𝑖𝑛 ′ 𝑠 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑆𝑜𝑢𝑟𝑐𝑒

𝑹 𝑳− 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒓𝒂𝒏𝒄𝒉𝒘𝒉𝒆𝒓𝒆𝒊𝒕𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒕𝒐 𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆𝒄𝒖𝒓𝒓𝒆𝒏𝒕

Equivalent Source, then its possible to determine the Load current, i.e. current between
terminals A & B
 Procedure
1.𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 𝑡h𝑒 𝑏𝑟𝑎𝑛𝑐h 𝑏𝑒𝑡𝑤𝑒𝑒𝑛𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠 𝐴 𝑎𝑛𝑑 𝐵

2.𝑇𝑒𝑚𝑝𝑜𝑟𝑎𝑟𝑖𝑙𝑦 𝑟𝑒𝑚𝑜𝑣𝑒 𝑡h𝑒𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠 𝐴 𝑎𝑛𝑑 𝐵 𝐴

𝐵
𝐴
𝑽 𝒕𝒉
𝐵

a
𝑹𝒕𝒉

6. 𝑅𝑒𝑐𝑜𝑛𝑛𝑒𝑐𝑡 𝑡h𝑒𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑏𝑟𝑎𝑛𝑐h)𝑏𝑒𝑡𝑤𝑒𝑒𝑛𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠 𝐴𝑎𝑛𝑑 𝐵

𝑰𝑳
7.𝐶𝑜𝑚𝑝𝑢𝑡𝑒 𝑡h𝑒𝑙𝑜𝑎𝑑𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼𝐿,𝑖.𝑒.𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴∧𝐵
 Example # 4 [Prove that I1=10A]
𝑬=𝟐𝟒𝑽 Ω 𝑹𝟐=𝟓Ω 𝒓 =𝟏Ω
A
A
I1 ? IL ?
E E E I L?
R1 R2 R1 R2 R2 R1

B
A B
A
I2
I 𝑰= 𝑰 𝟐
E R2 𝑽 𝒕𝒉=𝑰 𝟐 𝑹𝟐 𝒓 R2 𝑹 𝒕𝒉=
𝒓𝑹 𝟐
𝒓 + 𝑹𝟐
𝑽 𝒕𝒉=𝑬 − 𝑰𝒓
B B
A
𝑹𝒕𝒉=𝟎.𝟖𝟑𝟑Ω 𝑬 𝒕𝒉
𝑰 𝑳= =𝑰𝟏
𝑹𝒕𝒉+ 𝑹𝑳
𝟐𝟎
𝑰 𝑳= =𝟏𝟎 𝑨
𝑬 𝒕𝒉=𝟐𝟎𝑽 𝟎 . 𝟖𝟑𝟑+𝟏 . 𝟏𝟔𝟕
B
 R1

LECTURE # 6
E1 – Voltage Source

R2

E2 – Voltage Source

Superposition
R3
Theorem
R4
A-?

IA – Current Source
 Superposition Theorem

• In any linear network with more E1 R3

than one source (voltage
source/current source) the
resultant current flowing through
any branch is the sum of the I4
individual currents due to the
different sources considered
separately, while others are R1
R2 R4 IA
replaced by short circuits and
open circuit, or internal
resistances.
• The direction of the resultant E2
currents are determined by the
size of the and direction of
values of currents flowing along
the branch.
 Current due to Voltage Source E1
E1 R3
𝑰 𝑬 𝟏− 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬 𝟏
𝑬𝟏 𝐼 𝐸1
𝐈 𝐄 𝟏= 𝐼4𝐸1
𝑹 𝑬𝟏

E1 R3 R1 R4
R2
𝐼 𝐸1 𝐼4𝐸1
R4
R1 R2 𝑂𝑝𝑒𝑛𝐶𝑖𝑟𝑐𝑢𝑖𝑡

𝑆h𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡

𝑹 𝑬 𝟏− 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝑰 𝟒 𝑬 𝟏=

𝐈𝐄𝟏 𝒙𝑹𝟐
𝑹𝟐 ∗(𝑹𝟑+𝑹𝟒) 𝑹𝟐+ 𝑹𝟑+ 𝑹𝟒
𝑹 𝑬 𝟏=𝑹𝟏+
𝑹𝟐+𝑹𝟑+𝑹𝟒
 Current due to Voltage Source E2

𝑰 𝑬 𝟐− 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬 𝟐

𝑆h𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 𝑬𝟐
𝐈 𝐄 𝟐=
𝑹 𝑬𝟐
R3
𝐼4𝐸2
𝐼 𝐸2
R1 R4
R2
𝐈𝐄𝟐 𝒙𝑹𝟐
E2 𝑰 𝟒 𝑬 𝟐=
𝑹𝟐+ 𝑹𝟑+ 𝑹𝟒

𝑹 𝑬 𝟏− 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕

𝑹𝟏∗(𝑹𝟑+𝑹𝟒)
𝑹 𝑬 𝟐=𝑹𝟐+
𝑹𝟏+𝑹𝟑+𝑹𝟒
Current due to Voltage Source IA

𝑆h𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 R3
𝑰𝟒 𝑨
R1 R4 𝑰𝑨
R2

𝑆h𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡

𝑶𝒏𝒍𝒚 𝒂𝒑𝒑𝒍𝒚 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒆𝒗𝒊𝒅𝒆𝒓 𝒓𝒖𝒍𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆𝒃𝒓𝒂𝒏𝒄𝒉𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝑰 𝟒 𝑨

𝑰 𝟒 𝑨=𝑰𝑨𝒙
[ ( 𝑹𝟏 𝑹𝟐
𝑹𝟏+ 𝑹 𝟐 )
+ 𝑹 𝟑]

[ ( 𝑹𝟏 𝑹𝟐
𝑹 𝟏+ 𝑹 𝟐 )
+ 𝑹 𝟑]+ 𝑹 𝟒
Resultant Current Due to All Sources

𝑰𝟒 𝑨
𝑰𝟒𝑬𝟏

𝑰𝟒𝑬𝟐
𝑰𝟒

𝑰 𝟒=𝑰𝟒 𝑬 𝟏+𝑰𝟒 𝑬 𝟐+𝑰𝟒 𝑨

Success is hidden between the lines
 Example # 1

R3

I2=?
R1 R2 R4

IA

E1

𝑹𝟏=𝑹𝟒=𝟐Ω 𝑹𝟐=𝟒Ω, 𝑹𝟒=𝟑Ω

𝑰 𝑨=𝟖 𝑨 𝑬 𝟏=𝟏𝟔𝑽
 Example # 2

R3
I4=?
E2
I2=?
R1 R2 R4

IA

E1

𝑹𝟏=𝑹𝟒=𝟐Ω 𝑹𝟐=𝟒Ω, 𝑹𝟒=𝟑Ω

𝑬 𝟐=𝟔Ω
𝑰 𝑨=𝟖 𝑨 𝑬 𝟏=𝟏𝟔𝑽