Norton's theorem

Norton"s Theorem for DC Network

1. CIRCUIT THEORY


NORTON’S THEOREM

Friday, July 26, 2019 1 syed hasan saeed
SYED HASAN SAEED

2. NORTON’S THEOREM

REFERENCE BOOKS
•Introductory Circuit Analysis, Robert L. Boylested, Pearson Education,
Prentice Hall.
•Networks And Systems, Ashfaq Husain, Khanna Book Publishing Co (P)
Ltd. Delhi.
•Networks And Systems, A Sudhakr, Shyammohan S Palli, Tata McGraw
Hill, New Delhi.
•Network Analysis, M.E. Van Valkenburg, PHI Learning Private limited,
New Delhi.
•Circuit Analysis Principle and Applications, Allan H. Robbins &Wilhelm
C. Miller, DELMAR CENGAGE Learning, Indian Reprint.

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3. NORTON’S THEOREM
STATEMENT: Any linear, active, bilateral dc network having a number of
voltage sources and/or current sources with resistances can be replaced by
a simple equivalent circuit having single current source (I
N) in parallel
with a single resistance (R
N).
Where (I
N) is the known as Norton’s equivalent current through the terminal
a-b.
(R
N) is the Norton’s equivalent resistance viewed back into the network from
terminal a-b.

Note: independent voltage sources are short circuited and independent
current sources are open circuited. Dependent sources will remain in the
circuit for the calculation of Norton’s equivalent resistance.

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4. NORTON’S THEOREM
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Procedure for converting any circuit into Norton's equivalent circuit
Calculate Norton Current
Step 1: remove the load resistance R
L (through which current is required) and short
circuit it. Let terminals of load are labelled as a-b. Therefore a-b is the short
circuited.
Step 2: Find the current through the terminal a-b by applying KCL, KVL, Ohm’s
law or Superposition principle. This current is the short circuit current and it is
known as Nortons equivalent current (I
N).
Calculate Norton Resistance (equal to Thevinin resistance)
Step 3: Set all Independent voltage Sources as short circuit and Current Sources
open circuit. Dependent sources will not be changed
Step 4: Calculate the resistance as “seen” through the terminals a-b into the network.
This resistance is known as Norton’s equivalent resistance (R
N ).
Draw Equivalent Circuit
Step 5: Replace the entire network by Nortons equivalent current (I
N) in parallel
with Norton’s equivalent resistance (R
N ) and connect the load resistance R
L.

5. NORTON’S THEOREM
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R
L
Linear, Active,
Bilateral
Network
Linear, Active,
Bilateral
Network
R
L
a
b
I
L
R
L
Norton’s Equivalent Network
R
L
b
a
I
L
R
N
I
N

6. NORTON’S THEOREM
Example: Find the current through 3 ohm resistor by Norton’s Theorem for
the network shown in fig.1a






SOLUTION:
STEP 1: Calculation of R
N (calculation is same as R
th). Redraw the circuit by
removing the 3 ohm resistor and short circuit the voltage sources as
shown in fig. 1b




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Fig. 1a
6 ohm 1 ohm
3 ohm
12V 24V
R
1
R
2
R
3
a
b

7. NORTON’S THEOREM






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6 ohm 1 ohm
R
2
a
b
R
N
R
2
a
b
24V
12V
I
N
R
1
Fig. 1c
Fig. 1b
R
1 and R
2 are in parallel A16
1
12
6
24
III
21N
 




 0.857
16
16
RR
RR
R
21
21
N
Step2: Calculation of Norton’s Current I
N : Short circuit the terminals
a-b
and the current flow through a-b is I
N
I
1
I
2

8. NORTON’S THEOREM
Step2: Draw the Norton’s Equivalent Circuit:





Step3: Calculation of Current through R3, Reconnect R3 to Norton’s
Equivalent Circuit (Fig. 1e)


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I
N
16A 0.857 ohm R
N
I
N
=16A
0.857
ohm
R
N =
R3 =
3 Ohm
Apply Current divider rule A55.3
30.857
0.857
16I
L



I
L LN
N
NL
RR
R
II



9. Friday, July 26, 2019 syed hasan saeed 9



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