This document introduces a presentation on the superposition theorem and Norton's theorem given by six students: Mahmudul Hassan, Mahmudul Alam, Sabbir Ahmed, Asikur Rahman, Omma Habiba, and Israt Jahan. The superposition theorem allows analysts to determine voltages and currents in circuits with multiple sources by considering each source independently and then summing their effects. Norton's theorem represents a linear two-terminal circuit as an equivalent circuit with a current source in parallel with a resistor. The document provides examples of applying both theorems to solve circuit problems.

1. Welcome To Our Presentation
PRESENTED BY:
1.MAHMUDUL HASSAN - 152-15-5809
2.MAHMUDUL ALAM - 152-15-5663
3.SABBIR AHMED – 152-15-5564
4.ASIKUR RAHMAN – 152-15-5948
5.OMMA HABIBA – 151-15- 5975
6.ISRAT JAHAN – 152 -15 - 5956

2. Presentation On
Superposition theorem and
Norton theorem

3. Superposition Theorem :
 The superposition theorem extends the use of Ohm’s
Law to circuits with multiple sources.
 Definition :- The current through, or voltage across,
an element in a linear bilateral network equal to
the algebraic sum of the currents or voltages
produced independently by each source.
 The Superposition theorem is very helpful in determining
the voltage across an element or current through a
branch when the circuit contains multiple number of
voltage or current sources.

4. Condition:
 In order to apply the superposition theorem to a network, certain
conditions must be met :
1. All the components must be linear, for e.g.- the current is
proportional to the applied voltage (for resistors), flux linkage is
proportional to current (in inductors), etc.
2. All the components must be bilateral, meaning that the current is
the same amount for opposite polarities of the source voltage.
3. Passive components may be used. These are components such as
resistors, capacitors, and inductors, that do not amplify or rectify.
4. Active components may not be used. Active components include
transistors, semiconductor diodes, and electron tubes. Such
components are never bilateral and seldom linear.

5. Procedure for applying Superposition
Theorem:
 Circuits Containing Only Independent Sources
 Consider only one source to be active at a time.
 Remove all other IDEAL VOLTAGE SOURCES by SHORT CIRCUIT & all
other IDEAL CURRENT SOURCES by OPEN CIRCUIT.
Voltage source is
replaced by a Short
Circuit
Current source is
replaced by a
Open Circuit

6. Steps:
1) Select any one source and short all other voltage sources
and open all current sources if internal impedance is not
known. If known replace them by their impedance.
2) Find out the current or voltage across the required element,
due to the source under consideration.
3) Repeat the above steps for all other sources.
4) Add all the individual effects produced by individual sources
to obtain the total current in or across the voltage element.

7. Example:
Using the superposition theorem, determine the
voltage drop and current across the resistor 3.3K as
shown in figure below.
Step 1: Remove the 8V power supply from the original circuit, such that the new
circuit becomes as the following and then measure voltage across resistor.

8. Here 3.3K and 2K are in parallel, therefore resultant
resistance will be 1.245K.
Using voltage divider rule voltage across 1.245K will be
V1= [1.245/(1.245+4.7)]*5 = 1.047V
Step 2: Remove the 5V power supply from the original circuit such that
the new circuit becomes as the following and then measure voltage
across resistor.

9. Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be
1.938K.
Using voltage divider rule voltage across 1.938K will be
V2= [1.938/(1.938+2)]*8 = 3.9377V
Therefore voltage drop across 3.3K resistor is V1+V2 =
1.047+3.9377=4.9847 v

10. Circuit A of the sample circuit shown here has an
independent voltage source and an independent current
source. How do you find the output voltage vo as the
voltage across resistor R2?
Example : 2
Circuit A (with its two independent sources) breaks up into two simpler
circuits, B and C, which have just one source each.

11. Circuit B has one voltage source
because the current source was
replaced with an open circuit. Circuit
C has one current source because the
voltage source was replaced with a
short circuit.
For Circuit B, you can use the voltage
divider technique because its resistors,
R1 and R2, are connected in series with
a voltage source. So here’s the
voltage vo1 across resistor R2:

12. For Circuit C, you can use a current divider technique
because the resistors are connected in parallel with a current
source. The current source provides the following current i22
flowing through resistor R2:
You can use Ohm’s law to find the voltage output vo2 across resistor R2:
Now find the total output voltage across R2 for the two independent sources
in Circuit C by adding vo1 (due to the source voltage vs) and vo2 (due to the
source current is). You wind up with the following output voltage:

13. General Idea:
Nortons states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a current source
IN in parallel with a resistor RN, where IN is the short-circuit
current through the terminals and RN is the input or equivalent
resistance at the terminals when the independent sources are
turned off.
Norton’s Theorem:

14. 1. Find the norton current I(no) calculate the output current ,I(ab),with
a short circuit as the load. This is I(no).
2. Find the norton resistance R(no) when there are no dependent
sources there are two methods of determining the norton
impedence R(no)
(i) calculate the out put voltage , V(ab) when in open circuit
condition .R(no) equals this V(ab) divided by I(no).
(ii) Replace independent voltage with short circuit and independent
current sources with open circuits. This total resistance across the
output port is the norton impedence R(no)
Steps of Norton Theorem:

15. Conversion to a Thevenin’s
equivalent:

16. Mathematical problem: 1
Find the Norton equivalent circuit of the circuit at
terminals a-b.

17. Solution:
Solution:
We find in the same way we find
in the Thevenin equivalent
circuit. Set the independent
sources equal to zero. This leads to
the circuit from which we find RN
Thus,
RN = 5 || (8 + 4 +8) =5 || 20
=20 || 5
=20 *5/25
= 4

18. To find IN we short-circuit terminals
a and b, as shown in We ignore
the 5 ohm resistor because it
has been short-circuited.
Applying mesh analysis, we
obtain
i1 =2 A, 20-i2 -4i1 – 12=0
From these equations, we obtain
i2 =1 A=isc=IN

19. Mathematical problem: 2
To find the Nortons equivalent of the above circuit we firstly have to
remove the centre 40Ω load resistor and short out the terminals A and
B to give us the following circuit.

20. When the terminals A and B are shorted together the two resistors
are connected in parallel across their two respective voltage
sources and the currents flowing through each resistor as well as
the total short circuit current can now be calculated as:
If we short-out the two voltage sources and open circuit terminals A and B,
the two resistors are now effectively connected together in parallel. The value
of the internal resistor Rs is found by calculating the total resistance at the
terminals A and B giving us the following circuit.

21. Having found both the short circuit current, Is and equivalent internal
resistance, Rs this then gives us the following Nortons equivalent circuit.

22. Nortons equivalent circuit.
Ok, so far so good, but we now have to solve with the original 40Ω load
resistor connected across terminals A and B as shown below.

23. The voltage across the terminals A and B with the load
resistor connected is given as:
The voltage across the terminals A and B with the load resistor
connected is given as:
Then the current flowing in the 40Ω load resistor can be found as: