This document outlines various circuit analysis techniques including mesh analysis, nodal analysis, and network theorems. Section 1 discusses methods of analysis such as source conversion, mesh analysis using both general and format approaches, and nodal analysis using general and format approaches. Section 2 covers network theorems including superposition theorem, which is used to solve circuits by calculating the effect of individual sources and adding them together. Choosing an analysis method depends on whether the circuit has fewer meshes or nodes, as the goal is to generate the fewer simultaneous equations.

1. 1
Project
Subject :EE 201: Fundamentals of Electric Circuts
Prof. Sulaiman Almohaimeed
Student : Abdalmohsen Alabdali
ID:391110160

2. 2
Introduction
This project for first Methods of Analysis and how to solve circuit
and how to use mesh and nodel analysis and how to convert the
sources , second Network Theorems , Superposition theorem,
Thevenin’s theorem, Norton’s theorem, provide an opportunity to
determine the impact of a particular source or element on the
response of the entire system.
Table of content
Section 1: Methods of Analysis
1.1: Sourceconversion 3
1.2: Mesh analysis
1.2A: Mesh analysis (general approache) 7
1.2B: Mesh analysis (format approache) 12
1.3: nodel analysis
1.3A : nodel analysis (general approaches) 14
1.3B : nodel analysis (format approaches) 17
Sectionsummary 25
Section 2: Network Theorems
Section 2.1: Superposition theorem 26
Section 2.2: Thevenin’s theorem 28
Section 2.3: Norton’s theorem 31
Sectionsummary 33

3. 3
Setion 1: method of analysis
1.1: Source conversion
Basic idea
SOURCE CONVERSION : Convert a voltage source to current
source and vice versa makes the circuit simple.
Figure 1. An example of a DC source transformation. Notice
that the impedance Z is the same in both configurations. [1]
Note: the voltage source connected in series with a resistance,
and the current source is connected in parallel to the
resistance.
In reality, all sources—whether they are voltage sources or current
sources have some internal resistance . [2]

4. 4
FIG. 2 Practicalsources: (a) voltage; (b) current. [3]
Note: the internal resistance(RS , RP) must be there , the load connected to
the sources RL receive the same current, voltage, and power in both circuit .
Note: the resistance Rs=RP is same in each configuration.
Procedure
1-for voltage source,the voltage is determined by Ohm’s law E =IRP
And Rs in series .
2-For the current source determined by Ohm’s law: I =E/Rs , And Rp in
parallel.
3- check the polarity.
Note: here circuit the polarity of E same as current .

5. 5
Example:
[4]
Q1:a. Determine the current IL
b. Convert the voltage source to a current source.
c. Using the resulting current sourceof part (b), calculate the current through
the load resistor, and compare your answer to the result of part (a).
[3]
Solution:

6. 6
1.1A: CURRENT SOURCES IN PARALLEL
voltage sources of different terminal voltages cannot be placed in parallel
because of a violation of Kirchhoff’s voltage law. Similarly,current sources
of different valuescannotbe placed in series dueto a violation of
Kirchhoff’s current law.[3]
Procedure
1-Two or more currentsources in parallel replace it by a single current
source.
2-the currentin same direction add and opposite subtract.
3-RP equal to the total resistanceexcept R load .
Example:
Ex1:
[3]
Ex2:
Combining the parallel independent current sources into a single
equivalent source, we obtain the circuit:

7. 7
1.2:Mesh analysis (general and format approaches)
Introduction
The Mesh Current Method is well-organized methodfor solving a
circuit.(The other is the Node Voltage Method.)
The Mesh Current Method is basedon Kirchhoff's Voltage
Law (KVL).
1.2 A: Mesh analysis (general approaches)
Basic idea:
Mesh analysis: method is used to solve circuits for
the currents in the circuit.

8. 8
A loop is any closed path around acircuit. To trace a loop, you start at
any componentterminal, and trace a path through connected elements
untilyou get back to the starting point. A loop is allowed to go through
an elementjust one time . In the circuit above, there are three loops,
two solid loops, I, and II, and onedashed loop III all the way around the
outside.
If we trace the loopsin the clockwise direction, the three loopsin our
circuit go through
Loop I: V1 - R1 - R3 Loop II: R3 - R2 - V2
Loop III: (dashed loop):V1 - R1 - R2 - V2
A mesh is a restricted kind of loop; a meshis a loop that contains no
other loops. In the circuit above, loops , I, and II are meshes because
there are no smaller loops inside.The dashed loop is not a mesh,
because it contains two other loops. [5]
the number of mesh currents required to analyzea network will equal to
the number of “windows” of the configuration. Here “two-window”
network in figure .[2]

9. 9
Procedure
1- Assign a current to each mesh, using a clockwise direction (not
necessary in clockwise).
2- Indicate the polarity.
3- Write Kirchhoff's VoltageLaw around each mesh.
4- If two loop currents flow in opposite directions in a resistor, the voltage
goes in as R×(iloop1−iloop2)(if is in the samedirection plus instead of a
minus.).
How you find polarities?
When currentpass E (BATTERY ) currentin – sign currentout +
And when currentpass R(resistor ) currentin + sign currentout -.
EXAMPLE
[5]
Solution
Note : if I have two loop there is two equation.

10. 10
Supermesh Currents
What is Supermesh analysis?
Super mesh or Supermesh Analysis is a better technique
instead of using Mesh analysis to analysis such a complex
electric circuit or network, where two meshes have a current
source as a common element. ... Alternatively, KVL
(Kirchhoff's Voltage Law) is applied only to those meshes or
super meshes in the renewed circuit.[6]
Properties of a Supermesh
1. The current is not completelyignored.
provides the constraint equation necessaryto
solve for the mesh current.
2. Several current sources in adjacency form a
biggersupermesh.[7]
Note: use supermeshif the current sources in circuit without parallel
resistorso you cant convert to voltage source .
Procedure
1-removethe current source(changeit to open circuit )
2-solvethe loops as mesh
3- then useKCL
Example:
[7]

11. 11
Solve the circuit
VS=10V, IS=4AIS=4A, R1=2Ω R2=6Ω,R3=1Ω, R4=2Ω.
Solution::
Around the loop clockwise:
R1×(I2−I1)+R2×(I2)+R4×(I3)+R3×(I3−I1)=0.
As you can see, we were able to write the equation in one shot. That is why
the supermesh method is preferred.
Now, we have two equations: one for Mesh I and one for the supermesh. But
there are three unknowns: I1, I2 and I3. So we need another equation. The
third equation comes from the current source by writing KCL one of its
nodes. We choosethe nodewhich is not shared by third loop which is the
loop at the right hand side for this example. This way we minimize the
number of terms in the equation. Note that the current
of R2 andR4 are I2 and I3, respectively, but the terms for R2 and R3 are
more complicated because of I1 involvement.
Let's apply KCL for the right hand side node. I2 and IS are entering to the
node and I3 is leaving.
−I2−IS+I3=0−I2−IS+I3=0
Now we have all three equations:
−VS+R1×(I1−I2)+R3×(I1−I3)=0
R1×(I2−I1)+R2×(I2)+R4×(I3)+R3×(I3−I1)=0

12. 12
−I2−IS+I3=0
Let's substitute values:
VS=10V, IS=4A, R1=2Ω, R2=6Ω, R3=1Ω, R4=2Ω.
2(I1−I2)+(I1−I3)=10
2(I2−I1)+6I2+2I3+I3−I1=0
−I2+I3=4
3I1−2I2−I3=10
−3I1+8I2+3I3=0
−I2+I3=4
4.9166
I1=4.92A I2=0.25A I3=4.25A. [8]
1.2 B: Mesh analysis (format approaches)
Basic idea : examine a technique for writing the mesh
equations more rapidly and usually with fewer errors.[2]
Note :the format approach can be applied only to
networks in which all current sources have been
converted to their equivalent voltage source. [2]
Mesh(format ) Analysis Procedure
1- Assign a loop current in a clockwise direction.
2- Solve first current.
3- Take all resistance in first loop then multiply with first current .
4- Subtract with common resistance multiply with the other current
pass from it .
5- The equal to the volt
6-check the polarity
7-solve the second current same current one .
Note: number equations is equal to the number loops.

13. 13
Example
[3]
Solution:
I1=(8+6+2)I1-(2)I2=4v
I2=(7+2)I2-(2)I1=-9v
I2=-0.97A I1 =0.13A
1.3:nodel analysis (general and format approaches)
Basic idea: The Node Voltage Method is an organized methods of
analyzinga circuit. is based on Kirchhoff's Current Law (KCL).
Aim of Nodal Analysis:
The aim of nodal analysis is to determine the voltage at each node relative to the
reference node (or ground). Once you have done this you can easily work out anything
else you need.[9]
Definition: node voltage

14. 14
Weneed to definea new term: node voltage. When weusethe term node
voltage, wearereferringto the potential differencebetween two nodes
of a circuit.
Weselect one of the nodesin our circuit to be the reference node. Allthe
other nodevoltages are measured with respect to this onereference
node.
The referencenodeis almost always called the ground node, and it gets
a ground symbolin the schematic, The potential of the ground nodeis
defined to be 0V The potentials of all the other nodesare measured
relative to ground.[10]
1.3A : nodel analysis (general approaches)
Procedure
•Define a voltage at every node in the circuit
Write KCL at the nodes where the unknown voltages exist.
•Now, plug into these KCL equations with the unknown voltages,
remembering how Ohm’s Law works. In this case,I = (VH–VL)/R
because we are writing voltages for nodes,not just resistors.Since
current flows from a higher potential to a lower potential, the voltage
over a resistor that is connected to 2 nodes is just
V H -VL
•Other current and voltage sources must be factored in to either the
KCL equations or the unknown voltages. They sometimes actually
make the equations easier.
•Solve for the unknown voltages. [11]
example:

15. 15
solution:
[3]

16. 16
Example 2:
SUPERNODE:
voltage sources in a network that do not have a series
internal resistance to permit a conversion to a current
source.[3]
Procedure
Start as usual and assign a nodal voltage to each independent node
of the network, including each independent voltage source as if it
were a resistor or current source. Then mentally replace the
independent voltage sources with short-circuit equivalents, and
apply Kirchhoff’s current law to the defined nodes of the network.
Any node including the effect of elements tied only to other nodes
is referred to as a supernode (since it has an additional number of
terms). Finally, relate the defined nodes to the independent voltage
sources of the network, and solve for the nodal voltages.[2]

17. 17
Example:
[3]
1.3B : nodel analysis (format approaches)
nodal analysis can also be a very useful technique for solving
networks with only one source.[2]
Procedure
1-assign thenode
2- solve the first node
3-takeall resistanceconnected with node and multiply with voltage of the
node.
4- then subtracted with common resistancemultiply it with other node
equal to current
5-solvesecond node samesteps
Note :A major requirement, however, is that all voltage sources
must first be converted to current sources before the procedure is
applied.[2]

18. 18
Example 1:
Solution:
[12]
Note: the current in first eq because the current going
out the node .
Other example of nodel analysis
Solve the circuit by nodal analysis and find Va.

19. 19
Solution
a) Choose a reference node, label node voltages:
b) Apply KCL to each node:
Node 1:
−Is2+(V1−V2)/R3+(V1−V3)/R1=0→6V1−V2−5V3=5 (1)
Node 2:
Is3+(V2−V1)/R3+V2/R2=0→−4V1+9V2=−40 (2)
Node 3:
-Is1−Is3+(V3−V1)/R1=0→V3−V1=4 (3)
(1), (2) and (3) imply that V1=37v,V2=12v and V3=41.
c) Find the required quantities:

20. 20
If we apply KVL in the loop shown above:
−V1−Va+V2=0→Va=−25v .[13]
Example:V1, V2, V3 as extraordinary nodes, V3 is in reference to
(ground)
b. ApplyKCL at node 1 and 2
Node 1
Therefore

21. 21
Node 2
i4, i5, and i6
Therefore
Solve simultanousequations
Simultanousequation solution
. [14]

22. 22

23. 23

24. 24
Example:Aplyingnodalanalysisto find the nodevoltage V in the
followingcircuit
Solution
[14]

25. 25
Section summary
All sources –whether they are voltage or current –have some internal
resistance.
Source conversions are equivalent only at their
external terminals.
Source conversion: means change one sourceby an
equivalent source.
Current Source in Parallel
Connecting two or more current sources in parallel is equal to one current
source whose total current is given as sum of source currents.
What is the difference between loop and mesh?
A loop is a closed path in a circuit where two nodes are not traversed twice
except the initial point, which is also the final one. But in a loop other paths can
be included inside. ... Mesh: A mesh is a closed path in a circuit with no other
paths inside it. In other words, a loop with no other loops inside it .[15]
NODAL ANALYSIS weuseKCL ,MESH ANALYSIS WEUSEKVL
The Mesh method :find currents of the network.
nodal analysis:find method that the nodal voltages of a network.
If I haveresistor pass through 2 currenthow I can calculate ?
the current of the loop being examined plus or minus the other loop currents
as determined by their directions. If clockwise applications of Kirchhoff’s
voltage law, the other loop currents are always subtracted from the loop
current of the loop being analyzed.
Choosing a method
Now we have two efficient methodsfor analyzingcircuits, NodeVoltage
Method and Mesh CurrentMethod. Which is the best oneto usein a
given situation? To choose between the two methods, countthe number
of meshes in the circuit and comparethat to the number of nodes.

26. 26
Which number is smaller, meshes or nodes? It is usually best to choose
the method that generates fewer simultaneousequations. If the meshes
and nodesare the same, or nearly the same, you can choose the method
you understand thebest.[16]
Section 2: Network Theorems
Section 2.1: Superposition theorem
Basic idea: The superposition theorem is used to solve the
circuit where two or more sources are present and
connected (not in series or parallel) to calculate the effect of
individual sources then added toghter .
Procedure
To solve a circuit using superposition,the first step is stop all
source except one source.
 To stop a voltage source,replace with short circuit.
 To stop a current source, replace with open circuit.
Then analyze the resulting circuits. Do it again for all source.
The final result is the sum of individual results.
Note : only one source at a time.
Note : currents in the same direction are added, and current having the
oppositedirection are subtracted.

27. 27
FIG. Removing a voltage source and a currentsource to permit the
application of the superposition theorem. [3]
Example:
Using the superposition theorem, determine the voltage drop and current
across the resistor 3.3K as shown in the figure below.
Solution:

28. 28
Superposition cannotbe applied to power effects becausethe power is
the square of the voltage across a resistor or the current through
a resistor P=V2 /R or P=I2R .
Section 2.2:Thevenin’s theorem
Basic idea:
Thevenin’s Theorem : simplify the circuit no matter how complicated , any
circuit containing oneor morevoltages soruceand resistances can be
changeit to just one single voltage in series with a single resistance
connected with the load.

29. 29
[3]
The theorem says that the entire circuit inside the blue area change it to one
voltage sourceand one resistor as shown in Fig.
Note: the new circuit equal to the old one .
Note : If the change is done the voltage and the current through, resistor RL
will be same.
Note: R load is series with RT
Note: maybe the circuit currentsourceor voltage source .
Procedure
1-put terminal a and b around RL
2. Remove the load resistor RL
3-voltage sources arechange it to shortcircuits and currentsources to
open circuits.
4-calculate Rt from the terminal a and b(to know series or parallel )
5-return the source
6- calculate Et
7- return R load and redraw the circuit

30. 30
EXAMPLE :
[17]
Solution:-
Note:that there is no requirement that the Thévenin voltage have the same
polarity as the equivalent circuit.[2]

31. 31
Section 2.3: Norton’s theorem
Basic idea:
Norton's Theorem:simplifythe circuit even if its complicated to an
equivalent circuit with just single current source and parallel
resistance connected to a load.
Note : RN= RTh
Note: R load and RN parallel
Note: RN parallelwith In and parallel to load
[3]
The Norton and Thévenin equivalent circuits can also be found from
each other by using the source conversion.
Procedure
1-putterminala and b around RL
2. Removethe load resistor RL
3-voltagesources are changeit to shortcircuits and currentsources to open
circuits.

32. 32
4-calculateRN from the terminal a and b(to know series or parallel )
5-return thesource
6- calculate IN Note that this step is exactly oppositethe step in Thevenin’s
Theorem, where we replaced the load resistor with a (open circuit) here
shortcircuit
7- return R load and redrawthe circuit
Example:
Find RN, IN, the current flowing through and Load Voltage across the load resistor in fig by
using Norton’s Theorem.
[17]

33. 33
Solution:
Note : the polarity of the current same as the old source .
Section summary:
What is a Superposition Theorem?
The superposition theorem is a method for the Independent supplies
present in an electrical circuit (not in series or parallel )like voltage &
current and that is considered as one supply at a time.[18]
Superposition theorem can't be used for powercalculations due to
nonlinearity of power relationship with voltage and currents .[3]
Thevenin’s theorem :Any circuit (not in series or parallel )
can be reduce to Thevenin's equivalent circuit consist of
single voltage source and series resistance connected to a
load.

34. 34
Norton'sTheorem and Thevenin’s theorem have almost the same idea.
[3]
Norton'sand Thevenin’s circuit can easily interchange.
the difference betweenThevenin and Norton Theorem
Norton's theorem use current source,Thevenin's theorem use
voltage source. Norton's theorem use a resisterin parallel with the
source and parallel with load .Thevenin'stheorem use resistor in
series with the source and series with load.
Conclusion
In the end I hope the projectwas clear and all the idea explain very good
and you enjoy with it .

35. 35
References
[1]: https://blog.oureducation.in/source-conversion/
[2]: INTRODUCTORY CIRCUIT ANALYSIS
[3]: fromthe slide of the subject
[4]:http://electricala2z.com/electrical-circuits/source-transformation-example-problems-
solutions/
[5]:https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-
dc-circuit-analysis/a/ee-mesh-current-method
[6]:https://www.quora.com/What-is-supermesh
[7]:https://www.slideshare.net/jigneshprajapati9/planing-and-shaping?from_action=save
[8]: http://www.solved-problems.com/circuits/electrical-circuits-problems/resistive-
circuits/1226/mesh-analysis-supermesh/
[9]: http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/00300_Nodal.pdf
[10]: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-
dc-circuit-analysis/a/ee-node-voltage-method
[11]: https://studylib.net/doc/5729139/lecture-4--methods-of-analysis
[12]: https://electronics.stackexchange.com/questions/59227/format-approach-to-nodal-
analysis
[13]: http://www.solved-problems.com/circuits/electrical-circuits-problems/369/solving-nodal-
analysis-circuit-nodes/
[14]: https://www.cpp.edu/~elab/projects/project_06/index.html
[15]: https://www.differencebetween.com/difference-between-loop-and-vs-mesh/
[16]: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-
dc-circuit-analysis/a/ee-mesh-current-method
[17]: https://www.electricaltechnology.org/2014/01/thevenins-theorem.html
[18]: https://www.elprocus.com/superposition-theorem-example-problems-limitations-and-
applications/