Some Useful Techniques of Circuit Analysis

5833v1-0.333v2-0.2500v3 (1)
And the super-node:
3+ 25 = (v2- v1)/3 + (v3- v1)/4 + v2/1 + v3/ 5
	Or 
28 = -0.5833v1+1.3333v2+0.45v3 (2)
	And we have 
	v3- v2= 22V (3)
	Solving that we find v1= 1.071V
2.2. Mesh analysis
	Method:
Make certain that the network is a planar network. If it is non-planar, mesh analysis is not applicable.
Make a neat, simple circuit diagram. Indicate all element and source values. Resistance values are preferable to conductance values. Each source should have its reference symbol.
Assuming that the circuit has M meshes, assign a clockwise mesh current in each mesh, i1, i2,, iM.
If the circuit contains only voltage sources, apply KVL around each mesh. To obtain the resistance matrix if a circuit has only independent voltage source, equate the clockwise sum of all the resistor voltages to the counterclockwise sum of all the source voltages, and order the term, from i1 to iM. For each dependent voltage source present, relate the source voltage and the controlling quantity to the variables i1, i2,, iM, if they are not already in that form.
If the circuit contains current source, create a super-mesh for each current source that is common to two meshes by applying KVL around the larger loop formed by the branches that are not common to the two meshes; KVL need not be applied to a mesh containing a current source that lies on the perimeter of the entire circuit. The assigned mesh currents should not be changed. Relate each source current to the variables i1,i2,, iM, if it is not already in that form. 
Example:
Without current source
Determine i1, i2 in fig.
Apply KVL at mesh 1:
	-42 + 6i1 + 3(i1-i2) = 0 or 9i1-3i2=42 (1)
And the mesh 2:
	3(i2-i1) + 4i2 – 10 = 0 or -3i1 + 7i2 = 10 (2)
Solving 2 equations, we find i1= 6A, i2= 4A.
With current source (super- mesh)
Determine the i1,i2,i3
Here we note that a 7A independent current source is in the common boundary of 2 meshes. Mesh currents i1,i2,i3have already been assigned, and the current source leads us to create a super- mesh whose interior meshes i1and i3. 
Apply KVL in super-mesh(1-3): 
	-7 + 1(i1-i2) + 3(i3-i2) +1i3 = 0 or i1- 4i2 +4i3 = 0 (1)
And around mesh 2:
	1(i2-i1) + 2i2 + 3(i2-i3) =0 or -i1+6i2-3i3=0 (2)
	We have i1-i3=7 (3)
Solving 3 equations, we find i1= 9A, i2= 2.5A, i3= 2A
Problem:
Determine all 4 nodes voltage:
Naming: node 1: top left node, node 2: top right node, node 3: the bottom node of the 6V source, node4: top node of the 2Ω resistor
By inspection, v2= 5V
Forming super node with node 1&3 we have: 
At the super node: -2 = v3/1 + (v1-5)/10
At node 4: 2 = v4/2 + (v4-5)/4 
Rearranging, simplifying and collecting terms;
 v1+ 10v3 = -20 + 5= -15 and v1- v3=6 
We have 3 equations:
0.1v1+v3 = - 1.5 (1)
 0.75v4 = 3.25 (2)
 v1- v3 = 6 (3)
Solving (1), (2), (3), we find v1= 4.091V, v3=-1.909 V, v4=4.333V
Find the power supplied by the 2.2V source
Solution
We have 4 currents i1, i2,i3,i4 like figure:
We have super-mesh 3&4.
Apply KVL for super-mesh (3,4) we have:
-3 + 3i4 + 9(i4-i1) + 4(i3-i2) + 6i3+1i3 =0
For mesh 1:i1 = -4.5 A
For mesh 2 : 5 + 4(i2-i3) + 2.2 + 3i2 = 0
With super – mesh (3&4) we have equation:
i4-i3= 2A
Summary, we have 3 equations:
-4i2+ 11i3+ 12i4 = -37.5 (1)
7i2- 4i3 = -7.2 (2)
 -i3 + i4 = 2 (3)
	Solving 3 equations, we find i2=-2.83A,i3=-3.16A,i4=-1.16A
	So the power supplied by the 2.2V source is : 
P(2.2V)= UI2=2.2(i1-i2)2= 6.14 W
2.3. Linearity property 
 Linearity is the property of an element describing a linear relationship between cause (input, excitation) and effect (output, response) 
 Linearity property combines: 
+ Homogeneity (scaling) property: v = iR à kiR = kv
+Additivity property: v1= i1 R àv=v1+v2=(i1+i2)R
	 v2= i2 R
Ex: Resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and additivity properties. 
A linear circuit is one whose output is linearly related (or directly proportional) to its input. 
 A linear circuit consists of only linear elements, linear dependent sources, and independent sources. 
2.4. Superposition
If circuit has two or more independent sources, there several ways to determine the value of a specific variable(voltage,current):
-Use nodal or mesh analysis.
-Superposition approach: Determine the contribution of each independent source to the variable, and then add them up.
 The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents though) that element due to each independent source acting alone.
Superposition is not limited to circuit analysis but is applicable in many field where cause and effect bear a linear relation ship to one another.
Step to apply superposition principle:
 -Turn off all independent sources except one source (dependent sources are left intact):
-Replace voltage source by short circuit
-Replace current source by open circuit
Find the output (voltage or/and current) due to that active source (using nodal or mesh analysis)
-Repeat step 1&2 for each of the other independent sources.
Find the total contribution by adding algebraically all the contributions due to the independent sources.
Ex : Using the superposition theorem, find v0. 
Since there are two sources, let: v0=v01+v02 
Applying KVL to the loop gives: (3+5+2)i = 20ài = 2A à v01=2.2=4 V
Using current division gives:
 IR1= 8 2+3+55=4A à v02=2. 4=8 V
Finally,we find v0=v01+v02=4+8=12 V
2.5. Source transformation
Similar to the series-parallel combination and wye-delta transformation, source transformation is using to simplify circuits that bases on the concept of equivalence.
An equivalentcircuit is one whose v-I characteristics are identical with theoriginal circuit.
A source transformationis the process of replacing a voltage source vsin series with a resistor R by a current source isin parallel with a resistor R, or vice versa.
Notes: 
-The arrow of the current source is directed toward the positive terminal of the voltage source. 
- The source transformation is not possible when R = 0 (ideal voltage source) or R = ∞ (ideal current source). 
Ex 4.9: Use source transformation to find ix
We transform: 
Dependent voltage source: is= 2ix5=0.4ix
Applying KCL gives:
Ix=(4-0.4ix)=R2R1+R2= 43 - 0.4ix3
à3.4ix= 43 àix=0.39A
2.6. Thevenin’s and Norton’s theorem
Thevenin’s theorem
A linear two terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor R
Statement of theorem:
Given any linear circuit, rearrange it in the form of two networks A and B that are connected by two wires. A is the network to be simplified, B will be left untouched .
Disconnect network B. Define a voltage Voc as the voltage now appering across the terminals of network A
Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged
Connect an independent voltage source with value Voc in series with the inactive network. Do not complete the circuit; leave the two terminals disconnected
Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged
Example: Use Thevenin’s theorem to determine the Thevenin equivalent for that part of the circuit in figure to the left of RL 
Disconnect B, the voltage across the connecting terminal is Voc:
Voc =1263+6=8 (V)
The independent source has been killed,
We see 7Ω resistor connect in series with the parallel combination of 6 Ω and 3 Ω, so we have Thevenin equivalent resistance of network A
RTh =3×63+6+7 =9 (Ω)
The Thevenin equivalent then is Voc = 8V in series with a 9 Ω resistor:
Norton’s theorem
A linear two terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN
Statement theorem
Given any linear circuit, rearrange it in the form of two network A and B connected by two wires
Disconnect network B, short the terminals of A. define a current isc as the current now flowing through the shorted terminals of network A
Turn off every independent source in network A to form an inactive network
Connect an independent current source with value isc in parallel with the inactive network.
Connect network B to the terminals of the new network A
Example: Find the Norton equivalent circuits for the network faced by the 1k Ω resistor
At the beginning, same the Thevenin theorem, we have
Voc = 4V + 2mA x 2k Ω = 8V
Next, we kill all independent source
So we can determine RTh = 3 +2 = 5 k Ω
Isc =VocRth= 85=1.6A
Finally, we have Norton equivalent with Isc = 1.6A parallel with RTh = 5k Ω in network A:
Exercise 1: Determine the Thevenin and Norton equivalents of the network show in figure
Apply nodal analysis to determine VTh:
100 x 10-3 = Vx/250 + Voc/7.5 x 103 (1)
Ix = Vx/250 
Vx – Voc = 5ix
→ (49/50)Vx – Voc = 0 (2)
From (1) and (2), we find that 
Vx = 24.21 V
Voc = VTh = 23.72 V
In order to determine RTh , we inject 1 A into the port a,b:
Vab/7.5 x 103 + Vx/250 = 1 (1)
And Vx – Vab = 5ix = 5Vx/250 or
-Vab + 49/50Vx = 0 (2)
From (1) to (2), we find that Vab = 237.2 V.
So RTh = Vab/1 → RTh = 237.2 Ω
IN = VTh/RTh = 100mA
Finally, we have Thevenin equivalent with VTh = 23.72 V in series with RTh=237.2 Ω
Norton equivalent with IN = 100mA parallel with RTh = 237.2 Ω
Exercise 2: find the Norton equivalent of the network show on figure
We inject a current of 1 A into the port (arrow pointing up), select the bottom terminal as our reference terminal, and define the nodal voltage Vx across the 200-Ω resistor
Then, 1 = V1/100 + (V1 – Vx)/50 (1)
 -0.1 V1 = Vx/200 + (Vx - V1)/50 (2)
Which may be simplified to
 3 V1 - 2 Vx = 100
16 V1 + 5 Vx = 0
Solving, we find that V1 = 10.64 V, so RTH = V1/1(A) = 10.64Ω
Since, there are no independent sources present in original network, IN = 0