Week Nine: "Superposition"

What is Superposition Theorem?

       If a circuit has a two or more independent sources, one way to determine the value of a specific (voltage or current) is to nodal or mesh analysis. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition. The idea of superposition rests on the linearity property.

       The superposition principle states that the voltage across (or current through) an element in a linear circuit is the voltages across (or currents through) that element due to each independent source acting alone.

       The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However, to apply the superposition principle, we must keep two things in mind:

1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.
2. Dependent sources are left intact because they are controlled by circuit variables.

       With these in mind, we apply the superposition principle in three steps:
Steps to Apply Superposition Principle:

1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the independent sources.

Example:

Find the voltage Vx using superposition.

 

The voltage Vx can be found by solving three subproblems (one circuit for each source).

Subproblem 1:


       The first subproblem is shown below, where only one source is kept in the circuit. The current source is opened and the other voltage source is shorted.

       Now solve for Vx due to the 2V. Note that no current can flow through the 6 ohm resistor, since there is no return path. (Draw a supernode around 6, 7, and 8 to see this). Thus we can reduce the circuit to the following.

Now use a voltage divider to find Vx.

So Vx due to the 2V is (4/9)2V or around 0.89V.


Subproblem 2:
The second subproblem is shown below, where only one source is kept in the circuit. The current source is kept and the two voltage sources are shorted.

       Although we could do some simplification (the 7 and 8 are in parallel, and that parallel combination is then in series with the 6), this is not necessary. The 1A will flow through that combination regardless and then divide between the 4 and 5. To see this, redraw the circuit and imagine a supernode as shown:

       If 1A flows into the supernode to the right on the bottom, it must flow out of the supernode to the left from the top. Since 4 and 5 are in paralle, we can apply a current divider to find how much current goes through the 4 ohm (let's call it Ix).

       Note that current divider normally requires using conductances, but with just two parallel resistors, we can use the short cut. The resistance of the other branch goes on top of the fraction, and the total resistance goes in the bottom. Thus the current going left through the 4 ohm is (5/9)1A or around 0.56A. To find Vx, use Ohm's law, V=IR=-(0.56A)(4ohm). Thus Vx due to the 1A is -2.22V. The negative sign is because our Ix flowed into the negative side of Vx first.


Subproblem 3:

       The third subproblem is shown below, where only one source is kept in the circuit. The voltage source on the right is kept, the current source is opened, and the voltage source on the left is shorted.

       Note that no current can flow through the 6 ohm, as we can see by drawing a supernode:

       If current were flowing to the left through the 6, there is no other path for it to return. Thus the current must be zero. If the current into 4 and 5 is zero, then Vx due to the 3V must be 0. 

Final Solution:
To find the total Vx, add the answers from each subproblem:
Vx = 0.89V -2.22V + 0V = -1.33 V



Some Learnings:

• In using the Superposition, we replace the voltage source with a short circuit on its place, and an open circuit when we replace the current source.

• We turn off other independent sources except one source.

• After converting the circuit by superposition, we can use nodal and mesh analysis or any other techniques to get the value of the output voltage or current

• We add the two results to get the answer.

• Example: Vo = V1 + V2  or  Io = I1 + I2

Video:
       For more information, you can watch the video below!

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP - EE

Week Eight: "Linearity Property & Source Transformation"

What is a Linearity Property? 

Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property. The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called the response) is multiplied by the same constant. For a resistor, for example, Ohm’s law relates the input i to the output v,

V= iR.

If the current is increased by a constant k, then the voltage increases correspondingly by k, that is,

kiR = kV

 The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately. Using the voltage-current relationship of a resistor, if

V1 = i1R

and

V2 = i2R 

then applying (i1+i2) gives,

V = (i1+i2)R = i1R + i2R = V1 + V2

We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties.

In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources.

A Linear Circuit is one whose output is linearly related (or directly proportional)  to its input.


Example:

What is a Source Transformation?

We have noticed that series-parallel combination and wye-delta transfor-mation help simplify circuits. Source transformation is another tool for simplifying circuits. Basic to these tools is the concept of equivalence.

We recall that an equivalent circuit is one whose v-i characteristics areidentical with the original circuit.

Before, we saw that node-voltage (or mesh-current) equa- tions can be obtained by mere inspection of a circuit when the sources are all independent current (or all independent voltage) sources. It is therefore expedient in circuit analysis to be able to substitute a voltage source in series with a resistor for a current source in parallel with a resistor, or vice versa. Either substitution is known as a source transformation.

Figure 1:

 

A Source Transformation is the process of replacing a voltage source Vs in series with a resistor R by a current source Is in parallel with a resistor R, or vice versa.

Source transformation requires that,

Vs = IsR     or     Is = Vs/R

Source transformation also applies to dependent sources, provided we carefully handle the dependent variable. As shown in below, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa.

Figure 2:

 

Like the wye-delta transformation we studied in Chapter 2, a source transformation does not affect the remaining part of the circuit. When applicable, source transformation is a powerful tool that allows circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation.

1. Note from Fig.1 (or Fig.2) that the arrow of the current source is directed toward the positive terminal of the voltage source.
2.  Note that source transformation is not possible when R=0, which is the case with an ideal voltage source. However, for a practical, nonideal voltage source, R=0. Similarly, an ideal current source with R=∞ cannot be replaced by a finite voltage source. 

Example:

Some Learnings:

• Linearity Property talks about the voltage (v) and current (i), whereas voltage is directly proportional to the current, that is, when the voltage is increasing the current also increasing and vice  versa.
• Source Transformation is only applicable to simple circuits like the figures below:
  
  
  
  

  but is not applicable to complicated circuits like figures below:
  

  
  
• Source Transformation can be applied if and only if the voltage is in series with the resistor and/or the current is in parallel with the resistor.
• In Source Transformation when the polarity of the voltage source has a positive on top therefore the resulting current source is pointing upward and vice versa. Also when the voltage source is dependent therefore the resulting current is also dependent, same goes with independent sources.

Videos:

For more information, watch the videos below:

For Linearity Property:

For Source Transformation:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP - EE

Week Seven: "Mesh Analysis"

What is a Mesh Analysis?

Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it.

Nodal analysis applies KCL to find unknown voltages in a given circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches.

To understand mesh analysis, we should first explain more about what we
mean by a mesh.

In Figure below, for example, paths abefa and bcdeb

are meshes, but path abcdefa is not a mesh. The current through a mesh is
known as mesh current. In mesh analysis, we are interested in applying KVL to 

find the mesh currents in a given circuit.

In this section, we will apply mesh analysis to planar circuits that do not
contain current sources. In the next sections, we will consider circuits with 
current sources. In the mesh analysis of a circuit with n meshes, we take the 
following three steps.

Steps to Determine Mesh Currents:

1. Assign mesh currents i1,i2,...,into the n meshes.

1. Apply KVL to each of then meshes. Use Ohm’s law to express the
   voltages in terms of the mesh currents.

1. Solve the resulting n simultaneous equations to get the mesh currents.
   
   

Mesh Analysis Without Current Sources:

Example:

 

Now write KVL equations for each loop.

@Mesh 1:

-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0

then gather terms:

10i1 - 5i2 - 4 i3 -18V = 0

Note that the i1 term is positive, and all other current terms are negative (because 

they are all clockwise, all other panes will contribute a negative term). Let's do the 

other two panes with terms gathered up directly (write the total resistance of the 

loop multiplied by the mesh current that goes through that total resistance):

@Mesh 2:

-5i1 + 10i2 - 3i3 - 12 = 0

@ Mesh 3:

-4i1 -3i2 +9i3 = 0

Now solve the three equations in three unknowns:

i1 = 7.02A
i2 = 6.28A
i3 = 5.21A

The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current 

through the 18V is i1, or 7.02A. All the branch currents are shown below from a 

Pspice simulation:

 

Mesh Analysis With Current Sources:

Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what 
we encountered in the previous section, because the presence of the current 
sources reduces the number of equations. Consider the following two possible 
cases.

Case 1:

When a current source exists only in one mesh. We set

mesh current i equal to that current source.

Case 2:

When a current source exists between two meshes. We create a supermesh 
by excluding the current source and any elements connected in series with 
it.

Examples:

Case1:

First, define a mesh current arround each mesh (window pane) of the circuit. 
Define each one in a clockwise direction.

Now write KVL equations for each loop. On the first loop, we run into a problem. 
We do not know the voltage across the 5A source. (This is one of the difficulties 
you can run into when using mesh current.) We add a new unknown to handle this problem. We run into the same problem with the 2A source, so two new unknowns 
 (V5 and V2) are labeled:

Now we are ready to write the KVL equations.

@Mesh 1:

-V5 + 5(i1-i2) = 0

@ Mesh 2:

5(i2-i1)+10(i2-i4)+2(i2-i3) = 0

or

-5i1 + 17i2 - 2i3 = 0

@Mesh 3:

2(i3-i2) + V2 = 0

@Mesh 4:

10i4 + 4V + 10(i4-i2) = 0

or

-10i2 + 20i4 = -4

We have two extra unknowns, so we need two more equations. These can be 

found by examining the loops with the sources:

i1= 5A

i3 = -2A

We can then solve for the remaining unknowns:

i2 = 1.58A
i4 = 0.59A

We can then find the current through the 5 ohm resistor as:

I = i1 - i2 = 3.42A

The current through the 4V source is simply i4, or 0.59A.

Case 2:

Determine the values of the mesh currents, i1and i2, for this circuit.

Solution:

We can write one mesh equation by considering the current source. The current
source current is related to the mesh currents at by

(KCL)

 i1 - i2 = 1.5  or  i1 = 1.5 + i2

In order to write the second mesh equation, we must decide what to do about the
current source voltage. (Notice that there is no easy way to express the current
source voltage in terms of the mesh currents.) In this example, illustrate two 
methods of writing the second mesh equation.

Apply KVL to the supermesh corresponding to

the current source. Shown below in

blue, this supermesh is the perimeter of the two meshes that each contain the
current source.

 

 @ Supermesh:

9i1 + 3i2 + 6i2 - 12 = 0   or   9i1 + 9i2 = 12

Solving the equations gives:

i1 = 1.4167A   and   i1 = -83.3mA

Some Learnings:

• Mesh Analysis uses KVL to obtain unknown mesh currents, while Nodal Analysis uses KCL to obtain unknown nodal voltages.
• Mesh Analysis is only applicable to planar circuits. Planar circuits are circuits that can be drawn on a plane surface, with no wires crossing each other.
• Mesh Analysis is different from Loop Analysis. Loop Analysis is applicable to both planar or not. Both Mesh & Loop Analysis make use of Kirchhoff's 
• Voltage Law or KVL.
• Mesh Analysis is usually easier to use when the circuit is planar, compared to Loop Analysis.

Video:

       For more information, watch the video below:

 

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP - EE

Week Six: "Wye - Delta Transformations"

What is Wye - Delta Transformations?

     This week we talked about Wye - Delta Transformations. This topic is very important. Sometimes we are not sure in electric circuits  that the resistors are neither parallel or series. In many circuit applications, we encounter components connected together in one of two ways to form a three-terminal network: the “Delta,” or Δ (also known as the “Pi,” or π) configuration, and the “Y” (also known as the “T”) configuration.

     It is possible to calculate the proper values of resistors necessary to form one kind of network (Δ or Y) that behaves identically to the other kind, as analyzed from the terminal connections alone. That is, if we had two separate resistor networks, one Δ and one Y, each with its resistors hidden from view, with nothing but the three terminals (A, B, and C) exposed for testing, the resistors could be sized for the two networks so that there would be no way to electrically determine one network apart from the other. In other words, equivalent Δ and Y networks behave identically.

There are several equations used to convert one network to the other:

     Δ and Y networks are seen frequently in 3-phase AC power systems (a topic covered in volume II of this book series), but even then they're usually balanced networks (all resistors equal in value) and conversion from one to the other need not involve such complex calculations. When would the average technician ever need to use these equations?

     Solution of this circuit with Branch Current or Mesh Current analysis is fairly involved, and neither the Millman nor Superposition Theorems are of any help, since there's only one source of power. We could use Thevenin's or Norton's Theorem, treating R₃ as our load, but what fun would that be?

     If we were to treat resistors R₁, R₂, and R₃ as being connected in a Δ configuration (R_ab, R_ac, and R_bc, respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit:

After the Δ-Y conversion . . .

     If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration.

Some Learnings:

• “Delta” (Δ) networks are also known as “Pi” (π) networks.
• “Y” networks are also known as “T” networks.
• Δ and Y networks can be converted with the proper resistance equations. By “equivalent,” I mean that the two networks will be electrically identical as measured from the three terminals (A, B, and C).
• A bridge circuit can be simplified to a series/parallel circuit by converting half of it from a Δ to a Y network. After voltage drops between the original three connection points (A, B, and C) have been solved for, those voltages can be transferred back to the original bridge circuit, across those same equivalent points.

Video:

     For more examples & information, you can watch the video below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP