Instantaneous Power

The second chapter of this semester has been opened! Now we have discussed about instantaneous power. We were first had our insights about what instantaneous power is. Unlike DC power, instantaneous power is only present in an AC powered circuit. Alternating power or AC power by definition is a power that is changing by a definite amount over a period of time. And in every point in time, there is a particular designated power as well.




We begin our exploration of sinusoidal power calculations with the genaric circuit in the following figure. In here, v and i are steady-state sinusoidal signals. By using the passive sign convention (PSC), the power at any instant of time is given by: 

                                                                               p=vi

This equation describes instantaneous power. Recall that if the reference direction of the current is in the direction of the voltage rise, must be written with a minus sign. Instantaneous power is always measured in watts when the voltage is measured in volts and the current is measured in amperes. Two expressions of phase angles of v and i are written as


v=Vmcos(ωt+θv),v=Vmcos⁡(ωt+θv), (1.2)


i=Imcos(ωt+θi),i=Imcos⁡(ωt+θi), (1.3)



In these two expressions, θvθv is the voltage phase angle, and θiθi is the current phase angle.


While working in the sinusoidal steady state, a convenient reference for zero time may be chosen. Engineers who design systems that transfer large amounts of power have found it convenient to use a zero time that corresponds to the instant the current is passing through a positive maximum. By choosing such a reference time, a shift of both voltage and current by θiθi is required. Now, Eqs. 1.2 and 1.3 become



v=Vmcos(ωt+θv−θi)v=Vmcos⁡(ωt+θv−θi) (1.4)

i=Imcos(ωt)i=Imcos⁡(ωt) (1.5)



If Eqs. 1.4 and 1.5 are substituted into Eq. 1.1, the expression for the instantaneous power now becomes

p=VmImcos(ωt+θv−θi)cos(ωt)p=VmImcos⁡(ωt+θv−θi)cos⁡(ωt) (1.6)


Eq. 1.6 can be used to solve for average power the way it is; however, by applying a few simple trigonometric identities the instantaneous power equation can be simplified. Using cosine's product identity gives

cos(α)cos(β)=12cos(α−β)+12cos(α+β)cos⁡(α)cos⁡(β)=12cos⁡(α−β)+12cos⁡(α+β)



Letting
α=ωt+θv−θiα=ωt+θv−θi and
β=ωtβ=ωt provides


p=VmIm2cos(θv−θi)+VmIm2cos(2ωt+θv−θi)p=VmIm2cos⁡(θv−θi)+VmIm2cos⁡(2ωt+θv−θi) (1.7)



Lastly, using the cosine angle-sum identity


cos(α+β)=cos(α)cos(β)−sin(α)sin(β)cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)



to expand the second term on the right-hand side of Eq 1.7, which gives


p=VmIm2cos(θv−θi)+VmIm2cos(θv−θi)cos(2ωt)−VmIm2sin(θv−θi)sin(2ωt)p=VmIm2cos⁡(θv−θi)+VmIm2cos⁡(θv−θi)cos⁡(2ωt)−VmIm2sin⁡(θv−θi)sin⁡(2ωt) (1.8)

Relationship Between Current, Power, and Voltage

Figure 1.2 below depicts the relationship between i, v, and p, assuming that θv=60∘θv=60∘ and θi=0∘θi=0∘. The frequency of the instantaneous power is twice the frequency of the current or voltage. This depiction also follows from the second two terms on the right side of Eq. 1.8. This means that the instantaneous power goes through two complete cycles for every cycle of either the current or the voltage. If you look at Fig. 1.2, the instantaneous power can be negative for a portion of each cycle, even if the network between the terminals is passive. In a passive network, this negative power implies that the energy being stored in the inductors or capacitors is now being extracted. While the instantaneous power varies with time in the sinusoidal steady-state of a circuit, this causes some vibration in some motor-driven appliances. Due to this vibration in these appliances, resilient motor mountings are required to reduce any excessive vibration.

Average and Reactive Power

Eq. 1.8 can now be utilized to find the average power at the terminals of the circuit, as well as establish the concept of reactive power. Noting that the equation has three terms, it can be rewritten as

p=P+Pcos(2ωt)−Qsin(2ωt),p=P+Pcos⁡(2ωt)−Qsin⁡(2ωt), (1.9)


Where

Average (real) power
P=VmIm2cos(θv−θi)P=VmIm2cos⁡(θv−θi) (1.10)


Reactive power
Q=VmIm2sin(θv−θi)Q=VmIm2sin⁡(θv−θi) (1.11)



P is called the average power, and Q is called the reactive power. Average power is also known as real power, because it is the actual power in a circuit that is transformed from electric to nonelectric energy. The average power associated with sinusoidal signals is the average of the instantaneous power over one period, or


P=1T∫t0+Tt0pdt,P=1T∫t0t0+Tpdt, (1.12)




Where T is the period of the sinusoidal varying function. The bounds of the integral indicate that integration can be made at any convenient time
t0t0 and integration must be bounded exactly one period later. To grasp a better understanding of all the terms in Eq. 1.9 and the relationships they hold, we will need to examine the power in circuits that are purely resistive, purely inductive, and purely capacitive.


Purely Resistive Circuits

If a circuit between terminals is purely resistive, the current and voltage are in phase
(θv=θi)(θv=θi). Thus, Eq. 1.9 can be reduced to



p=P+Pcos(2ωt)p=P+Pcos⁡(2ωt) (1.13)



This is referred to as the instantaneous real power. Figure 1.3 is a graph of the instantaneous real power for a purely resistive circuit, assuming
ω=377rad/sω=377rad/s. The average power, P, is the average of p, over one period. This can be seen by looking at the graph where P=1 for the circuit. From the Fig 1.3, instantaneous real power can never be negative; in other words, power cannot be removed from a purely resistive network. While the power cannot be removed, it is however, dissipated in the form of thermal energy.

Purely Inductive Circuits

Now, if the circuit between the terminals is purely inductive, the current and voltage are out of phase by 90∘.90∘.The current of the circuit lags the voltage by 90∘90∘ (θi=θv−90∘).(θi=θv−90∘). The instantaneous power equation can be reduced to


p=−Qsin(2ωt)p=−Qsin⁡(2ωt) (1.14)



In this purely inductive circuit, the average power is zero. This means that no transformation of energy from electric to nonelectric energy takes place. The power at the terminals is continually exchanged between the circuit and the power source driving the circuit at a frequency of
2ω.2ω. What this means, is that when p is positive, energy is stored in the magnetic fields associated with the inductive elements, and when p is negative, energy is being removed from the magnetic fields.


Power associated with purely inductive circuits is known as the reactive power Q. Reactive power comes from the characterization of an inductor as a reactive element. To differentiate between average power and reactive power, units watt (W) for average power and var (volt-amp reactive, or VAR) for reactive power are used. Figure 1.4 depicts the instantaneous power for a purely inductive circuit, assuming ω=377rads/sω=377rads/sand Q = 1 VAR.

Purely Capacitive Circuits

In this purely capacitive circuit, the current and voltage are
90∘90∘ out of phase with each other. In this case, the current leads the voltage by exactly
90∘90∘ (θi=θv+90∘)(θi=θv+90∘). The expression of this instantaneous power is given by


p=−Qsin(2ωt)p=−Qsin⁡(2ωt) (1.15)



In this circuit, there is no transformation of energy from electric to nonelectric energy because the average power is zero. In a purely capacitive circuit, the power is continually transferred between the source delivering power and to the electric field associated with the capacitive elements. Figure 1.5 depicts the instantaneous power for a purely capacitive circuit, assuming ω=377rads/sω=377rads/s and Q = -1 VAR.

Understanding the Power Factor

This angle (θv−θi)(θv−θi) has a significant role in computing both the average and the reactive power and is known as the power factor angle. Taking the cosine of this angle gives what is known as the power factor, shortened to pf, and taking the sine of this angle is known as the reactive factor, shortened to rf. This can be denoted as:


pf=cos(θv−θi)pf=cos⁡(θv−θi) (1.16)


rf=sin(θv−θi)rf=sin⁡(θv−θi) (1.17)


To completely describe the power factor angle, either lagging power factor or leading power factor terms are used. If the power factor lags, the current lags voltage (i.e. an inductive load is present). On the other hand, if the power factor leads, the current leads voltage (i.e. a capacitive load is present).

Wew. That was a hell of a research! I think my power is drained too. Got to go to sleep then!

Thévenin's Theorem with dependent sources (AC Analysis)

Thévenin's Theorem is a lot work itself, how much more if we had to add dependent sources. The introduction of dependent sources complicates the calculation and process in solving for the thevenins impedance (Zth).

Having dependent source/s twitches the process of Thévenin's theorem. 

1. Determine Voc (Vth) the usual way
2. Determine Zth:
- Remove all independent sources
- Place an assumed value of voltage at the open circuit
- Determine the current supplied by the assumed voltage source
- Solve for Zth using the formula:
Zth = (Assumed Voltage) / (Current Supplied by the Voltage)

Solving for Vth was already discussed on our previous entries, the problem however is with solving for the required thevenins impedance. This is tricky but we will try our best to express and try to make it a brief easy to understand entry.

Suppose we have this circuit.

As stated, the first process in solving for Zth is by removing all independent sources. Also for the following figure, we already removed the load impedance.

Nicely done eh? Now we have to place an assumed value of voltage at the open circuit.  And determine the current supplied by the assumed voltage source. In this case it is Io.

Usually when solving for the current in any given element on a circuit. Mesh analysis is the way to go. However, we also should consider if there is a better route around. This is where all of our learnings in circuits come into play. 

If we already acquired the current supplied by the assumed voltage source, which we should be, we can now get the thevenins impedance (Zth) (finally) using this equation. 

And tadaa! Now we have our Zth and Vth (which is solved in orthodox manner). The equivalent circuit should now be. 

From this point, We can now easily calculate the voltage from a to b simply by using voltage division.

We still think that this method is of greater hassle, because you will have multiple solutions, minimum of three different approach (for Vth, Zth, Vload) in order to find the required value. Although it is efficient that we learn different approach in order to become flexible when faced with different kinds or circuits.

Good? news! This will be our last topic for midterms. There's too much work to do. We have to double time! We worked on this path, we must stay and continue to venture forward. To never give up no matter how many tries (T.T) we must take. 

Norton's Theorem (AC Analysis)

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel impedance connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current. As with Thevenin's Theorem, everything in the original circuit except the load impedance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton) and Norton impedance (ZNorton). As before, the first step is to identify the load impedance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through Z1 is strictly a function of B1's voltage and Z1's impedance: amps (I=E/Z). Likewise, the current through Z3 is now strictly a function of B3's voltage and Z3's impedance: amps (I=E/Z). To calculate the Norton impedance (ZNorton), we do the exact same thing as we did for calculating Thevenin impedance (ZThevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total impedances from one load connection point to the other: Now our Norton equivalent circuit looks like this:

Thévenin's Theorem (AC Analysis)

Thévenin's Theorem for AC circuits with sinusoidal sources is very similar to the theorem we have learned for DC circuits. The only difference is that we must consider impedance instead of resistance. Concisely stated, Thévenin's Theorem for AC circuits says:

Any two terminal linear circuit can be replaced by an equivalent circuit consisting of a voltage source (VTh) and a series impedance (ZTh).

In other words, Thévenin's Theorem allows one to replace a complicated circuit with a simple equivalent circuit containing only a voltage source and a series connected impedance. The theorem is very important from both theoretical and practical viewpoints.
It is important to note that the Thévenin equivalent circuit provides equivalence at the terminals only. Obviously, the internal structure of the original circuit and the Thévenin equivalent may be quite different. And for AC circuits, where impedance is frequency dependent, the equivalence is valid at one frequency only.

Using Thévenin's Theorem is especially advantageous when:


We want to concentrate on a specific portion of a circuit. The rest of the circuit can be replaced by a simple Thévenin equivalent.
We have to study the circuit with different load values at the terminals. Using the Thévenin equivalent we can avoid having to analyze the complex original circuit each time.

We can calculate the Thévenin equivalent circuit in two steps:

1. Calculate ZTh. Set all sources to zero (replace voltage sources by short circuits and current sources by open circuits) and then find the total impedance between the two terminals.
2. Calculate VTh. Find the open circuit voltage between the terminals.

Source Transformation (AC Analysis)

Source transformation is simplifying a circuit solution, especially with mixed sources, by transforming a voltage into a current source, and vice versa. Finding a solution to a circuit can be difficult without using methods such as this to make the circuit appear simpler. Source transformation is an application of Thévenin's theorem and Norton's theorem.


Performing a source transformation consists of using Ohm's law to take an existing voltage source in series with a resistance, and replace it with a current source in parallel with the same resistance. Remember that Ohm's law states that a voltage on a material is equal to the material's resistance times the amount of current through it (V=IR). Since source transformations are bilateral, one can be derived from the other. [2] Source transformations are not limited to resistive circuits however.

They can be performed on a circuit involving capacitors and inductors, as long as the circuit is first put into the frequency domain. In general, the concept of source transformation is an application of Thévenin's theorem to a current source, or Norton's theorem to avoltage source.

Specifically, source transformations are used to exploit the equivalence of a real current source and a real voltage source, such as a battery. Application of Thévenin's theorem and Norton's theorem gives the quantities associated with the equivalence. Specifically, suppose we have a real current source I, which is an ideal current source in parallel with an impedance. If the ideal current source is rated at I amperes, and the parallel resistor has an impedance Z, then applying a source transformation gives an equivalent real voltage source, which is ideal, and in series with the impedance. This new voltage source V, has a value equal to the ideal current source's value times the resistance contained in the real current source. The impedance component of the real voltage source retains its real current source value.



In general, source transformations can be summarized by keeping two things in mind:
· Ohm's Law
· Impedance's remain the same

Source transformation also is one of the easiest way to solve for the wanted values. However, it is also not applicable to every circuit, and also it requires a lot of redrawing of circuit. Being unable to have the talent of good drawing capabilities, it is quite a downfall. 

Superposition (AC Analysis)

The superposition theorem for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances.

To ascertain the contribution of each individual source, all of the other sources first must be "turned off" (set to zero) by:

1. Replacing all other independent voltage sources with a short circuit (thereby eliminating difference of potential i.e. V=0; internal impedance of ideal voltage source is zero (short circuit)).

2. Replacing all other independent current sources with an open circuit (thereby eliminating current i.e. I=0; internal impedance of ideal current source is infinite (open circuit)).

This procedure is followed for each source in turn, and then the resultant responses are added to determine the true operation of the circuit. The resultant circuit operation is the superposition of the various voltage and current sources.

The superposition theorem is very important in circuit analysis. It is used in converting any circuit into its Norton equivalent or Thevenin equivalent.

The theorem is applicable to linear networks (time varying or time invariant) consisting of independent sources, linear dependent sources, linear passive elements (resistors, inductors, capacitors) and linear transformers.


Another point that should be considered is that superposition only works for voltage and current but not power. In other words the sum of the powers of each source with the other sources turned off is not the real consumed power. To calculate power we should first use superposition to find both current and voltage of each linear element and then calculate the sum of the multiplied voltages and currents.

It is one of the easiest process in order to calculate the wanted value. However, it is not applicable to all circuits. That is saddening but we should just accept such. That is life, we have to accept and move on.

Mesh Analysis (AC Analysis)

MESH ANALYSIS:

Another method of analyzing circuits and the counterpart of nodal analysis is the mesh analysis. Also known as loop analysis or the mesh-current method. A mesh is a loop which does not contain any other loops within it. And we can recall that a loop in an electronic circuit is a closed path with no node passed more than once. The current through a mesh is known as mesh current.

Mesh analysis makes use of Kirchhoff’s Voltage Law (KVL) in forming equations to solve for unknown currents in a given circuit. If it’s Vx for nodal analysis, it’s Ix for mesh analysis. Don’t you find it ironic that nodal analysis applies KCL to find unknown voltages, while mesh analysis applies KVL to find unknown current? Though mesh analysis isn’t quite as general as nodal analysis because it is only applicable to planar circuits**.

**one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar.

Here are the steps in determining mesh currents:

Assign the meshes found in the circuit as mesh currents i1, i2, … , in.
Apply KVL to each of the assigned meshes to form equations. Use Ohm’s law (V=IR) to express the voltages in terms of the mesh currents.
Solve the formed equations from step 2 to get the value of the unknown mesh currents.
Let’s take this very familiar circuit for this tutorial:

Assuming that all the values for the elements are already given. And that we are supposed to determine the value of the voltage across the capacitor Z3 (Vz3).

Again, as mentioned above, a mesh is a loop which does not contain any other loops within it. By looking at this simple circuit, we can already say that there are two meshes present in the circuit. The bigger loop outside the circuit cannot be considered as a mesh, because it contains loops within it. The direction of the mesh current can either be clockwise or counterclockwise and it will not affect the validity of the solution, as long as you follow the rules in mesh analysis. And we will encounter those rules as we proceed in solving this circuit.

I have assumed that both meshes i1 and i2 loop at a clockwise direction. You can have meshes at different directions (one goes clockwise, another goes counter) but for ethicality purposes, it’s better to have them at the same direction.

To determine Vz3, we have to first get the current that passes through the capacitor Z3. Then we apply Ohm’s law to solve for Vz3. If you look at Z3, you can observe that there are two currents that are passing through it but not in the same direction as each other: i1 and i2. We can also see that i2 passes through a current source Is but in a different position. This means we can conclude that i2 = -Is.

In nodal analysis, the voltage of a non-reference node directly connected to a voltage source is already equal to that source. Unless the node is connected to another voltage source, then that’d be a supernode. In mesh analysis, the current of a mesh that passes through a current source is already equal to that source (positive if they go the same direction, otherwise negative). Unless the mesh shares the current source with another mesh, then that’d be a supermesh.

So, now that we know mesh i2 = -Is, we’ll only need the current of mesh i1. And we can form its equation using KVL this way:

@mesh i1:

+Vs = 0

In nodal analysis, a current source is considered a constant* and its sign depends on its direction. In mesh analysis, the voltage source is the one considered as a constant* and its sign depends on the direction of the mesh that enters it.

*unless it’s a dependent source.

In this case, Vs is positive since mesh i1 came out of the positive terminal of the source (this is one of the rules I was talking about earlier). Moving on as mesh i1 passes through the resistor Z1..

+Vs – (i1 · Z1) = 0

You might think that the i1 · Z1 is quite familiar. That is actually just Ohm’s law, where V = I · R. So, it’s Vz1 = i1 · Z1. See the resemblance? Now as mesh i1 passes through the capacitor Z3..

+Vs – (i1 · Z1) – Z3(i1 – i2) = 0

As you can see from the last expression of the equation, mesh i1 is subtracted by mesh i2 and is multiplied to Z3. This is because both meshes go through the capacitor but with a difference in direction. Maybe you’d add the meshes if they go to the same direction, but i’ve never tried it before since i’m ethical in solving my circuits.

..and since i2 = -Is..

+Vs – (i1 · Z1) – Z3(i1 – (-Is)) = 0

..or..

+Vs – (i1 · Z1) – Z3(i1 + Is) = 0

..simplifying further..

Vs – (i1 · Z1 ) – (i1 · Z3) – (Is · Z3) = 0

Vs – (Is · Z3) = (i1 · Z1) + (i1 · Z3)

Vs – (Is · Z3) = i1(Z1 + Z3)

[Vs – (Is · Z3)] / (Z1 + Z3) = i1

i1 = [Vs – (Is · Z3)] / (Z1 + Z3)

And now that we have solved i1, we substitute it to:

Vz3 = Z3 · (i1 – i2)

..or..

Vz3 = Z3 · (i1 – Is)

Tadaa! And that is how you do mesh analysis, without the presence of a supermesh. I hope you were able to comprehend all that. If you have some questions, leave them below!

Nodal Analysis (AC Analysis)

Nodal analysis is one of the many methods in solving and finding a specific value of a parameter in electronic circuit analysis. The aim of using nodal analysis is to determine the voltage in each node that’s relative to the reference node, which is the ground GND where voltage is equal to 0. This means that all the other nodes present in the circuit are referred to as the non-reference nodes; the ones that has voltage you are trying to solve for. Depends, of course, if you do need the voltage present in them or not.

Let’s do a quick recap about the parts of an electronic circuit; A node is a point of connection between two or more branches. A branch represents a single element such as a voltage source, or a resistor, etc. And a loop is any closed path in a circuit.

Here are the steps on how to determine node voltages:

Determine the nodes of the circuit, and then select a node as the reference node (ground GND). Then assign the non-reference nodes to voltages V1, V2, Vx, or whatever you feel comfortable with.
Apply KCL (Kirchhoff’s Current Law) to each of the non-reference nodes. Use Ohm’s law (V=IR) to express the branch currents in terms of node voltages. I use the shortcut method, though the same principles are still applied.
Solve the resulting simultaneous equations formed from the non-reference nodes to obtain the unknown node voltages.
Always remember that the number of equations formed should be equal to the number of unknowns. So, taking this simple circuit as an example:

This sample circuit has an AC voltage source, a current source, a resistor, an inductor, and a capacitor. We can also observe that there are four nodes present in this circuit. Assuming that we need to find the voltage across the capacitor, what node will we choose as our reference node GND that will make the problem easier? Note that the lesser the unknowns, the easier the problem will be. It’s like the unknowns determine the difficulty of the problem.

So, there are four nodes. If we were to select the top-left node as the GND, V1 as the top-mid node, V2 as the top-right node, and V3 as the bottom node, it would mean that we can get the voltage across the capacitor with V1 – V3 where V3 = -Vs (voltage source, since the negative terminal of the voltage source is connected to node V3). This would be a fine option but it’s kind of — maybe, unethical — to have the GND connected to the positive terminal of the voltage source.

If we were to select the top-mid node as the GND, V1 as the top-right, and the same position for V2 and V3, then a supernode (formed by enclosing a voltage source, either dependent or independent, connected between two non-reference nodes and any elements connected in parallel with it) would be present which adds more difficulty in solving the circuit. Same situation goes if the top-right node is the GND.

But if we were to select the bottom node as the GND, and V1, V2, and V3 as the top-left, top-mid, and top-right nodes, respectively, then we can get the voltage across the capacitor with V2 – GND = V2 – 0 = V2 (since current flows from a higher potential to a lower potential in a resistor), while V1 = Vs (since the positive terminal of the voltage source is connected to node V1). That makes two unknowns (V2 and V3), though we only need to solve for V2. And it’s more ethical compared to having the GND on top of the circuit.

Now to label the nodes and elements, assuming that all the elements’ respective values are given and have been converted to their equal impedances already. If you want to know how to solve for the impedance in a resistor, inductor, and a capacitor,

There we go! So, what we are trying to find is the voltage across the capacitor Z3 and we have discussed earlier that Vz3 = V2, since Vz3 = V2 – GND and that the voltage at the GND is equal to 0. Now to form the equation at node V2 using the shortcut method. Brace yourself for I am about to make up names that i’ll be using to better explain the shortcut method.

@node V2:

V2( ) = 0

To form the equation at node V2, you must locate V2 (imagine V2 as a person or an animal or any object) and look around its surroundings. And I mean the lines or pathways that are connected to it. In this case, there are three pathways connected to V2 (path to V1, to V3, and to GND). And each path is connected to an element (resistor, inductor, and capacitor), or what I will be naming as a bridge. And this is how you start the equation:

V2( 1/bridge1 + 1/bridge2 + 1/bridge3) = 0

..or..

V2( 1/z1 + 1/z2 + 1/z3 ) = 0

It’s like connecting impedances in parallel. Continuing to the path across one of the bridges, you’ll encounter another node (it could either be the GND or not), or what i’ll be naming as your neighbour. This is how neighbours are treated in the equation:

V2( 1/z1 + 1/z2 + 1/z3 ) – neighbour1/bridge1 – neighbour2/bridge2 – neighbour3/bridge3 = 0

..or..

V2( 1/z1 + 1/z2 + 1/z3 ) – V1/z1 – V3/z2 – GND/z3 = 0

..and since V1 = Vs, and GND = 0..

V2( 1/z1 + 1/z2 + 1/z3 ) – Vs/z1 – V3/z2 – 0 = 0

..and since Vs/z1 is a constant, we transpose it to the other side..

V2( 1/z1 + 1/z2 + 1/z3 ) – V3/z2 = Vs/z1

And that’s it for the first equation, with V2 and V3 as unknowns. As I mentioned earlier, the number of unknowns should be equal to the number of equations. So, we are going to need one more equation, and that’d be the equation at node V3.

@node V3:

V3( ) = 0

Following the same procedures with the bridges and neighbours, we get–

V3( 1/z2 ) – V2/z2 = 0

As you can see, node V3 is connected to a current source. A current source in nodal analysis is considered as a constant unless it’s a dependent source. When the current source’s direction is away from the node, you add the current to the equation. If its direction is towards the node, you subtract the current to the equation. And since the current source in the sample circuit is directed towards node V3, it goes like this:

V3( 1/z2 ) – V2/z2 – Is = 0

..and since Is is already a given constant, we transpose it to the other side..

Vs( 1/z2 ) – V2/z2 = Is

And there’s your second equation. Now you can solve for Vz3 by fusing the two equations using matrices, substitution, elimination, or whatever method you know. Though our professor requires us to use the matrix method.

Phasor (AC Analysis)

Phasor   is a complex number that represents the amplitude and phase of a sinusoid. Phasor in our understanding is the simplified form of the sinusoidal function. Which means it can be used provided that the angular frequencies if faced with two or more sinusoidal input functions, are all equal. Also it is simpler to input in our calculators, there is a saying that I made up, although it might exist already, that simpler is better. Hoho.

PHASOR:

It has three forms:

 

Although we rarely to never use the exponential form, learning its existence can also help in some type with different specification for different approach and derivation in order to provide the correct solution of circuit models. 

On the other hand, polar form is what we often use in solving because it shows the angle of the sinusoid. Which means we can actually track down if our answers lags or leads the other values. 

There are rules in relating two phasors with each other: 

Time domain and phasor domain: 

Time domain is the general expression of the sinusoid, while phasor domain is a more simplified one. 

Rules for transforming Time domain into Phasor domain:

1. Time domain must be in cosine function
2. Vm or the magnitude of the function must be positive.

- amplitude and phase difference fare two principal concerns in the study of voltage and current sinusoids.
- phasor will be defined form the cosine-function in all our proceeding study. If a voltage or expression is in the form of a sine, it will be changed to cosine (function) by substracting from the phase.

Phasor is also called a complex number, which we are about to take in Advance Mathematics. We are grateful because we can double the learnings we get from both subject and apply each others' learnings with one another, if that make sense.

SINUSOIDS (AC Analysis)

SINUSOIDS 

Here we are, the basic of all learnings we are about to pass through in this program. Since we are dealing with alternating power (AC analysis), we have to first have the foundation of the structures and understand its components. Because unlike direct current (DC), alternating current (AC) is oscillating and periodic, which mean it is in a form of a sinusoid, a signal that has the form of the sine or cosine function. Sinusoidal functions are the basis for study of all periodic functions. This periodic pattern contains several sinusoidal functions.

Amplitude - highest and lowest point of the wave.

Period (T) - the amount of time the wave has to consume in order to get back to a specified point of origin.

As mentioned, the period (T) of the periodic function is the time of one
complete cycle or the number of seconds per cycle. The reciprocal of
this quantity is the number of cycles per second, known as the cyclic
frequency f of the sinusoid. Thus, 

However, sinusoids in AC analysis will have an angle. Thus:

Showing a figure that will clearly illustrate and show the lags and leads of the two given sinusoids.

A sinusoid can be expressed in either sine or cosine form. When
comparing two sinusoids, it is expedient to express both as either sine
or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:

And there we have it! The start of it all, understanding the alternating electricity (AC) is just the beginning. We know, however far we will get, there is always something more to learn :)

Start of the new chapter of Electrical Circuits: Impedances (AC Analysis)

IMPEDANCES:

Let’s start the lessons from here — one of the major (or maybe minor) changes when it comes to analyzing an AC circuit from a DC circuit: the impedance.

Now, the impedance of an electronic circuit is the ratio of the phasor voltage V to the phasor current I. It is declared with the variable Z and is measured in Ohms (Ω). In some ways, you could say that the impedance acts the same as the resistance in a circuit, but not completely.

Resistors, capacitors, and inductors are the three elements that cause impedance. Though you have to convert the capacitance C and the inductance L to get their equivalent impedances Z, by applying these formulas:

RESISTORS: 

 Z=R

INDUCTORS: 

Z=j•ω•L

Z=s•L , s=j•ω

CAPACITORS: 

Z=1/(j•ω•C)

Z=1/s•C , s=j•ω

From the stated formulas above, you can observe that the resistance of a resistor can already be considered as its impedance. While for inductors, you have to multiply its inductance by j (imaginary number*) and ω (angular frequency). And lastly for capacitors, you have to divide 1 by the capacitance, j, and ω.

*equal to the square root of -1.

Here are some important notes to remember for inductors and capacitors when it comes to analyzing circuits at DC and high frequencies:

Inductors are considered as short circuit at DC equivalent circuits, while as open circuit at high frequency. The opposite goes for the capacitors, since they are considered as open circuit at DC, while as short circuit at high frequency.

Don’t forget that impedances Z are also measured in Ohms (Ω) like resistances R. So, when you solve for impedances in series and in parallel, all you have to do is to follow the same formulas used when solving for resistors in series and in parallel.

SERIES IMPEDANCES: 

Zeq=Z1+Z2 + ... +Zn

PARALLEL IMPEDANCES: 

1/Zeq = 1/Z1 + 1/Z2 + … + 1/Zn

That’s how impedances look like in a schematic circuit diagram. Just rectangles. Pretty creative, huh?

In solving for the voltage V and current I, you will still apply the same formulas but replacing the resistance R with impedance Z. So, they would now go like:

V = I · Z

I = V / Z

Though for solving power P in AC circuit analysis, it’s a different story compared to solving it in DC. We’ll deal with that sooner or later. Or not.

And that’s about everything there is in the basics of analysing the impedance in an electronic circuit! Though its only the start for another chapter of our life learning the field we chose to walk.

Inductor

An inductor, also called a coil or reactor, is a passive two-terminal electrical component which resists changes in electric current passing through it. It consists of a conductor such as a wire, usually wound into a coil. When a current flows through it, energy is stored temporarily in a magnetic field in the coil.

HOW INDUCTOR WORKS:
•The number of coils – More coils means more inductance.
•The material that the coils are wrapped around (the core)
•The cross-sectional area of the coil – More area means more inductance.
•The length of the coil – A short coil means narrower (or overlapping) coils, which means more inductance.

Putting iron in the core of an inductor gives it much more inductance than air or any non-magnetic core would. The standard unit of inductance is the henry. In a capacitor, the formula is expressed as “the current in a circuit is in proportion to the time rate of change of the voltage across it.

i=c(dv/dt)

In series and parallel capacitors, they are combined in the same way as conductances. An inductor formula states that the voltage across it is directly equitable to the rate of change of the current through the circuit.

The formula is written like this:

v = L(di/dt)

Capacitors

Capacitors are components designed to take advantage of this phenomenon by placing two conductive plates (usually metal) in close proximity with each other. There are many different styles of capacitor construction, each one suited for particular ratings and purposes. For very small capacitors, two circular plates sandwiching an insulating material will suffice. For larger capacitor values, the "plates" may be strips of metal foil, sandwiched around a flexible insulating medium and rolled up for compactness. The highest capacitance values are obtained by using a microscopic-thickness layer of insulating oxide separating two conductive surfaces. In any case, though, the general idea is the same: two conductors, separated by an insulator. The schematic symbol for a capacitor is quite simple, being little more than two short, parallel lines (representing the plates) separated by a gap. Wires attach to the respective plates for connection to other components. An older, obsolete schematic symbol for capacitors showed interleaved plates, which is actually a more accurate way of representing the real construction of most capacitors

When a voltage is applied across the two plates of a capacitor, a concentrated field flux is created between them, allowing a significant difference of free electrons (a charge) to develop between the two plates:

As the electric field is established by the applied voltage, extra free electrons are forced to collect on the negative conductor, while free electrons are "robbed" from the positive conductor. This differential charge equates to a storage of energy in the capacitor, representing the potential charge of the electrons between the two plates. The greater the difference of electrons on opposing plates of a capacitor, the greater the field flux, and the greater "charge" of energy the capacitor will store. Just as Isaac Newton's first Law of Motion ("an object in motion tends to stay in motion; an object at rest tends to stay at rest") describes the tendency of a mass to oppose changes in velocity, we can state a capacitor's tendency to oppose changes in voltage as such: "A charged capacitor tends to stay charged; a discharged capacitor tends to stay discharged." Hypothetically, a capacitor left untouched will indefinitely maintain whatever state of voltage charge that its been left it. Only an outside source (or drain) of current can alter the voltage charge stored by a perfect capacitor:

Practically speaking, however, capacitors will eventually lose their stored voltage charges due to internal leakage paths for electrons to flow from one plate to the other. Depending on the specific type of capacitor, the time it takes for a stored voltage charge to self-dissipate can be a long time (several years with the capacitor sitting on a shelf!). When the voltage across a capacitor is increased, it draws current from the rest of the circuit, acting as a power load. In this condition the capacitor is said to becharging, because there is an increasing amount of energy being stored in its electric field. Note the direction of electron current with regard to the voltage polarity:

Conversely, when the voltage across a capacitor is decreased, the capacitor supplies current to the rest of the circuit, acting as a power source. In this condition the capacitor is said to be discharging. Its store of energy -- held in the electric field -- is decreasing now as energy is released to the rest of the circuit. Note the direction of electron current with regard to the voltage polarity:

An obsolete name for a capacitor is condenser or condensor. These terms are not used in any new books or schematic diagrams (to my knowledge), but they might be encountered in older electronics literature. Perhaps the most well-known usage for the term "condenser" is in automotive engineering, where a small capacitor called by that name was used to mitigate excessive sparking across the switch contacts (called "points") in electromechanical ignition systems.

 

The Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum. This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance. Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):

 

With this value of load resistance, the dissipated power will be 39.2 watts: The Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance. Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, "output impdance" : "load impedance" is known as damping factor, typically in the range of 100 to 1000. [rar] [dfd] Maximum power transfer does not coincide with the goal of lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.

Norton's Theorem

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots). Contrasting our original example circuit against the Norton equivalent: it looks something like this:

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current. As with Thevenin's Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton). As before, the first step is to identify the loadresistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R1 is strictly a function of B1's voltage and R1's resistance: 7 amps (I=E/R). Likewise, the current through R3 is now strictly a function of B2's voltage and R3's resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (INorton) in our equivalent circuit: Remember, the arrow notation for a current source points in the direction opposite that of electron flow. Again, apologies for the confusion. For better or for worse, this is standard electronic symbol notation. Blame Mr. Franklin again! To calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other: Now our Norton equivalent circuit looks like this:

Thevenin's Theorem

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of “linear” is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits.
Thevenin's Theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit (called the “load” resistor) is subject to change, and re-calculation of the circuit is necessary with each trial value of load resistance, to determine voltage across it and current through it. Let's take another look at our example circuit:

Let's suppose that we decide to designate R2 as the “load” resistor in this circuit. We already have four methods of analysis at our disposal (Branch Current, Mesh Current, Millman's Theorem, and Superposition Theorem) to use in determining voltage across R2 and current through R2, but each of these methods are time-consuming. Imagine repeating any of these methods over and over again to find what would happen if the load resistance changed (changing load resistance is verycommon in power systems, as multiple loads get switched on and off as needed. the total resistance of their parallel connections changing depending on how many are connected at a time). This could potentially involve a lot of work!
Thevenin's Theorem makes this easy by temporarily removing the load resistance from the original circuit and reducing what's left to an equivalent circuit composed of a single voltage source and series resistance. The load resistance can then be re-connected to this “Thevenin equivalent circuit” and calculations carried out as if the whole network were nothing but a simple series circuit:

The “Thevenin Equivalent Circuit” is the electrical equivalent of B1, R1, R3, and B2 as seen from the two points where our load resistor (R2) connects.
The Thevenin equivalent circuit, if correctly derived, will behave exactly the same as the original circuit formed by B1, R1, R3, and B2. In other words, the load resistor (R2) voltage and current should be exactly the same for the same value of load resistance in the two circuits. The load resistor R2 cannot “tell the difference” between the original network of B1, R1, R3, and B2, and the Thevenin equivalent circuit of EThevenin, and RThevenin, provided that the values for EThevenin and RThevenin have been calculated correctly.
The advantage in performing the “Thevenin conversion” to the simpler circuit, of course, is that it makes load voltage and load current so much easier to solve than in the original network. Calculating the equivalent Thevenin source voltage and series resistance is actually quite easy. First, the chosen load resistor is removed from the original circuit, replaced with a break (open circuit):

Next, the voltage between the two points where the load resistor used to be attached is determined. Use whatever analysis methods are at your disposal to do this. In this case, the original circuit with the load resistor removed is nothing more than a simple series circuit with opposing batteries, and so we can determine the voltage across the open load terminals by applying the rules of series circuits, Ohm's Law, and Kirchhoff's Voltage Law:

The voltage between the two load connection points can be figured from the one of the battery's voltage and one of the resistor's voltage drops, and comes out to 11.2 volts. This is our “Thevenin voltage” (EThevenin) in the equivalent circuit: