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.