Power in Balanced System

This requires that the analysis be done in the time domain. For a Y - connected load, the phase voltages are

                  Van = √2 Vp coswt ,                         Vbn = √2 Vp cos(wt - 120),                                 Vcn = √2 Vp cos(wt + 120)  

where the factor √2 is important because Vp has been defined as the rms value of the phase voltage. If Zy = Z∠Ɵ, the phase currents lag behid their corresponding phase voltage by Ɵ.

                
  ia = √2Ip cos(wt - Ɵ),                        

  ib = √2Ip cos(wt - Ɵ - 120)                    
  ib = √2Ip cos(wt - Ɵ + 120)

TOTAL INSTANTANEOUS POWER

p = pa + pb + pc = VANia + VBNib + VCNic

    = 2VpIp [ coswt cos(wt - Ɵ) + cos(wt  - 120) cos(wt - Ɵ - 120) + cos(wt + 120) cos(wt - Ɵ + 120) ]

p = 3VpIp cosƟ

INSTANTANEOUS POWER PER PHASE IS:

p = VpIp cosƟ

REACTIVE POWER PER PHASE IS:

Qp = VpIp sinƟ

APPARENT POWER PER PHASE IS:

Sp = VpIp

COMPLEX POWER PER PHASE IS :

Sp = Pp + jQp = Vp Ip*

Balance Three-Phase Circuit

The voltages in the three-phase power system are produced by a synchronous generator. In a balanced system, each of the three instantaneous voltages have equal amplitudes but separated from the other voltages by a phase angle of 120 . The three voltages (or phases) are typically labeled a, b and c. The common reference point for the three phase voltages is designated as the neutral connection and is labeled as n. We may define either a positive phase sequence (abc) or a negative phase an bn cn sequence (acb) as shown below. The three sources Van , Vbn and Vcn are designated as the line-to-neutral voltages in the three-phase system.

LINE TO LINE VOLTAGES

 First, the wye - connected voltages Van, Vbn, and Vcn are respectively between lines a, b, c, and the neutral line n. These are called phase voltages. The voltages are said to be balanced phase voltages if the voltages sources have the same amplitude and frequency w and are out of phase by 120degrees. 

                                                                Van = Vp∠0

                                                                Vbn = Vp∠-120

                                                                Vcn = Vp∠-240 = Vp∠120

     where Vp is the effective or rms value of the phase voltage. this is known as abc sequence or positive sequence, Van leads Vbn, which turn leads Vcn. This rotates counterclockwise. For negative or acb sequence, Van leads Vcn, ehich turn leads Vbn. this rotates clockwise direction.

AC Power Analysis

Instantaneous Power

The instantaneous power changes with time and it is therefore difficult to measure. The average power is more convenient to measure. For sinusoidal signals or any other periodic signal we compute average power over one period using the following relation:

• Product of time-domain voltage 

      and time-domain current
           p(t) = v(t) i(t)

• Determine maximum value
• Transients
• Steady-state

ƒ

• v(t) = Vmcos(ωt+θ) and i(t) = Imcos(ωt +ϕ)

ƒ

• p(t) = Vm Imcos(ωt+θ) cos(ωt +ϕ) =

            = ½ Vm Imcos(θ-ϕ) + ½ Vm Imcos(2ωt + θ+ ϕ)

where:

p(t) = instantaneousƒ,    two parts,     ƒaverage,     ƒperiodic

 ½ Vm Imcos(θ-ϕ)  =  constant,     ƒindependent of t,     ƒthe "average",     ƒwantedƒ,    active or real

+ ½ Vm Imcos(2ωt + θ+ ϕ) = periodicƒ,     period is ½T,     ƒaverage is zero,     ƒunwanted

EXAMPLE:

1. Voltage across impedance (V= 100∠25V and Z= 50∠55Ω). Determine active power absorbed.

I   =V / Z
    = 2 ∠25-55°
    = 2 ∠-30°
I   = 1.7 - j A

Z= 28.7 + j41.9 Ω

P   = ½ 2228.7
     = ½ (1.72+12) 28.7
P   = 57.4 W

EFFECTIVE OR RMS VALUE

Effective or RMS Value

The RMS Value is a constant itself which depending on the shape of the function i(t).

The effective value of a periodic current is equal to dc current that delivers the same average power to a resistor as the periodic current.

The RMS Value of a sinusoid i(t) = Imcos(wt) is:

                                                              I rms =  Im 

                                                                           √2

                                             

                                                           P = Irms Vrms cos(θv-θi)

• Take note: Supply is always Vrms, unless stated that it is maximum voltage.
  
  

   

   Apparent Power and Power Factor

  
  
  • Apparent Power S, is theproduct of the RMS values of voltage and current.
  • It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts.
  
  
  
  
                                             P = Vrms Irms cos(θv-θi) = S cos(θv-θi)
                                  where:
                                             S - the apparent power
                                             (θv-θi) - the power factor, PF 
            
                                                         PF = P/S = (θv-θi)
• PF is the cosine of the phase difference between the voltage and current. Also, angle of the load impedance. 

                                                                 P = S 0.9

                                                   where: 

                                                             P - active power (W)

                                                             S - apparent power (VA)

                                                            0.9 - reactive power (VAR)

Purely resistive load (R) - θv-θi = 0

                                        - PF = 1

                                        - P = S

Purely reactive load (L or C) - θv-θi = ±90

                                              - PF =0

                                               - P = 0

Resistive and Inductive load (R and L) - θv-θi > 0

                                                                - P = S x PF

                                                                - 0 < PF < 1, lag

Resistive and Capacitive load (R and C) - θv-θi < 0

                                                                  - 0 < PF < 1, lead

                     

                                                 

REVIEW:

• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0deg and 90deg.

COMPLEX POWER

S = 1/2 VI* = Vrms Irms ∠ (θv-θi)

         

                  V = Vm∠θv           I = Im∠θpf

S =  Vrms Irms cos(θv-θi) + jVrms Irms sin(θv-θi)

S = P +jQ

P: is the average power in watts delivered to a load and it is the only useful power.

Q: is the reactive power exchange between the source and the reactive part of the load. It is measured in VAR.

• Q = 0, for resistive loads (unity PF)
• Q < 0, for capacitive loads (leading PF)
• Q > 0, for inductive loads (lagging PF)

Maximum and Average Power

The maximum power transfer theorem for DC circuit, we can determine the condition for an AC load to absorb maximum power in an AC circuit. For an AC circuit, both the thevenin impedance and the load can have a reactive component. Although these reactances do not absorb any average power, they will limit the circuit current unless the load reactances cancels the reactance of the thevenin impedance. For the maximum power transfer, the thevenin and load reactances must be equal in magnitude but opposite in sign.

If the load is purely real, then RL = √(Rth)^2 + (Xth)^2 = |Zth|

Zth = Rth + jXth    ;     ZL = RL + jXL

Prove that:

XL = -Xth    &        RL = Rth

SOLUTION:

P = 1/2 (I)^2 (RL)

I = VTH/Zth + ZL

P = 1/2 |Vth/(Zth+ZL)|^2 (RL)

P = 1/2 |Vth/(Rth+jXth)+(RL+jXL)|^2 (RL)

P = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| (RL)

dP = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| d(RL) + 1/2 (RL) |Vth^2 (dRL) (-2) (Rth + jXth + RL + jXL)^-3|

dP/dRL = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| + 1/2 (RL) |Vth^2 (-2) (Rth + jXth + RL + jXL)^-3| = 0

          (1/2 Vth^2)                -    RL                 (Vth^2)                  = 0

(Rth + jXth + RL + JXL)^2                (Rth + jXth + RL + jXL)^3

1   -                RL                  =   0

2      Rth + jXth + RL +jXL

1    =                      RL               

2              Rth + jXth + RL +jXL

Rth + jXth + RL + jXL = 2RL

RL = Rth + j (Xth + XL)

RL - Rth = j (Xth +XL)

(RL - Rth) - j (Xth +XL)  =  0

-j (Xth +XL) = 0

Xth = -XL

XL = -Xth

RL - Rth = 0

RL = Rth

Thevenin's And Norton's Theorem

Any combination of sinusoidal AC sources and impedances with two terminals can be replaced by a single voltage source e and a single series impedance z. The value of e is the open circuit voltage at the terminals, and the value of z is e divided by the current with the terminals short circuited. In this case, that impedance evaluation involves a series-parallel combination.

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 Z_Th. 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 V_Th. Find the open circuit voltage between the terminals. 

Norton's Theorem, already presented for DC circuits, can also be used in AC circuits. Norton's Theorem applied to AC circuits states that the network can be replaced by a current source in parallel with an impedance.

We can calculate the Norton equivalent circuit in two steps:

1. Calculate Z_Th. 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 I_Th. Find the short circuit current between the terminals.

Superposition on AC circuit

Since we already studied the Superposition Theorem in DC on the previous chapter, and so on this article will talk about Superposition Theorem on AC circuit.

We have already studied the superposition theorem for DC circuits. In this chapter we will show its application for AC circuits.

The superposition theorem states that in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently. The theorem is valid for any linear circuit. The best way to use superposition with AC circuits is to calculate the complex effective or peak value of the contribution of each source applied one at a time, and then to add the complex values. This is much easier than using superposition with time functions, where one has to add the individual time functions.

To calculate the contribution of each source independently, all the other sources must be removed and replaced without affecting the final result.

When removing a voltage source, its voltage must be set to zero, which is equivalent to replacing the voltage source with a short circuit.

When removing a current source, its current must be set to zero, which is equivalent to replacing the current source with an open circuit. 

Nodal Analysis And Mesh Analysis In AC circuit

Since we know that Kirchhoff's Law is applicable to the AC circuit. We will also apply the Nodal Analysis and Mesh Analysis in analyzing an ac circuits.

STEPS TO ANALYZE AC CIRCUITS:
1. Transform the circuit to the phasor or frequency domain.
2. Solve the problem using circuit techniques(Nodal/Mesh analysis)
3. Transform the resulting phasor to the time domain.


~We've been through about Nodal Analysis and Mesh Analysis in my past blogs. We will just recall it.

Nodal Analysis provide a general procedure for analyzing circuits using node voltages as the circuit variables.

Steps to Determine Node Voltages:

1. Select a node as the reference node, Assign voltages v1, v2, . . . . . , 

vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.

2.Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express currents in terms of node voltages.

3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

Nodal Analysis with Voltage Sources

Case 1: If the voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalized node or super node, we apply both KCL and KVL to determine the node voltages.

Case 2: if a voltage source is connected between the reference node and a non-reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source.

~In this case we will solve a problem in which the three(capacitor,conductor,resistor) elements are included. Also with the rectangular form and polar form since the phasor and frequency domain was being applied. 


Example:

The unknown for this problem is Io, in order to get Io we must transform first the inductor and capacitor into the impedance. 

@node Vo,

But Io= (25-Vo) / 2000, so substitute to get the Vo and also the answer must be in polar form.

We can get now the Io using Ohm's Law.

MESH ANALYSIS;

A Mesh is a loop that does not contain any other loop within it.

STEPS TO DETERMINE MESH CURRENTS:
1. Assign mesh currents I1, I2,... In to the n meshes.
2. Apply KVL to each  of the n meshes. Use ohm's law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.

~We  will apply mesh to ac;

Example: 

 

Source:

- Fundamental of Electric Circuits by Alexander and Sadiku

- Google

- Youtube

Phasor Relationship For Circuit Elements

Impedance and Admittance

Impedance Z – of a circuit is the ratio of the phasor voltage to the phasor current, measured in Ohms(Ω).

               Z=V/I                    or           V=ZI

Admittance Y – is the reciprocal of impedance, measured in siemens(s).

               Y=1/Z = I/V

                              Impedances and admittances of passive elements.

Elements	Impedance	Admittance
R (Ω)	Z=R	Y=1/R
L (H)	Z= jωL	Y=1/jωL
C (F)	Z=1/jωC	Y=jωC

That three elements(R,L,C) only the Resistances are real and the rest were imaginary and the reactances. The Capacitor is the positive reactance while the Inductor is the negative reactance.

Example:

We can transform the imaginary elements into an impedance by the use of shown equations.

1H

Z= jωL = j1x10 = j10

1F

Z=1/jωC = 1/j10x1 = -j0.1

In order to proceed and get the unknown values, we must first identify if what concept should be done in this problem. Since, circuit analysis was all about analyzing the different character of a circuit in any ways, we will use the basic law which we’ve been through in a few months “The Kirchhoff’s Law” was already discuss here in the past topics. We will just recall some important matters.

Kirchhoff’s Law has two parts, the Kirchhoff’s voltage Law and Kirchhoff’s current Law.

Kirchhoff's voltage law, states that the algebraic sum of all the voltages around a closed circuit equals zero. 

Kirchhoff's current law, states that the algebraic sum of all the currents entering and leaving a node is equal to zero. 

Looking at the circuit, the KVL must be applied since the sum of all the voltage drops in a closed circuit will equal the voltage source if only we’ll combine all of the impedances in order to have a single loop. And here we can apply the time domain converted into phasor domain.

Z= 1 + (1/j10 + 1/-j0.1 + 1/1)^-1 = 1.01010 – j0.1 = 1.015∠ -5.653

We can solve now the unknown which is the current using Ohm’s Law;

I=V/R = 2∠0/ 1.015∠ -5.653

I = 1.9704∠5.653 = 1.9704cos(10t+5.65) A

SINUSOIDS AND PHASORS

A Sinusoid is a signal that has the form of the sine or cosine function.

                              There are two parts of sinusoid, the Sinusoidal Current and Sinusoidal Voltage. Sinusoidal current is usually referred to as alternating current. Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called ac circuit. 

Sinusoidal voltage,

v(t) = Vmsinωt

where;

Vm= the amplitude of the sinusoid

ω = the angular frequency in radian/s

ωt = the argument of the sinusoid

Sample equation to determine it's label;

  6cos(200t + 15° )

Amplitude- 6

Phase angle- 15°

Angular Frequency- 200t

A Phasor – is a complex number that represents the amplitude and phase of a sinusoid.

2 PHASES

* IN PHASE

*OUT OF PHASE

IN PHASE,

The same;

*Time

*Period

*Frequency

OUT OF PHASE,

It's either have the same amplitude or not.

~Sinusoids are easily expressed in terms of phasors, in which are more convenient to work with than sine and cosine function.

Sinusoid-Phasor Transformation 

Time Domain representation	Phasor Domain Representation
Vmcos(ωt + ɸ )	Vm ∠ ɸ
Vmsin(ωt + ɸ )	Vm ∠ ɸ - 90 °
Imcos(ωt + 0 )	Im∠0
Imsin(ωt + 0 )	Im∠ 0 - 90 °

To transform the Time domain into the Phasor domain, the time domain is in the rectangular form of

 z = x + jy,

where in x is the real part of z and y is the imaginary part.

The equation going to polar form for the amplitude;

Square root of x squared plus y squared

For the Phase angle;

arctan(y divided by x) 

Example; 

5 + j2
= 5.39∠ 21.80°

Since all of these was all about the currents and voltages, in getting the value of each of them we go through graphing in a sinusoidal form. So between that two, there must be leading and lagging. 

Looking at the figure, the voltage leads the current since leading is when a sinusoid peaks first in time and it is closer to the reference axis. And the current here is lagging.

Thevenin's and Norton'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).

• Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.
• Steps to follow for Thevenin's Theorem:
• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

Example:

This figure is an example that can be solved by the Thevenin's Theorem. Since R2 is a load, it can be remove temporarily in order to get the RTH and VTH. :) 

SUPERPOSITION Theorem

The superposition theorem for electrical circuit 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)).

Example:

 Find R2;

The figure has two(2) voltage source, in order to get the voltage across R2, we need to get first the voltage that being supplied on the other loop by deactivating/turning off the other source.

As you can see, the B2 is already turned off, therefore we can solve it using the voltage division principle.
-The Voltage division can be seen on the past blog for more information!

After that, the B1 must be turned off and B2 is now on in order to get the voltage on the other loop.

 

 I'm sure, we can get now the total voltage that R2 have.

TAKE NOTE: When there is Three(3) or more sources in a circuit, only 1 must be turned on and the rest is disable.

Mesh Analysis

A Mesh is a loop that does not contain any other loop within it.

~ a loop can be a mesh, but a mesh can't be a loop.

Same as the nodal analysis, a mesh analysis have also a steps in getting the equation but they differ for some aspects which is the Nodal analysis talks about the nodal voltages while the Mesh analysis talks about mesh currents. 

~ a mesh current is quite similar to the Branch Current method in that it uses simultaneous equations, Kirchhoff's Voltage Law, and Ohm's Law to determine unknown currents in a network.

Why is it QUITE SIMILAR TO THE BRANCH CURRENT?
-  it is quite similar, then of course it is also quite different but it depends on how the mesh is being ISOLATED.


There are steps in determining mesh currents same as nodal analysis, there are steps to determine nodal voltage in order to form an equation.

STEPS TO DETERMINE MESH CURRENTS:
1. Assign mesh currents I1, I2,... In to the n meshes.
2. Apply KVL to each  of the n meshes. Use ohm's law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.

Example:

As you observe the figure, the current flow counter clockwise but wherever the mesh currents will flow, it's direction is arbitrary and does not affect the validity of the solution.

~ as a class, we are more prefer to have the mesh currents direction clockwise because for us it is more convenient and easier to analyze.

Mesh analysis with current sources

Case 1:

When a current source exists only in one mesh

Case 2:

When a current source exists between two meshes and that is SUPERMESH.

What is supermesh?

A SUPERMESH results when two meshes have current source in common.

For more information about mesh, just watch this:

Chapter III. Continuation of Nodal Analysis and Wye-Delta Transformation

As we go through the "Nodal Analysis" topic which includes the KCL and KVL and the SUPERNODE, we come up with a lot of problem solving then we have two method in solving/analyzing the circuit, which is the short cut method and the long way method.

The short cut method,

•  The nodal voltage must be determined.

•   it gives the adjacent of the resistors into voltage.

The long way method, 

• The nodal voltages must be determined.

• The flow current of the current must be assigned.

 

#REMINDERS
     ~ The current leaving is positive, and the current entering is negative.

WYE - DELTA TRANSFORMATION

The wye - delta transform, also written wye-delta and also known by many other names, is a mathematical technique to simplify the analysis of an electrical network.

•  wye    -   Y
• delta    -   Δ

~ I think why the wye-delta transformation are made because usually in a circuit(complicated design) have a form of  Y and Δ in order to get the total resistance.

What is Nodal Analysis

Nodal Analysis provide a general procedure for analyzing circuits using node voltages as the circuit variables.

Steps to Determine Node Voltages:

1. Select a node as the reference node, Assign voltages v1, v2, . . . . . , 

vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.

2.Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express currents in terms of node voltages.

3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

Nodal Analysis with Voltage Sources

Case 1: If the voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalized node or super node, we apply both KCL and KVL to determine the node voltages.

Case 2: if a voltage source is connected between the reference node and a non-reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source.

~ In every different kind of circuits when in comes to analyzing and solving, we should practice the steps of determining the node voltages and must able to observe if what kind of case the circuit is.

To understand more, watch this video :)

There are some instances that a voltage source is in between to the two non-reference node in a loop, that's SUPERNODE.
In circuit theo, a super-node is a theoretical construct that can be used to solve a circuit. This is done by viewing a voltage source on a wire as a point source voltage in relation to other point voltages located at various nodes in the circuit, relative to a ground node assigned a zero or negative charge.

The application that needs to study:
- KVL/KCL (on my CHAPTER 2 blog)
- Cramer's rule