BLOG 23 - FUNDAMENTAL CIRCUIT

Complex Power

In power system analysis the concept of Complex Power is frequently used to calculate the real and reactive power.

This is a very simple and important representation of real and reactive power when voltage and currentphasors are known. Complex Power is defined as the product of Voltage phasor and conjugate of current phasor. See Fig-A

Let voltage across a load is represented by phasor V  and  current through the load is I.

If S is the complex power then,

                 S = V . I^*

V is the phasor representation of voltage and I* is the conjugate of current phasor. 

So if  V is the reference phasor then V can be written as |V| ∠0.

(Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage)

Let current lags voltage by an angle φ, so  I = | I | ∠-φ

(current phasor makes -φ degrees with real axis)

                            I^*=  | I | ∠φ

So,

                           S = |V|  | I | ∠(0+φ) =  |V|  | I | ∠φ

(For multiplication of phasors we have considered polar form to facilitate calculation)

Writting the above formula for S in rectangular form we get

                           S =  |V|  | I | cos φ  +  j  |V|  | I | sin φ 

             The real part of complex power S is |V| | I | cos φ which is the real power or average power and the imaginary part  |V| | I | sin φ is the reactive power. 

             So,              S = P + j Q

             Where        P = |V| | I | cos φ    and    Q = |V| | I | sin φ

It should be noted that S is considered here as a complex number. The real part P is average power which is the average value, where as imaginary part is reactive power which is a maximum value. So I do not want to discuss further and call S as phasor. If you like more trouble I also advise you to read my article about phasor or some other articles on phasor and complex numbers.

Returning to our main point, from the above formula it is sure that P is always more than zero. Q is positive when φ is positive or current lags voltage by φ degrees. This is the case of inductive load. We previously said that inductance and capacitance do not consume power. The power system engineers often say about reactive power consumption and generation. It is said that inductive loads consume reactive power and capacitors produce reactive power. This incorrect terminology creates confusion.

The fact is that most of the loads are inductive and they unnecessarily draw more current from source. Although in each cycle both inductance and capacitance draw power from the source and return same amount of power to the source but the behavior of inductance and capacitance are opposing to each other. When capacitors are connected in parallel to inductive load the power requirement of inductive load is supplied by capacitor in half cycle and in next half cycle the reverse happens. Depending upon the values of capacitor this power requirement of inductance in the load may be fully or partially satisfied. If partially satisfied the rest will be drawn from the distant source. By properly selecting the capacitance the maximum value of reactive power (Q) drawn from the distant source (or returned to the distant source) is reduced. This reduction in reactive power results in reduction of line current so the reduction of losses in transmission line and improvement in voltage at load end.

Power Triangle

Returning to the complex power formula, P, Q and S are represented in a power triangle as shown in figure below.

S is the hypotenuse of the triangle, known as Apparent Power. The value of apparent power is |V|| I |

or    |S| = |V|| I |

 It is measured in VoltAmp or VA.

P is measured in watt and Q is measured in VoltAmp-Reactive or VAR. In power systems instead of these smaller units larger units like Megawatt, MVAR and MVA is used.

The ratio of real power and apparent power is the power factor of the load.

power factor = Cos φ = |P| / |S|

                      = |P| / √(P ²+Q ²)

The reactive power Q and apparent power S are also important in power system analysis. As just shown above the control of reactive power is important to maintain the voltage within the allowed limits. Apparent power is important for rating the electrical equipment or machines.

Total Power of Parallel Circuits

In real world the loads are usually connected in parallel. Here we will show the total power consumed by parallel branches. See Figure-C. It has two branches.

First we have to draw the individual power triangles for each branch. Next the power triangles are arranged back to back keeping real power in positive x direction as shown. The total power consumed is obtained by connecting starting point O to the tip of last triangle. This is actually the result of addition of complex numbers.

If   S₁ = P₁ +j Q₁

     S₂ = P₂ +j Q₂

       

     Then,  S = S₁+  S₂  

   or    S = (P₁+  P₂  ) + j  (Q₁+  Q₂ )

         P = P₁+  P₂ 

        Q = Q₁+  Q₂ 

_{In the above diagram} S₁  P_{1 ,}   Q_{1 and} φ ₁ correspond to branch1 and S₂  P_{2 ,}   Q_{2 and} φ ₂  correspond to branch2.

S, P, Q and φ correspond to total power consumption as seen by the generator

LEARNING:

I learn that the complex power is the product of the complex effective voltage and the complex effective conjugate current. In our notation here, the conjugate is indicated by an asterisk Complex power can also be computed using the peak values of the complex voltage and current, but then the result must be divided by 2. Note that complex power is applicable only to circuits with sinusoidal excitation because complex effective or peak values exist and are defined only for sinusoidal signals. The unit for complex power is VA.The power triangle graphically shows the relationship between real (P), reactive (Q) and apparent power (S).

BLOG 22 - FUNDAMENTAL CIRCUIT

 Apparent Power in AC Circuits

Apparent power is the power that appears to the source because of the circuit impedance. Since the impedance is the total opposition to ac, the apparent power is that power the voltage source "sees." Apparent power is the combination of true power and reactive power. Apparent power is not found by simply adding true power and reactive power just as impedance is not found by adding resistance and reactance.

To calculate apparent power, you may use either of the following formulas:
 

 

For example, find the apparent power for the circuit shown in figure (22)
 

Recall that current in a series circuit is the same in all parts of the circuit.

Power Factor

The POWER FACTOR is a number (represented as a decimal or a percentage) that represents the portion of the apparent power dissipated in a circuit.

If you are familiar with trigonometry, the easiest way to find the power factor is to find the cosine of the phase angle Ө. The cosine of the phase angle is equal to the power factor.

You do not need to use trigonometry to find the power factor. Since the power dissipated in a circuit is true power, then:

 

If true power and apparent power are known you can use the formula shown above.

Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

 

Since current in a series circuit is the same in all parts of the circuit, I_R equals I_Z. Therefore, in a series circuit,

For example, to compute the power factor for the series circuit shown in figure (22), any of the above methods may be used.

Another method:

If you are familiar with trigonometry you can use it to solve for angleӨ and the power factor by referring to the tables in appendices V and VI.

LEARNING

I learn about the apparent power and power factor. The apparent power is a combination of both reactive power and true power.Power factor is the ratio between the kW and the kVA drawn by an electrical load where the kW is the actual load power and the kVA is the apparent load power and it is a measure of how efficiently the load current is being converted into useful work output and more particularly is a good indicator of the effect of the load current on the efficiency of the supply system.

BLOG 21 - FUNDAMENTAL CIRCUIT

RMS VALUE or Effective

WHAT IS RMS VALUE?

When dealing with Alternating Voltages (or currents) we are faced with the problem of how we represent the signal magnitude. One easy way is to use the peak values for the waveform. Another common method is to use the effective value which is also known by its more common expression of Root Mean Square or simply the RMS value.

Note that the RMS value is not the same as the average of all the instantaneous values. The ratio of the RMS value of voltage to the maximum value of voltage is the same as the ratio of the RMS value of current to the maximum value of current. Most multi-meters, either voltmeters or ammeters, measure RMS value assuming a pure sinusoidal waveform. For finding the RMS value of non-sinusoidal waveform a “True RMS Multimeter” is required.

Having now determined the RMS value of an alternating voltage (or current) waveform, in the next tutorial we will look at calculating the “Average” value VAV of an alternating voltage and finally compare the two.

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• THERE TWO TYPE BASIC METHOD IN RMS VALUE

RMS Voltage Graphical Method

Whilst the method of calculation is the same for both halves of an AC waveform, for this example we will consider only the positive half cycle. The effective or RMS value of a waveform can be found with a reasonable amount of accuracy by taking equally spaced instantaneous values along the waveform.

The positive half of the waveform is divided up into any number of “n” equal portions or mid-ordinates and the more mid-ordinates that are drawn along the waveform, the more accurate will be the final result. The width of each mid-ordinate will therefore be no degrees and the height of each mid-ordinate will be equal to the instantaneous value of the waveform at that time along the x-axis of the waveform.

Graphical Method

 

Each mid-ordinate value of a waveform (the voltage waveform in this case) is multiplied by itself (squared) and added to the next. This method gives us the “square” or Squared part of the RMS voltage expression. Next this squared value is divided by the number of mid-ordinates used to give us the Mean part of the RMS voltage expression, and in our simple example above the number of mid-ordinates used was twelve (12). Finally, the square root of the previous result is found to give us the Root part of the RMS voltage.

Then we can define the term used to describe an RMS voltage (VRMS) as being “the square root of themean of the square of the mid-ordinates of the voltage waveform” and this is given as:

 

and for our simple example above, the RMS voltage will be calculated as:

 

So lets assume that an alternating voltage has a peak voltage (Vpk) of 20 volts and by taking 10 mid-ordinate values is found to vary over one half cycle as follows:

 

Voltage	6.2V	11.8V	16.2V	19.0V	20.0V	19.0V	16.2V	11.8V	6.2V	0V
Angle	18o	36o	54o	72o	90o	108o	126o	144o	162o	180o

The RMS voltage is therefore calculated as:

 

Then the RMS Voltage value using the graphical method is given as: 14.14 Volts.

RMS Voltage Analytical Method

The graphical method above is a very good way of finding the effective or RMS voltage, (or current) of an alternating waveform that is not symmetrical or sinusoidal in nature. In other words the waveform shape resembles that of a complex waveform. However, when dealing with pure sinusoidal waveforms we can make life a little bit easier for ourselves by using an analytical or mathematical way of finding the RMS value.

A periodic sinusoidal voltage is constant and can be defined as V(t) = Vm.cos(ωt) with a period of T. Then we can calculate the root-mean-square (rms) value of a sinusoidal voltage (V(t)) as:

 

Integrating through with limits taken from 0 to 360o or “T”, the period gives:

 

Dividing through further as ω = 2π/T, the complex equation above eventually reduces down too:

RMS Voltage Equation

Then the RMS voltage (VRMS) of a sinusoidal waveform is determined by multiplying the peak voltage value by 0.7071, which is the same as one divided by the square root of two ( 1/√2 ). The RMS voltage, which can also be referred to as the effective value, depends on the magnitude of the waveform and is not a function of either the waveforms frequency nor its phase angle.

From the graphical example above, the peak voltage (Vpk) of the waveform was given as 20 Volts. By using the analytical method just defined we can calculate the RMS voltage as being:

VRMS = Vpk x 0.7071 = 20 x 0.7071 = 14.14V

 

Note that this value of 14.14 volts is the same value as for the previous graphical method. Then we can use either the graphical method of mid-ordinates, or the analytical method of calculation to find the RMS voltage or current values of a sinusoidal waveform. Note that multiplying the peak or maximum value by the constant 0.7071, ONLY applies to sinusoidal waveforms. For non-sinusoidal waveforms the graphical method must be used.

LEARNING:

The r.m.s. value of an a.c. supply is the steady d.c. which would convert electrical energy to thermal energy at the same rate in a given resistance. In RMS  VALUE we need to use the formula of Irms = I/square root of 2 and Vrms = V/square root of 2. The RMS value have two type of basic method the graphical and analytical method.

BLOG 20 - FUNDAMENTAL CIRCUIT

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:

If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:

Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:

If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverterloading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).

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. 

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.

Learning:

I laearn in maximum average power is  states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power. The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency.

BLOG 19 - FUNDAMENTAL CIRCUIT

Instantaneous power, Average power, real and reactive power

DC Circuit

As long as our analysis is restricted to Direct Current(DC) circuit the power consumed by the resistance load is the product of voltage across the resistance and current flowing through the resistance. It is really simple.

P = V . I

The power consumed by the load is the product of voltage across the load and current drawn by the load (Fig-A). Or the Power supplied by the DC source (battery/cell) is the product of voltage across the cell and current supplied by the cell. Both are equal in our example figure(considering ideal battery of zero internal resistance). The law of energy conservation implies power supplied by the source must be same as power consumed by the circuit. In DC circuit case instantaneous power is same as average power.

AC Circuit

In AC circuit analysis, what is this power that we talk about. The main problem is that the AC voltage and current varies sinusoidally with time. Moreover the presence of circuit reactive elements like Inductor and capacitor shift the current wave with respect to voltage wave (angle of phase difference). 

Power is rate at which energy is consumed by load or produced by generator. Whether it is DC circuit or AC circuit, the value of instantaneous power is obtained by multiplying instantaneous voltage with instantaneous current. If at any instant of time t the voltage and current values are represented by sine functions as

         v = V_m  sin ωt  

          i = I_m  sin (ωt-φ)

V_m and I_m  are the maximum values of the sinusoidal voltage and current. Here ω=2 π f

f is the frequency and ω is the angular frequency of rotating voltage or current phasors. It should be clear that for a power system f is usually 50 or 60 Hz

φ is the phase difference between the voltage and current.

As we said the instantaneous power is the product of instantaneous voltage and current, if we name instantaneous power as p then

p = v.i =  V_m  sin ωt  .  I_m  sin (ωt-φ)

         or  p = V_m I_m  sin ωt  sin (ωt-φ)

Applying trigonometric formula 2.sin A.sin B = cos(A-B) - cos (A+B)  we get

It can be written as

This is the equation of instantaneous power

In the Fig-C is drawn all the three waves corresponding to v, i and p. Graphically also we can get the value of instantaneous power (p) at any instant of time t by simply multiplying the value of current i and voltage v at that particular instant t. (You can verify that in the diagram p is negative when either v or i is negative otherwise p is positive. See the points where p is zero). In the graph we have shown horizontal axis as angle φ instead of time t for easy visualization. It should be clear that both way it is correct.

Clearly the instantaneous power p is composed of two terms. The first term is constant because for a given load the phase angle φ is fixed. It does not change unless the load is changed.  The second term  is varying with time sinusoidally due to the presence of the term cos (2ωt-φ). Look that the instantaneous power frequency is twice the frequency of voltage or current.

So the instantaneous power in a single phase circuit varies sinusoidally.

The instantaneous power,  p = constant term + sinusoidal oscillating term.

In one complete period the average of oscillating term is zero.

Then what is the average power within a given time, say one Time Period of the wave?

It is the constant term.

Here is another way to think about the average power.

Just observe that the instantaneous power is negative for a small time. For any time interval you just find the total +ve area A+ (above horizontal-axis (blue line) and below p curve) and total -ve area A- (below horizontal axis and  above p curve). The net area is obtained by subtracting A- from A+. By dividing this net area ( by the time interval T_i  we get the average power(P). You can do this using calculus. What you will ultimately get is only the first term in the above formula for instantaneous power p.

In still another way it is easier to realize that the formula for instantaneous power p has a constant term  (V_m.I_m / 2) cos φ and the other sinusoidal term (V_m.I_m / 2) cos (2 wt - φ). Actually p is the oscillating power which oscillates about the average constant term  (V_m.I_m / 2) cos φ .

So the average power is

The above formula can be written as

Or,

here,

 

  

V and I are the phasor representation of RMS values* of voltage and current sinusoids. The symbols |V| and |I| are  the magnitudes of phasors V and I. (See at the buttom for definition of RMS value).

This above formula is your favorite formula for useful power that we are most concerned about. This average power formula is used to find the power consumed by the load. The monthly electric energy bill at home is based on this power. The engineers and technicians in power or electrical industry simply use the term power instead of average power. So whenever we simply call power it means average power.

Of course the instantaneous power is oscillating in nature. As we already said it does not oscillates about the horizontal-axis rather about the average power P (cyan color horizontal line). 

P will be zero when cos φ =0 or  φ  = 90 degree, that is when the phase angle between voltage and current waves is 90 degrees. It is only when the load is pure inductive or capacitive. In this case the second term only remains in the instantaneous power formula.

From the above figure for some time the power becomes negative that means the load supply energy to source for this period. This is due to the presence of reactive element in load.

The above formula for instantaneous power can be written in another form. This form actually is an attempt to distinguish the oscillating reactive power from the instantaneous power formula.   Rearranging the terms in equation for instantaneous power above we get

                      p = |V| | I | cos φ (1-cow2ωt) - |V| | I | sin φ sin2ωt

In this equation the first term |V| | I | cos φ (1-cow2ωt) is oscillatory whose average value is |V| | I | cos φ. We already talked about this average power. 

The second term |V| | I | sin φ sin2ωt which is also oscillatory but with zero average value. The maximum value of this term is |V| | I | sin φ. This is the so called Reactive power. So Reactive power is the maximum value of a oscillatory power that is repeatedly drawn from the source and again returned to the source within each cycle. So the average of this reactive power is zero.

The average power P is called as Real Power. It is also sometimes called active power.

              Real power = P = |V| | I | cos φ

It is usually written as P = VI cos φ. But it should be remembered that V and I are the rms values of voltage and current. For example when we say single phase 220 volt AC it means the rms value of voltage is 220 volts ( it is not maximum value of voltage sinusoid)

              Reactive power = Q = |V| | I | sin φ

Real power is measured in Watt and the reactive power is measured in VAR (VoltAmpereReactive). In power sector these units are too small so real power is measured in Megawatt (MW) and reactive power in Megavar (MVAR). The letter R at the end denotes reactive power.

Many times students and practicing engineers are confused about the average power (often simply called power). They think that what they get by multiplying RMS voltage and RMS current is RMS power. No that is wrong. There is no RMS power. RMS power has no meaning or not defined. (Also see definition of RMS value, below at the end). It is average power or real power or true power.

Learnings

I learn about the average power is describes the power in a circuit that is transformed form electric to nonelectric energy. I also understand that the reactive power is associated with the energy stored in an inductor or capacitor. This energy that is exchanged back and forth between the source and the load and it is not converted into nonelectric energy.In solving the average power and reactive power we need to use the given formula and perform it.I also learn about the instantaneous is that electric power in an AC circuit is given by P=VI where V and I are the instantaneous voltage and current.