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.

BLOG 18 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE THEVENIN THEOREM IN AC CIRCUIT.

WHAT IS THEVENIN THEOREM IN AC CIRCUIT?

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 impedanceinstead 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 (V_Th) and a series impedance (Z_Th). 

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 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. 

EXAMPLE

In this circuit, the load is the series-connected RL and CL. These load components are not part of the circuit whose equivalent we are seeking. Find the current in the load using the Norton equivalent of the circuit.

v₁(t) = 10 cos wt V; v₂(t) = 20 cos (wt+30°) V; v₃(t) = 30 cos (wt+70°) V;

v₄(t) = 15 cos (wt+45°) V; v₅(t) = 25 cos (wt+50°) V; f = 1 kHz.

First find the open circuit equivalent impedance Z_eq by hand (without the load).

Numerically

 Z_N = Z_eq = (13.93 - j5.85) ohm. 

Below we see TINA's solution. Note that we replaced all the voltage sources with short circuits before we used the meter.

 

Now the short-circuit current:

The calculation of the short-circuit current is quite complicated. Hint: this would be a good time to use Superposition. An approach would be to find the load current (in rectangular form) for each voltage source taken one at a time. Then sum the five partial results to get the total.

We will just use the value provided by TINA: 

i_N(t) = 2.77 cos (w×t-118.27°) A

 

Putting it all together (replacing the network with its Norton equivalent, reconnecting the load components to the output, and inserting an ammeter in the load), we have the solution for the load current that we sought:

 

By hand calculation, we could find the load current using current division:

Finally

I = (- 0.544 - j 1.41) A

and the time function

i(t) = 1.51 cos (w×t - 111.1°) A

LEARNING

•     In solving thevenin theorem first we find the ZTH and VTH.
•     In finding ZTH we need to turn off all the sourcing.
•     We can use the series and parallel in solving the ZTH.
•     In solvong the VTH we can use the mesh and nodal analysis.
•     We can use the basic law in solving the VTH.

BLOG 17 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE SOURCE TRANSFORMATION IN AC CIRCUIT.

WHAT IS SOURCE TRANSFORMATION IN AC CIRCUIT?

   -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 a voltage 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 V=I*Z. 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

Impedances remain the same

LEARNING

• In solving the source transformation, if the voltage source change to current source  use the formula of I = V / Z.
• In solving the source transformation, if the current source change to the voltage source use the formula of V =  I Z.
• Then we can use the kirchhoff's voltage law and we can use the basic law in solving the source transformation.

BLOG 16 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE SUPERPOSITION THEOREM IN AC CIRCUIT.

WHAT IS SUPERPOSITION THEOREM IN AC CIRCUIT.

    -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.

• 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. 

• Superposition Theorem 

•  Since AC circuits are linear, the superposition theorem applies to AC circuits the same way it applies to dc circuits. 

•  The theorem becomes important if the circuit has sources operating different frequencies. 

• In this case, we must have a different frequency‐domain circuits for each frequency. 

•  The total response must be obtained by adding the individual responses in the time domain. 

•  İts incorrect to try to add the responses in the phasor or frequency domain.

EXAMPLE

Notice that both sources have the same frequency: we will only work in this chapter with sources all having the same frequency. Otherwise, superposition must be handled differently.

Find the currents i(t) and i₁(t) using the superposition theorem.

First substitute an open circuit for the current source and calculate the complex phasors I', I1' due to the contribution only from VS.

The currents in this case are equal: 

I' = I₁' = V_S/(R_i + R₁ + j*w*L) = 50/(120+j2*p*4*10⁵*10^-5) = 0.3992-j0.0836

I' = 0.408 e^j 11.83° A

Next substitute a short-circuit for the voltage source and calculate the complex phasors I'', I1'' due to the contribution only from IS.

In this case we can use the current division formula:

I'' = -0.091 - j 0.246 A

and

I₁^" = 0.7749 + j 0.2545 A

The sum of the two steps:

I = I' + I" = 0.3082 - j 0.3286 = 0.451 e^- j 46.9° A

I₁ = I₁^" + I₁' = 1.174 + j 0.1709 = 1.1865 e^j 8.28° A 

The time functions of the currents: 

i(t) = 0.451 cos (w×t - 46.9°) A

i₁(t) = 1.1865 cos (w×t + 8.3°) A

LEARNING

•   I learn about the ac superposition theorem is we need to turn of the one of the independent source.
•   In Alternating Current it has they same procedure in Direct Current.
•   When removing source, we short the voltage source and the current source we open it.
•   We can use the mesh and nodal analysis to solve the superposition

BLOG 15 - ELECTRICAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE MESH AND NODAL ANALYSIS FOR AC CIRCUITS.

NODE POTENTIAL AND MESH CURRENT METHOD IN AC CIRCUITS

---

In this chapter we will demonstrate these methods by two examples.In the previous chapter, we've seen that the use of Kirchhoff's laws for AC circuit analysis not only results in many equations (as too with DC circuits), but also (due to the use of complex numbers) doubles the number of unknowns. To reduce the number of equations and unknowns there are two other methods we can use: the node potential and the mesh (loop) currentmethods. The only difference from DC circuits is that in the AC case, we have to work with complex impedances (or admittances) for the passive elements and complex peak or effective (rms) values for the voltages and currents.

Let's first demonstrate the use of the node potentials method.

Example 1

Find the amplitude and phase angle of the current i(t) if R = 5 ohm; L = 2 mH; C₁ = 10 mF; C₂ = 20 mF; f = 1 kHz; v_S(t) = 10 cos wt V and i_S(t) = cos wt A

Here we have only one independent node, N₁ with an unknown potential: j = v_R = v_L = v_C2 = v_IS . The best method is the node potential method.

The node equation:

Express j_M from the equation:

 

Now we can calculate I_M (the complex amplitude of the current i(t)):

 A

The time function of the current:

i(t) = 0.3038 cos (wt + 86.3^°) A

Using TINA

{Solution by TINA's Interpreter}  om:=2000*pi; V:=10; Is:=1; Sys fi (fi-V)*j*om*C1+fi*j*om*C2+fi/j/om/L+fi/R1-Is=0 end; I:=(V-fi)*j*om*C1; abs(I)=[303.7892m] radtodeg(arc(I))=[86.1709]

Now an example of the mesh current method

Click/tap the circuit above to analyze on-line or click this link to Save under Windows

Example 2

Find the current of the voltage generator V = 10 V, f = 1 kHz, R = 4 kohm, R₂ = 2 kohm, C = 250 nF, L = 0.5 H, I = 10 mA, v_S(t) = V cosw t, i_S(t) = I sinw t

Although we could again use the method of node potential with only one unknown, we will demonstrate the solution with the mesh current method.

Let's first calculate the equivalent impedances of R₂,L (Z₁) and R,C (Z₂) to simplify the work: and

ws

We have two independent meshes (loops).The first is: v_S, Z₁ and Z₂ and the second: i_S and Z₂. The direction of the mesh currents are: I₁ clockwise, I₂ counterclockwise.

The two mesh equations are: V_S = J₁*(Z₁ + Z₂) + J₂*Z₂ J₂ = I_s

You must use complex values for all the impedances, voltages and currents.

The two sources are: V_S = 10 V; I_S = -j*0.01 A.

We calculate the voltage in volts and the impedance in kohm so we get the current in mA.

Hence:

j₁(t) = 10.5 cos (w×t -7.1°) mA

Solution by TINA:

{Solution by TINA's Interpreter}  Vs:=10; Is:=-j*0.01; om:=2000*pi; Z1:=R2*j*om*L/(R2+j*om*L); Z2:=R/(1+j*om*R*C); Sys I Vs=I*(Z1+Z2)+Is*Z2 end; I=[10.406m-1.3003m*j] abs(I)=[10.487m] radtodeg(arc(I))=[-7.1224]

Finally, let's check the results using TINA.

LEARNING

I learn in this topic that you need to follow the step in DC circuit and follow the super node cases and mesh cases. There Since KCL is valid for phasors, we can analyze AC circuits by NODAL analysis. Determine the number of nodes within the network. Pick a reference node and label each remaining node with a subscripted value of voltage: V1, V2 and so on. Apply Kirchhoff’s current law at each node except the reference.  Assume that all unknown currents leave the node for each application of Kirhhoff’s current law  Solve the resulting equations for the nodal voltages. For dependent current sources: Treat each dependent current source like an independent source when Kirchhoff’s current law is applied to each defined node.  However, once the equations are established, substitute the equation for the controlling quantity to ensure that the unknowns are limited solely to the chosen nodal voltages. FOR MESH ANALYSIS Since KVL is valid for phasors, we can analyze AC circuits by MESH analysis.

THIS BLOG CONTAIN ABOUT THE SINUSOIDAL AND PHASE.

Sinusoidal Functions and Circuit Analysis

The sinusoidal functions (sine and cosine) appear everywhere, and they play an important role in circuit analysis. The sinusoidal functions provide a good approximation for describing a circuit’s input and output behavior not only in electrical engineering but in many branches of science and engineering.

The sinusoidal function is periodic, meaning its graph contains a basic shape that repeats over and over indefinitely. The function goes on forever, oscillating through endless peaks and valleys in both negative and positive directions of time. Here are some key parts of the function:

• The amplitude V_A defines the maximum and minimum peaks of the oscillations.
• Frequency f₀ describes the number of oscillations in 1 second.
• The period T₀ defines the time required to complete 1 cycle.

The period and frequency are reciprocals of each other, governed by the following mathematical relationship:

Here is a cosine function you can use as the reference signal:

You can move sinusoidal functions left or right with a time shift as well as increase or decrease the amplitude. You can also describe a sinusoidal function with a phase shift in terms of a linear combination of sine and cosine functions. Here is a cosine function and a shifted cosine function with a phase shift of π/2.

Phase shifts in a sinusoidal function

A signal that’s out of phase has been shifted left or right when compared to a reference signal:

Right shift: When a function moves right, then the function is said to be delayed. The delayed cosine has its peak occur after the origin. A delayed signal is also said to be a lag signalbecause the signal arrives later than expected.

Left shift: When the cosine function is shifted left, the shifted function is said to be advanced. The peak of the advanced signal occurs just before the origin. An advanced signal is also called a lead signal because the lead signal arrives earlier than expected.

Here are examples of unshifted, lagged, and lead cosine functions.

To see what a phase shift looks like mathematically, first take a look at the reference signal:

At t = 0, the positive peak V_A serves as a reference point. To move the reference point by time shift T_S, replace the t with (t – T_S):

where

The factor ϕ is the phase shift (or angle). The phase shift is the angle between t = 0 and the nearest positive peak. You can view the preceding equation as the polar representation of the sinusoid. When the phase shift is π/2, then the shifted cosine is a sine function.

Express the phase angle in radians to make sure it’s in the same units as the argument of the cosine (2πt/T₀ – ϕ). Angles can be expressed in either radians or degrees; make sure you use the right setting on your calculator.

When you have a phase shift ϕ at the output when compared to the input, it’s usually caused by the circuit itself.

Expand a sinusoidal function and find Fourier coefficients

The general sinusoid v(t) involves the cosine of a difference of angles. In many applications, you can expand the general sinusoid using the following trigonometric identity:

Expanding the general sinusoid v(t) leads to

The terms c and d are just special constants called Fourier coefficients. You can express the waveform as a combination of sines and cosines as follows:

The function v(t) describes a sinusoidal signal in rectangular form.

If you know your complex numbers going between polar and rectangular forms, then you can go between the two forms of the sinusoids. The Fourier coefficients c and d are related by the amplitudeV_A and phase ϕ:

If you go back to find V_A and ϕ from the Fourier coefficients c and d, you wind up with these expressions:

The inverse tangent function on a calculator has a positive or negative 180° (or π) phase ambiguity. You can figure out the phase by looking at the signs of the Fourier coefficientsc and d. Draw the points c and d on the rectangular system, where c is the x-component (or abscissa) and d is the y-component (or ordinate).

The ratio of d/c can be negative in Quadrants II and IV. Using the rectangular system helps you determine the angles when taking the arctangent, whose range is from –π/2 to π/2.

Connect sinusoidal functions to exponentials with Euler’s formula

Euler’s formula connects trig functions with complex exponential functions. The formula states that for any real number θ, you have the following complex exponential expressions:

The exponent jθ is an imaginary number, where j = √-1.

The imaginary number j is the same as the number i from your math classes, but all the cool people use j for imaginary numbers because i stands for current.

You can add and subtract the two preceding equations to get the following relationships:

These equations say that the cosine and sine functions are built as a combination of complex exponentials. The complex exponentials play an important role when you’re analyzing complex circuits that have storage devices such as capacitors and inductors.

The Phasor 

In the last tutorial, we saw that sinusoidal waveforms of the same frequency can have aPhase Difference between themselves which represents the angular difference of the two sinusoidal waveforms. Also the terms “lead” and “lag” as well as “in-phase” and “out-of-phase” were used to indicate the relationship of one waveform to the other with the generalized sinusoidal expression given as: A(t) = Am sin(ωt ± Φ) representing the sinusoid in the time-domain form.

But when presented mathematically in this way it is sometimes difficult to visualise this angular or phase difference between two or more sinusoidal waveforms so sinusoids can also be represented graphically in the spacial or phasor-domain form by a Phasor Diagram, and this is achieved by using the rotating vector method.

Basically a rotating vector, simply called a “Phasor” is a scaled line whose length represents an AC quantity that has both magnitude (“peak amplitude”) and direction (“phase”) which is “frozen” at some point in time.

A phasor is a vector that has an arrow head at one end which signifies partly the maximum value of the vector quantity ( V or I ) and partly the end of the vector that rotates.

Generally, vectors are assumed to pivot at one end around a fixed zero point known as the “point of origin” while the arrowed end representing the quantity, freely rotates in an anti-clockwise direction at an angular velocity, ( ω ) of one full revolution for every cycle. This anti-clockwise rotation of the vector is considered to be a positive rotation. Likewise, a clockwise rotation is considered to be a negative rotation.

Although the both the terms vectors and phasors are used to describe a rotating line that itself has both magnitude and direction, the main difference between the two is that a vectors magnitude is the “peak value” of the sinusoid while a phasors magnitude is the “rms value” of the sinusoid. In both cases the phase angle and direction remains the same.

The phase of an alternating quantity at any instant in time can be represented by a phasor diagram, so phasor diagrams can be thought of as “functions of time”. A complete sine wave can be constructed by a single vector rotating at an angular velocity of ω = 2πƒ, where ƒ is the frequency of the waveform. Then a Phasor is a quantity that has both “Magnitude” and “Direction”. Generally, when constructing a phasor diagram, angular velocity of a sine wave is always assumed to be: ω in rad/s. Consider the phasor diagram below.

Phasor Diagram of a Sinusoidal Waveform

As the single vector rotates in an anti-clockwise direction, its tip at point A will rotate one complete revolution of 360o or 2π representing one complete cycle. If the length of its moving tip is transferred at different angular intervals in time to a graph as shown above, a sinusoidal waveform would be drawn starting at the left with zero time. Each position along the horizontal axis indicates the time that has elapsed since zero time, t = 0. When the vector is horizontal the tip of the vector represents the angles at 0o, 180o and at 360o.

Likewise, when the tip of the vector is vertical it represents the positive peak value, ( +Am ) at 90o orπ/2 and the negative peak value, ( -Am ) at 270o or 3π/2. Then the time axis of the waveform represents the angle either in degrees or radians through which the phasor has moved. So we can say that a phasor represent a scaled voltage or current value of a rotating vector which is “frozen” at some point in time, ( t ) and in our example above, this is at an angle of 30o.

Sometimes when we are analysing alternating waveforms we may need to know the position of the phasor, representing the Alternating Quantity at some particular instant in time especially when we want to compare two different waveforms on the same axis. For example, voltage and current. We have assumed in the waveform above that the waveform starts at time t = 0 with a corresponding phase angle in either degrees or radians.

But if a second waveform starts to the left or to the right of this zero point or we want to represent in phasor notation the relationship between the two waveforms then we will need to take into account this phase difference, Φ of the waveform. Consider the diagram below from the previous Phase Difference tutorial.

Phase Difference of a Sinusoidal Waveform

The generalised mathematical expression to define these two sinusoidal quantities will be written as:

The current, i is lagging the voltage, v by angle Φ and in our example above this is 30o. So the difference between the two phasors representing the two sinusoidal quantities is angle Φ and the resulting phasor diagram will be.

Phasor Diagram of a Sinusoidal Waveform

The phasor diagram is drawn corresponding to time zero ( t = 0 ) on the horizontal axis. The lengths of the phasors are proportional to the values of the voltage, ( V ) and the current, ( I ) at the instant in time that the phasor diagram is drawn. The current phasor lags the voltage phasor by the angle, Φ, as the two phasors rotate in an anticlockwise direction as stated earlier, therefore the angle, Φ is also measured in the same anticlockwise direction.

If however, the waveforms are frozen at time t = 30o, the corresponding phasor diagram would look like the one shown on the right. Once again the current phasor lags behind the voltage phasor as the two waveforms are of the same frequency.

However, as the current waveform is now crossing the horizontal zero axis line at this instant in time we can use the current phasor as our new reference and correctly say that the voltage phasor is “leading” the current phasor by angle, Φ. Either way, one phasor is designated as the reference phasor and all the other phasors will be either leading or lagging with respect to this reference.

Phasor Addition

Sometimes it is necessary when studying sinusoids to add together two alternating waveforms, for example in an AC series circuit, that are not in-phase with each other. If they are in-phase that is, there is no phase shift then they can be added together in the same way as DC values to find the algebraic sum of the two vectors. For example, if two voltages of say 50 volts and 25 volts respectively are together “in-phase”, they will add or sum together to form one voltage of 75 volts.

If however, they are not in-phase that is, they do not have identical directions or starting point then the phase angle between them needs to be taken into account so they are added together using phasor diagrams to determine their Resultant Phasor or Vector Sum by using the parallelogram law.

Consider two AC voltages, V1 having a peak voltage of 20 volts, and V2 having a peak voltage of 30 volts where V1 leads V2 by 60o. The total voltage, VT of the two voltages can be found by firstly drawing a phasor diagram representing the two vectors and then constructing a parallelogram in which two of the sides are the voltages, V1 and V2 as shown below.

Phasor Addition of two Phasors

By drawing out the two phasors to scale onto graph paper, their phasor sum V1 + V2 can be easily found by measuring the length of the diagonal line, known as the “resultant r-vector”, from the zero point to the intersection of the construction lines 0-A. The downside of this graphical method is that it is time consuming when drawing the phasors to scale. Also, while this graphical method gives an answer which is accurate enough for most purposes, it may produce an error if not drawn accurately or correctly to scale. Then one way to ensure that the correct answer is always obtained is by an analytical method.

Mathematically we can add the two voltages together by firstly finding their “vertical” and “horizontal” directions, and from this we can then calculate both the “vertical” and “horizontal” components for the resultant “r vector”, VT. This analytical method which uses the cosine and sine rule to find this resultant value is commonly called the Rectangular Form.

In the rectangular form, the phasor is divided up into a real part, x and an imaginary part, y forming the generalised expression  Z = x ± jy. ( we will discuss this in more detail in the next tutorial ). This then gives us a mathematical expression that represents both the magnitude and the phase of the sinusoidal voltage as:

Definition of a Complex Sinusoid

So the addition of two vectors, A and B using the previous generalised expression is as follows:

Phasor Addition using Rectangular Form

Voltage, V2 of 30 volts points in the reference direction along the horizontal zero axis, then it has a horizontal component but no vertical component as follows.

• • Horizontal Component = 30 cos 0o = 30 volts
• • Vertical Component = 30 sin 0o = 0 volts
• This then gives us the rectangular expression for voltage V2 of:  30 + j0

Voltage, V1 of 20 volts leads voltage, V2 by 60o, then it has both horizontal and vertical components as follows.

• • Horizontal Component = 20 cos 60o = 20 x 0.5 = 10 volts
• • Vertical Component = 20 sin 60o = 20 x 0.866 = 17.32 volts
• This then gives us the rectangular expression for voltage V1 of:  10 + j17.32

The resultant voltage, VT is found by adding together the horizontal and vertical components as follows.

• VHorizontal = sum of real parts of V1 and V2 = 30 + 10 = 40 volts
• VVertical = sum of imaginary parts of V1 and V2 = 0 + 17.32 = 17.32 volts

Now that both the real and imaginary values have been found the magnitude of voltage, VT is determined by simply using Pythagoras’s Theorem for a 90o triangle as follows.

Then the resulting phasor diagram will be:

Resultant Value of VT

Phasor Subtraction

Phasor subtraction is very similar to the above rectangular method of addition, except this time the vector difference is the other diagonal of the parallelogram between the two voltages of V1 and V2 as shown.

Vector Subtraction of two Phasors

This time instead of “adding” together both the horizontal and vertical components we take them away, subtraction.

The 3-Phase Phasor Diagram

Previously we have only looked at single-phase AC waveforms where a single multi-turn coil rotates within a magnetic field. But if three identical coils each with the same number of coil turns are placed at an electrical angle of 120o to each other on the same rotor shaft, a three-phase voltage supply would be generated. A balanced three-phase voltage supply consists of three individual sinusoidal voltages that are all equal in magnitude and frequency but are out-of-phase with each other by exactly 120o electrical degrees.

Standard practice is to colour code the three phases as Red, Yellow and Blue to identify each individual phase with the red phase as the reference phase. The normal sequence of rotation for a three phase supply is Red followed by Yellow followed by Blue, ( R, Y, B ).

As with the single-phase phasors above, the phasors representing a three-phase system also rotate in an anti-clockwise direction around a central point as indicated by the arrow marked ω in rad/s. The phasors for a three-phase balanced star or delta connected system are shown below.

Three-phase Phasor Diagram

The phase voltages are all equal in magnitude but only differ in their phase angle. The three windings of the coils are connected together at points, a1, b1 and c1 to produce a common neutral connection for the three individual phases. Then if the red phase is taken as the reference phase each individual phase voltage can be defined with respect to the common neutral as.

Three-phase Voltage Equations

If the red phase voltage, VRN is taken as the reference voltage as stated earlier then the phase sequence will be R – Y – B so the voltage in the yellow phase lags VRN by 120o, and the voltage in the blue phase lags VYN also by 120o. But we can also say the blue phase voltage, VBN leads the red phase voltage, VRN by 120o.

One final point about a three-phase system. As the three individual sinusoidal voltages have a fixed relationship between each other of 120o they are said to be “balanced” therefore, in a set of balanced three phase voltages their phasor sum will always be zero as:  Va + Vb + Vc = 0

LEARNING

In this topic I learn about the phasor are Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC waveform. Phasor diagrams can be drawn to represent more than two sinusoids. They can be either voltage, current or some other alternating quantity but the frequency of all of them must be the same while in the sinusoidal Directly finding the steady-state response without solving the differential equation. According to the characteristics of steady-state response, the task is reduced to finding two real numbers, i.e. amplitude and phase angle, of the response. The waveform and frequency of the response are already known.

THIS BLOG CONTAIN ABOUT THE IMPEDANCE, IMPEDANCE COMBINATION AND ADMITTANCE.

IMPEDANCE

In quantitative terms, it is the complex ratio of the voltage to the current in an alternating current (AC) circuit. Impedance extends the concept of resistance to AC circuits, and possesses both magnitude and phase, unlike resistance, which has only magnitude. When a circuit is driven with direct current (DC), there is no distinction between impedance and resistance; the latter can be thought of as impedance with zero phase angle.Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied.

It is necessary to introduce the concept of impedance in AC circuits because there are two additional impeding mechanisms to be taken into account besides the normal resistance of DC circuits: the induction of voltages in conductors self-induced by the magnetic fields of currents (inductance), and the electrostatic storage of charge induced by voltages between conductors (capacitance). The impedance caused by these two effects is collectively referred to as reactance and forms the imaginary part of complex impedance whereas resistance forms thereal part.

The symbol for impedance is usually Z and it may be represented by writing its magnitude and phase in the form |Z|∠θ. However, cartesian complex number representation is often more powerful for circuit analysis purposes.

The term impedance was coined by Oliver Heaviside in July 1886.^[1][2] Arthur Kennelly was the first to represent impedance with complex numbers in 1893.^[3]

Impedance is defined as the frequency domain ratio of the voltage to the current.^[4] In other words, it is the voltage–current ratio for a singlecomplex exponential at a particular frequency ω. In general, impedance will be a complex number, with the same units as resistance, for which the SI unit is the ohm (Ω). For a sinusoidal current or voltage input, the polar form of the complex impedance relates the amplitude and phase of the voltage and current. In particular,

• The magnitude of the complex impedance is the ratio of the voltage amplitude to the current amplitude.
• The phase of the complex impedance is the phase shift by which the current lags the voltage.

The reciprocal of impedance is admittance (i.e., admittance is the current-to-voltage ratio, and it conventionally carries units of siemens, formerly called mhos).

Complex impedance

Impedance is represented as a complex quantity  and the term complex impedance may be used interchangeably; the polar form conveniently captures both magnitude and phase characteristics,

where the magnitude  represents the ratio of the voltage difference amplitude to the current amplitude, while the argument  (commonly given the symbol ) gives the phase difference between voltage and current.  is the imaginary unit, and is used instead of  in this context to avoid confusion with the symbol for electric current. In Cartesian form,

where the real part of impedance is the resistance  and the imaginary part is the reactance .

Where it is required to add or subtract impedances the cartesian form is more convenient, but when quantities are multiplied or divided the calculation becomes simpler if the polar form is used. A circuit calculation, such as finding the total impedance of two impedances in parallel, may require conversion between forms several times during the calculation. Conversion between the forms follows the normal conversion rules of complex numbers.

Ohm's law

The magnitude of the impedance  acts just like resistance, giving the drop in voltage amplitude across an impedance  for a given current . The phase factor tells us that the current lags the voltage by a phase of  (i.e., in the time domain, the current signal is shifted  later with respect to the voltage signal).

Just as impedance extends Ohm's law to cover AC circuits, other results from DC circuit analysis such as voltage division, current division,Thévenin's theorem, and Norton's theorem can also be extended to AC circuits by replacing resistance with impedance.

Complex voltage and current

In order to simplify calculations, sinusoidal voltage and current waves are commonly represented as complex-valued functions of time denoted as  and .^[7][8]

Impedance is defined as the ratio of these quantities.

Substituting these into Ohm's law we have

Noting that this must hold for all , we may equate the magnitudes and phases to obtain

The magnitude equation is the familiar Ohm's law applied to the voltage and current amplitudes, while the second equation defines the phase relationship.

Validity of complex representation

This representation using complex exponentials may be justified by noting that (by Euler's formula):

The real-valued sinusoidal function representing either voltage or current may be broken into two complex-valued functions. By the principle of superposition, we may analyse the behaviour of the sinusoid on the left-hand side by analysing the behaviour of the two complex terms on the right-hand side. Given the symmetry, we only need to perform the analysis for one right-hand term; the results will be identical for the other. At the end of any calculation, we may return to real-valued sinusoids by further noting that

Phasors[edit]

Main article: Phasor (electronics)

A phasor is a constant complex number, usually expressed in exponential form, representing the complex amplitude (magnitude and phase) of a sinusoidal function of time. Phasors are used by electrical engineers to simplify computations involving sinusoids, where they can often reduce a differential equation problem to an algebraic one.

The impedance of a circuit element can be defined as the ratio of the phasor voltage across the element to the phasor current through the element, as determined by the relative amplitudes and phases of the voltage and current. This is identical to the definition from Ohm's law given above, recognising that the factors of  cancel.

Device examples

The impedance of an ideal resistor is purely real and is referred to as a resistive impedance:

In this case, the voltage and current waveforms are proportional and in phase.

Ideal inductors and capacitors have a purely imaginary reactive impedance:

the impedance of inductors increases as frequency increases;

the impedance of capacitors decreases as frequency increases;

In both cases, for an applied sinusoidal voltage, the resulting current is also sinusoidal, but in quadrature, 90 degrees out of phase with the voltage. However, the phases have opposite signs: in an inductor, the current is lagging; in a capacitor the current is leading.

Note the following identities for the imaginary unit and its reciprocal:

Thus the inductor and capacitor impedance equations can be rewritten in polar form:

The magnitude gives the change in voltage amplitude for a given current amplitude through the impedance, while the exponential factors give the phase relationship.

Deriving the device-specific impedance's

What follows below is a derivation of impedance for each of the three basic circuit elements: the resistor, the capacitor, and the inductor. Although the idea can be extended to define the relationship between the voltage and current of any arbitrary signal, these derivations will assume sinusoidal signals, since any arbitrary signal can be approximated as a sum of sinusoids through Fourier analysis.

Resistor

For a resistor, there is the relation:

This is Ohm's law.

Considering the voltage signal to be

it follows that

This says that the ratio of AC voltage amplitude to alternating current (AC) amplitude across a resistor is , and that the AC voltage leads the current across a resistor by 0 degrees.

This result is commonly expressed as

Capacitor

For a capacitor, there is the relation:

Considering the voltage signal to be

it follows that

And thus

This says that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is , and that the AC voltage lags the AC current across a capacitor by 90 degrees (or the AC current leads the AC voltage across a capacitor by 90 degrees).

This result is commonly expressed in polar form, as

or, by applying Euler's formula, as

Inductor

For the inductor, we have the relation:

This time, considering the current signal to be:

it follows that:

And thus:

This says that the ratio of AC voltage amplitude to AC current amplitude across an inductor is , and that the AC voltage leads the AC current across an inductor by 90 degrees.

This result is commonly expressed in polar form, as

or, using Euler's formula, as

Generalised s-plane impedance

Impedance defined in terms of jω can strictly only be applied to circuits which are driven with a steady-state AC signal. The concept of impedance can be extended to a circuit energised with any arbitrary signal by using complex frequency instead of jω. Complex frequency is given the symbol s and is, in general, a complex number. Signals are expressed in terms of complex frequency by taking the Laplace transform of the time domain expression of the signal. The impedance of the basic circuit elements in this more general notation is as follows:

Element	Impedance expression
Resistor
Inductor
Capacitor

For a DC circuit this simplifies to s = 0. For a steady-state sinusoidal AC signal s = jω.

Resistance vs reactance[edit]

Resistance and reactance together determine the magnitude and phase of the impedance through the following relations:

In many applications the relative phase of the voltage and current is not critical so only the magnitude of the impedance is significant.

Resistance

Resistance  is the real part of impedance; a device with a purely resistive impedance exhibits no phase shift between the voltage and current.

Reactance[edit]

Reactance  is the imaginary part of the impedance; a component with a finite reactance induces a phase shift  between the voltage across it and the current through it.

A purely reactive component is distinguished by the sinusoidal voltage across the component being in quadrature with the sinusoidal current through the component. This implies that the component alternately absorbs energy from the circuit and then returns energy to the circuit. A pure reactance will not dissipate any power.

Capacitive reactance

A capacitor has a purely reactive impedance which is inversely proportional to the signal frequency. A capacitor consists of two conductors separated by an insulator, also known as a dielectric.

At low frequencies a capacitor is open circuit, as no charge flows in the dielectric. A DC voltage applied across a capacitor causes charge to accumulate on one side; the electric field due to the accumulated charge is the source of the opposition to the current. When the potential associated with the charge exactly balances the applied voltage, the current goes to zero.

Driven by an AC supply, a capacitor will only accumulate a limited amount of charge before the potential difference changes sign and the charge dissipates. The higher the frequency, the less charge will accumulate and the smaller the opposition to the current.

Inductive reactance

Inductive reactance  is proportional to the signal frequency  and the inductance .

An inductor consists of a coiled conductor. Faraday's law of electromagnetic induction gives the back emf  (voltage opposing current) due to a rate-of-change of magnetic flux density  through a current loop.

For an inductor consisting of a coil with  loops this gives.

The back-emf is the source of the opposition to current flow. A constant direct current has a zero rate-of-change, and sees an inductor as a short-circuit (it is typically made from a material with a low resistivity). An alternating current has a time-averaged rate-of-change that is proportional to frequency, this causes the increase in inductive reactance with frequency.

Total reactance

The total reactance is given by

so that the total impedance is

Combining impedances

The total impedance of many simple networks of components can be calculated using the rules for combining impedances in series and parallel. The rules are identical to those used for combining resistances, except that the numbers in general will be complex numbers. In the general case however, equivalent impedance transforms in addition to series and parallel will be required.

Series combination

For components connected in series, the current through each circuit element is the same; the total impedance is the sum of the component impedances.

Or explicitly in real and imaginary terms:

Parallel combination

For components connected in parallel, the voltage across each circuit element is the same; the ratio of currents through any two elements is the inverse ratio of their impedances.

Hence the inverse total impedance is the sum of the inverses of the component impedances:

or, when n = 2:

The equivalent impedance  can be calculated in terms of the equivalent series resistance  and reactance .^[9]

Admittance

In electrical engineering, admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the inverse of impedance. The SI unit of admittance is the siemens (symbol S). Oliver Heaviside coined the term admittance in December 1887.^[1]

Admittance is defined as

where

Y is the admittance, measured in siemens

Z is the impedance, measured in ohms

The synonymous unit mho, and the symbol ℧ (an upside-down uppercase omega Ω), are also in common use.

Resistance is a measure of the opposition of a circuit to the flow of a steady current, while impedance takes into account not only the resistance but also dynamic effects (known as reactance). Likewise, admittance is not only a measure of the ease with which a steady current can flow, but also the dynamic effects of the material's susceptance to polarization:

where

•  is the admittance, measured in siemens.
•  is the conductance, measured in siemens.
•  is the susceptance, measured in siemens.
• 

Conversion from impedance to admittance

Parts of this article or section rely on the reader's knowledge of the complex impedance representation of capacitors and inductors and on knowledge of the frequency domainrepresentation of signals.

The impedance, Z, is composed of real and imaginary parts,

where

• R is the resistance, measured in ohms
• X is the reactance, measured in ohms

Admittance, just like impedance, is a complex number, made up of a real part (the conductance, G), and an imaginary part (the susceptance, B), thus:

where G (conductance) and B (susceptance) are given by:

The magnitude and phase of the admittance are given by:

where

• G is the conductance, measured in siemens
• B is the susceptance, also measured in siemens

Note that (as shown above) the signs of reactances become reversed in the admittance domain; i.e. capacitive susceptance is positive and inductive susceptance is negative.

LEARNING

In this topic i learn about this is you need to follow the formula and use it.Larger C more movement of charge smaller resistance Higher f more movement of charge approach short-circuit.DC circuit Resistance (R) ≠ conductance (G) AC circuit Impedance (Z)≠ admittance (Y) Reactance (X) ≠ susceptance (B) BUT, note sign inversion for imaginary numbers:BC = 1/XC, but due to imaginary numbers:jBC = -j/XC