BALANCED WYE- WYE CONNECTION

Balanced Wye - Wye Connection

• The wye-wye arrangement requires a neutral connection between the source and the primary of the transformers not only to assure balanced line-to-neutral voltage, but also to provide a path for the third-harmonic component in the exciting current of the transformers. Without the primary neutral connection, serious unbalances in the line-neutral voltage may result from (a) unequal exciting admittances among the three transformers and (b) unbalanced line-to-neutral loads in the secondary. Furthermore, if the third harmonics are suppressed in the exciting current, large third harmonic components may appear in the line-to-neutral voltages. The wye-wye arrangement is, therefore, a four-wire system if balanced voltages are to be assured. The delta-delta arrangement is, on the other hand, a 3-wire system.

In the three phase circuit arrangements displayed, each leg is shown with an impedance. If the impedances are identical on each phase, then the load is said to be balanced. A three-phase source is balanced when each leg produces equal magnitude voltage, with phase difference between any two phases.

POWER FACTOR CORRECTION

Power Factor Correction

The process of increasing power factor without altering the voltage or the current to the original load is called Power Factor Correction. 

Alternatively, power factor correction may be viewed as the addition of a reactive element usually a capacitor in parallel with the load in order to make the power factor closer to unity.An inductive load is modeled as a series of combination of an inductor and a resistor. 

The effect of adding a capacitor can be illustrated using either the power triangle or the phasor diagram of the currents involved. Adding the capacitor has caused the phase angle between the supplied voltage and the current to reduce from ϴ1 to ϴ2, 

Thereby increasing the power factor.


Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made completely in phase with the voltage, implying unity power factor.


Capacitive Power Factor correction (PFC) is applied to electric circuits as a means of minimizing the inductive component of the current and thereby reducing the losses in the supply.

The introduction of Power Factor Correction capacitors is a widely recognized method of reducing an electrical load, thus minimizing wasted energy and hence improving the efficiency of a plant and reducing the electricity bill.It is not usually necessary to reach unity, ie Power Factor 1, since most supply companies are happy with a PF of 0.95 to 0.98

COMPLEX POWER AND POWER TRIANGLE

Complex Power and Power Triangle

Complex power S is the product of the voltage and the complex conjugate of the current.

S = VrmsIrms Cos (θv - θi) + jVrms Sin (θv - θi) 

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 | ∠φSo,

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 

Power Triangle


POWER TRIANGLE HAS THREE SIDES:-
1-APPARENT POWER (VA)
2-REACTIVE POWER (VAR)
3-REAL POWER (W)
THE RELATIONSHIP B/W THESE CAN BE EXPRESSED BY REPRESENTING QUANTITIES AS VECTORS.
REAL POWER IS HORIZONTAL VECTOR.
REACTIVE POWER IS VERTICAL POWER.
APPARENT POWER IS HYPOTENUSE OF RT.ANGLED TRIANGLE.

(apparent power)2 = (real power)2 + (reactive power)2

RATIO OF REAL POWER TO APPARENT POWER IS CALLED POWER FACTOR.
PF=W / VA 

APPARENT POWER AND POWER FACTOR

APPARENT POWER & POWER FACTOR

• Apparent factor S is the product of RMS values of voltage & current.

• It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in WATTS.

S = VrmsIrms 

• The apparent power is so called because it seems apparent that the power should be the voltage-current product, by analogy with dc resistive circuits. 

• Power factor is the cosine of the phase difference between the voltage & current. It is also the cosine of the angle of the load impedance if V is the voltage across the load and I is the current through it. 

Pf = P / S = cos ( ɸv-ɸi )

Maximum Average Power Transfer

MAXIMUM AVERAGE POWER TRANSFER

In maximizing the power directed by a power-supplying resistive network to a load RL. Representing the circuit by its Thevenin equivalent, we proved that the maximum power would be delivered to the load if the load resistance RL = RTH. An ac circuit is connected to a load ZL and is represented by its Thevenin equivalent. The load is usually represented by an impedance. 

In rectangular form the Thevenin Impedance ZTH, and the load impedance ZL are 

ZTH = RTH + jXTH

ZL = RL + jXL

The maximum average power can be transferred to the load if 

XL = - XTH

RL = RTH

It will give us a maximum average power as;

Pmax = (VTH)^2 / 8RTH

This means that for maximum average power transfer to a purely resistive load, the load impedance is equal to the magnitude of the Thevenin impedance.

Instantaneous & Average Power

Instantaneous & Average Power

Instantaneous power is the power at any instant of time. In Ac circuit the instantaneous electric power is given by

P= VI

but these quantities are continuously varying. Instantaneous Power p(t) is the power, p(t)= u(t)*i(t). It is the product of the time functions of the voltage and current. 

This definition of instantaneous power is valid for signals of any waveform. The unit for instantaneous power is VA. 

Where; P(t) = 1/2 VmIm cos(ɸv-ɸi) + 1/2 VmIm cos(2wt + ɸv + ɸi)

Instantaneous power is at positive values if power is absorbed by the circuit. It will be on the negative values if power is absorbed by the source. 

Average power is the average of instantaneous power over one period. Defined as;

P = 1/2 VmIm cos(ɸv-ɸi) 

Characteristics

1. P IS NOT TIME DEPENDENT

2. ɸv=ɸi, purely resistive load

3. ɸv-ɸi = 90, purely reactive load. 

THEVENIN EQUIVALENT CIRCUITS WITH DEPENDENT SOURCES

Thevenin's Equivalent circuits with dependent sources

Procedures in Solving Thevenin's Theorem with dependent sources:

1. Determine Vth the usual way

2. Determine Zth

3. Remove all independent sources

4. Place an assumed value of voltage at the open circuit

5. Determine the current supplied by the assumed voltage source

6. Solve for the Zth using the formula where Zth=Vo/Io

Example:

Find the thevenin equivalent of the circuit in the figure below as seen from therminals a-b.

To find Vth we apply KCL at node 1 in the figure below

15=Io+ O.5Io  => Io=10A

Applying KVL to the loop on the right-hand side in the figure above we obtain

-Io(2-j4)+0.5Io(4+j3) + Vth=0

or

Vth=10(2-j4) - 5(4+j3) = -j55

Vth= 55angle -90 V

To Obtain Zth, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals a-b as shown in figure b at the node, KCL gives

3=Io+0.5Io  =>  Io=2A

Applying KVL to the outer loop in figure b gives

Vs=Io(4+j3+2-j4)=2(6-j)

The Thevenin impedance is

Zth= Vs/Is =2(6-j)/3 =4-j0.6667 ohms

Thevenin and Norton Equivalent Circuits AC circuit

Thevenin and Norton Equivalent Circuits

Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a thevenin equivalent circuit is depicted where a linear circuit is replace by a voltage source in series with an impedance. The norton equivalent is a linear circuit replaced by a current source in parallel with an impedance. Two equivalent circuits are related as

The thevenin or Norton equivalent circuit must be determined at each frequency. This leads to entirely different equivalent circuits, one for each frequency, not one equivalent circuit with equivalent sources and equivalent impedances.

Example:

Obtain the thevenin equivalent at terminals a-b of the circuit 

We must find Zth by setting the voltage source to zero. As shown in the figure below, the 8ohm resistance is now in parallel with the -j6 reactance, so that their combination gives

Z1=-j6ll8=-j6(8)/8-j6=2.88-j3.84ohms

Similarly, the 4ohm resistance is in parallel with the j12 reactance and their combination gives

z2=4llj12=j12(4)/4+j12=3.6+j1.2ohms

The thevenin impedance is the series combination of Z1 and Z2 that is:

Zth=Z1+Z2=6.48-j2.64ohms

To find Vth consider the circuit in the figure below. Currents I1 and I2 are obtained as

I1=120angle75/8-j6 A

I2=120angle75/4+j12 A

Applying KVL around loob gives

Vth-4I2+(-j6)I1=0

or

Vth=4I2+j6I10=(480angle75)/4+j12)+(720angle75+90)/8-j6

=37.95angle3.43+72angle201.87

Vth=37.95angle220.31 V

Example:

Obtain current Io using Norton's Theorem

Our first objective is to find the Norton equivalent at terminals a-b. Zn is found in the same way as Zth. We set the sources to zero as shown in the figure above. As evident from the figure, the (8-j2) and (10+j4) impedances are short-circuited, so that 

Zn=5ohms

To get In, we short-circuit terminals a-b as in figure below and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. 

For Mesh 1,

-j40+(18+j2)I1-(8-j2)I2-(10+j4)I3=0

(a) finding Zn,   (b) finding Vn,  (c) Calculating Io

For the supermesh,

(13-j2)I2+(10+j4)I3-(18+j2)I1=0

at node a, due to the current source between meshes 2 and 3,

I3=I2+3

Adding Equations gives

-j40+5I2=0  => I2=j8

From eq where 

I3=I2+3=3+j8

The norton current is 

IN=I3=(3+j8) A

Io= 5/5+20+j15 IN = 3+j8/5+j3=1.465angle 38.48 A