The Three Phase Circuits

THE 'Y' CONNECTION

GENERAL DEFINITIONS:

• The conductors connected to the three points of a three-phase source or load are called lines.
• The three components comprising a three-phase source or load are called phases.
• Line voltage is the voltage measured between any two lines in a three-phase circuit.
• Phase voltage is the voltage measured across a single component in a three-phase source or load.
• Line current is the current through any one line between a three-phase source and load.
• Phase current is the current through any one component comprising a three-phase source or load.
• In balanced “Y” circuits, line voltage is equal to phase voltage times the square root of 3, while line current is equal to phase current.

THE DELTA CONNECTION

For Delta Circuits:

Eline = Ephase

Iline = Square root of 3(Iphase)

THE DELTA - DELTA CONNECTION

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• All sources are represented by a set of balanced 3-phase variables
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• Loads are three phase with equal Z's (impedances)
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• Line impedance are equal in all three phases

• All three variables have the same amplitude
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• All three variables are in phase 120^o 

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• All three variables have the same frequency
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Power in AC circuits

     

   

    Electricity was discovered by many experts at different ages, and is not invented. Until in our modern time, electricity, indeed, is really on a state of people's demand. In other words,the higher of technology, the more electricity is needed. To acquire electricity, we need an electric power.

     
       Power in AC Circuits has any different expressions to be analyzed clearly and understand its concepts. It has an instantaneous power. Across the element, it is the product of an instantaneous voltage and an instantaneous current flowing through it. 

      
The energy flow in the system are expressed in the following:

    Remember that S is the complex power. Otherwise, the S length is the apparent power. Reactive Power has the reason values established(inductor and capacitor). We know that in an induction, it creates magnetic field by a coil of wire. On the other hand, in capacitor, has a magnetic field on which these two only comes from reactive power.  Active Power is equal to the product of current and voltage in a root mean square(rms) on which it is a cosine of each angles. 

Then,

 P = Vrms Irms cos(Vangle - Iangle) 

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THE POWER FACTOR DEFINED & CORRECTION

What is p.f.? It is a power factor. Power factor is the ratio between the active power(P) and apparent power(S). Therefore, p.f. = P/S or p.f. =  cos(Vangle - Iangle). As what I've observed lately, getting a p.f. value should be near or exactly or ahead to 0.8. 

Take note that, getting a load value must be identified if:

• it is positive, it is an inductive. And it is lagging.
• it is negative, it is capacitive. And it is leading.
• it is purely resistive, it means that the voltage and current are both  "in phase."

Example 1.0:Find the power factor when: 

• The wattmeter is reading 1.5 kW. And
• Ammeter reading 7.53 A rms. at 240 V rms  

At first, we're going to calculate the apparent power by multiplying both V rms and I rms.

S = EI
S = (240)(7.53)
S = 1807.2 kVA

Since we've got the exact value of an apparent power, it is now time to find the power factor.

Solution: Using the ratio of True Power and a Complex Power.

Thus, 

p.f. = P/S
p.f. = 1.5kW / 1807.2kVA
p.f. = 0.8

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THE POWER TRIANGLE

In this diagram, it shows the functions of this right triangle figure. Side by side, it has different power expressions. Reactive power is placed in an opposite. The Active power or "True Power", is placed in an adjacent and the Complex power is placed in a hypotenuse.If we turn back on Trigonometry, the principle "sohcahtoa" is most useful in such calculations. 

Source Transformation

In this topic, another AC analysis included is a Source Transformation. It is when you transform the voltage source to the current source, or via vice versa. To obtain the solution by transformation, use the Ohm's law. And when you obtained a value for example,of a current source from the voltage source, any elements to series like resistors, capacitors or inductors should be placed parallel to the source. 

Considering this circuit example:

You can even set it in a fullscreen. :)

Superposition Theorem

   The strategy used in the superposition theorem is to eliminate all but one source of power within a network at a time, by using series/parallel analysis to determine voltage drops(and/or currents) within the network separately. 

   The values are all "superimposed" on top of each other when once the voltage drops and/or currents have been determined for each power source working separately.

Thevenin's Theorem

Any combination of impedances and sources with two terminals, it can be replaced by a single voltage source e and a single series impedance z. The value of e is an open circuit voltage at the terminals, and z value is divided by the current on which it is with the terminals in a short-circuit. This involves the series/parallel combination.

Considering this circuit example:

Mesh Analysis

The Kirchorff's Voltage Law (KVL) is the main basis of the Mesh Analysis. It is either the current only exist in one mesh or when the current source is in between of two meshes. And these are the two cases in solving the mesh analysis.  


To understand such cases, let's have an example.
Considering this circuit:

Here are the steps in solving:
1. Analyze circuit.
2. Look at the figure. I have assigned I1 for the first current loop and I2 for the second current loop.
3. Since that the second loop has the current source of 2 amperes,therefore,I conclude that:

                              I2 = -2 Amps

4. Solve Mesh 1 to get I1:
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12/_0 + 3I1 - j2(I1-I2) = 0
12/_0 + 3I1 - j2I1 - j2I2 = 0
12/_0 + I1(3-j2) - j2I2 = 0
I1(3-j2)-j2I2 = -12/_0
I1(3-j2)-j2(-2/_0) = -12/_0
I1(3-j2)+j4 = -12/_0
I1 (3-j2) = -12/_0 - j4

I1 = 3.508/_-127.87 Amps


Itotal = 4.99/_-146.30 Amps
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V = IR
V = (4.99/_-146.30)(j2)
V = 9.98/_-56.307 Volts




Considering this circuit:

Here are the steps in solving:
1. Analyze the circuit.
2. In this case, the current source is in between of two meshes. Then it will be changed by an open circuit. See the figure below.

 3. Solve for Supermesh by using KVL:

-j2I1 - 3I2 = 0 ----------> Equation 1

• Get I2 by KCL: 

I2 = 2/_0 + I1  ----------> Equation 2

• Substitute Eq. 2 in 1:

-j2I1 - 3(2/_0 + I1) = 0

-j2I1 - 6 - 3I1 = 0

I1(-3-j2) = 6

I1 = 1.664/_146.309 A

• for I2:

I2 = 2/_0 + 1.664/_146.309

I2 = 1.109/_56.309 Amps

• Solve for V:

V = IR

V = 1.109/_56.309(3)

V = 3.328/_56.309 Volts

Impedance and its Components

       An impedance of the circuit is the total effective resistance or the opposition measure to the passage of current in an AC voltage source supply. It's unit is ohms and is given V = IZ. Resistance in the form of ohm's law, reamins constant in DC or AC circuits and is different in capacitor and an inductor. The Capacitor in DC circuit, as well as in an inductor is an open circuit,while in AC voltage source, the capacitive resistance is called reactance or Xc, on the other hand, inductive resistance is equal to the reactance or XL.