Balanced Three-Phase Circuits

Initially we explored the idea of three-phase power systems by connecting three voltage sources together in what is commonly known as the “Y” (or “star”) configuration. This configuration of voltage sources is characterized by a common connection point joining one side of each source. 

Three-phase “Y” connection has three voltage sources connected to a common point.

If we draw a circuit showing each voltage source to be a coil of wire (alternator or transformer winding) and do some slight rearranging, the “Y” configuration.

Three-phase, four-wire “Y” connection uses a “common” fourth wire.

The three conductors leading away from the voltage sources (windings) toward a load are typically called lines, while the windings themselves are typically called phases. In a Y-connected system, there may or may not be a neutral wire attached at the junction point in the middle, although it certainly helps alleviate potential problems should one element of a three-phase load fail open, as discussed earlier.

Three-phase, three-wire “Y” connection does not use the neutral wire.

When we measure voltage and current in three-phase systems, we need to be specific as to where we’re measuring. Line voltage refers to the amount of voltage measured between any two line conductors in a balanced three-phase system. With the above circuit, the line voltage is roughly 208 volts.Phase voltage refers to the voltage measured across any one component (source winding or load impedance) in a balanced three-phase source or load. For the circuit shown above, the phase voltage is 120 volts. The termsline current and phase current follow the same logic: the former referring to current through any one line conductor, and the latter to current through any one component.

Y-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents. If the Y-connected source or load is balanced, the line voltage will be equal to the phase voltage times the square root of 3:

However, the “Y” configuration is not the only valid one for connecting three-phase voltage source or load elements together. Another configuration is known as the “Delta,” for its geometric resemblance to the Greek letter of the same name (Δ). Take close notice of the polarity for each winding.

Three-phase, three-wire Δ connection has no common.

At first glance it seems as though three voltage sources like this would create a short-circuit, electrons flowing around the triangle with nothing but the internal impedance of the windings to hold them back. Due to the phase angles of these three voltage sources, however, this is not the case.

One quick check of this is to use Kirchhoff’s Voltage Law to see if the three voltages around the loop add up to zero. If they do, then there will be no voltage available to push current around and around that loop, and consequently there will be no circulating current. Starting with the top winding and progressing counter-clockwise, our KVL expression looks something like this:

Indeed, if we add these three vector quantities together, they do add up to zero. Another way to verify the fact that these three voltage sources can be connected together in a loop without resulting in circulating currents is to open up the loop at one junction point and calculate voltage across the break.

Voltage across open Δ should be zero.

Starting with the right winding (120 V ∠ 120o) and progressing counter-clockwise, our KVL equation looks like this:

Sure enough, there will be zero voltage across the break, telling us that no current will circulate within the triangular loop of windings when that connection is made complete.

Having established that a Δ-connected three-phase voltage source will not burn itself to a crisp due to circulating currents, we turn to its practical use as a source of power in three-phase circuits. Because each pair of line conductors is connected directly across a single winding in a Δ circuit, the line voltage will be equal to the phase voltage. Conversely, because each line conductor attaches at a node between two windings, the line current will be the vector sum of the two joining phase currents. Not surprisingly, the resulting equations for a Δ configuration are as follows:

Example:

The load on the Δ source is wired in a Δ.

With each load resistance receiving 120 volts from its respective phase winding at the source, the current in each phase of this circuit will be 83.33 amps:

• REVIEW:
• 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.
• 
• In balanced Δ circuits, line voltage is equal to phase voltage, while line current is equal to phase current times the square root of 3.
• 
• Δ-connected three-phase voltage sources give greater reliability in the event of winding failure than Y-connected sources. However, Y-connected sources can deliver the same amount of power with less line current than Δ-connected sources.
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• Phase sequence is the time order in which the the voltages pass through their respective maximum values. The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time.
• 
  
• Balanced Wye-Wye Connection

• Balanced Wye-Delta Connection

• Balanced Delta-Delta Connection

• Balanced Delta-Wye Connection

Power Factor Correction

Power factor is the ratio between the useful (true) power (kW) to the total (apparent) power (kVA) consumed by an item of a.c. electrical equipment or a complete electrical installation. It is a measure of how efficiently electrical power is converted into useful work output.

The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.

Alternatively, power factor correction may be viewed as the addition of a reactive element ( usually a capacitor) in parallel with load in order to make the power factor closer to unity.

Conservation of AC Power

The complex, real and reactive powers of all the sources equal the respective sums of the complex, real and reactive powers of the individual loads.

Regardless of how circuit elements are connected, the total complex power delivered is equal to the complex power absorbed by the elements.

S_T = S₁ + S₂ + S₃ + S_N.....

Complex Power and Reactive Power

Complex Power

The Complex Power S 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.

Reactive Power

Reactive power Q is the imaginary part of the complex power. It is given in units of volt-amperes reactive (VAR). Reactive power is positive in an inductive circuit andnegative in a capacitive circuit. This power is defined only for sinusoidal excitation. The reactive power doesn't do any useful work or heat and it is the power returned to the source by the reactive components (inductors, capacitors) of the circuit.

Apparent Power and Power Factor

Apparent Power

The apparent Power ( in VA) is the product of the rms values of voltage and current. It is measured in colts-amperes or VA to distinguish it form the average power, which is measured in watts. 
Apparent power is a measure of alternating current (AC) power that is computed by multiplying the root-mean-square (rms) current by the root-mean-square voltage. In a direct current (DC) circuit, or in an AC circuit whose impedance is a pure resistance, the voltage and current are in phase.

Power Factor

The power factor is the cosine of the phase difference between voltage and current. It is the cosine of the angle of the load impedance.
 The power factor is the ratio of the real power that is used to do work and the apparent power that is supplied to the circuit. The power factor can get values in the range from 0 to 1.

In sinusoidal, power factor is

• θ - ϕ is the angle the voltage leads the current: PF angle

MAXIMUM AVERAGE POWER TRANSFER

A simple loop circuit used to illustrate the derivation of the maximum power transfer theorem as it applies to circuits operating in the sinusoidal steady state.

  

To get the maximum average power transferred from a source to a load, the load impedance should be equal to the conjugate of the Thevenin equivalent impedance of the network or circuit.

Instantaneous and Average Power

Instantaneous Power

Instantaneous power p(t) absorbed by an element is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it.

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

From the formula above, this shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase difference between the voltage and current. The second part is a sinusoidal function whose frequency is 2ω, which is twice the angular frequency of the voltage or current.

Average Power

P can be defined in two ways: as the real part of the complex power or as the simple average of the instantaneous power. The second definition is more general because with it we can define the instantaneous power for any signal waveform, not just for sinusoidal.

The instantaneous power changes with time and therefore difficult to measure. The average power is more convenient to measure.

NOTE: A resistive load absorbs power at all times, while a reactive load ( L or C) absorbs zero average power.

Thevenin's Theorem and Norton's Theorem

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

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

 Since Thevenin’s and Norton’s Theorems are two equally valid methods of reducing a complex network down to something simpler to analyze, there must be some way to convert a Thevenin equivalent circuit to a Norton equivalent circuit, and vice versa (just what you were dying to know, right?). Well, the procedure is very simple.

You may have noticed that the procedure for calculating Thevenin resistance is identical to the procedure for calculating Norton resistance: remove all power sources and determine resistance between the open load connection points. As such, Thevenin and Norton resistances for the same original network must be equal. Using the example circuits from the last two sections, we can see that the two resistances are indeed equal:

Considering the fact that both Thevenin and Norton equivalent circuits are intended to behave the same as the original network in supplying voltage and current to the load resistor (as seen from the perspective of the load connection points), these two equivalent circuits, having been derived from the same original network should behave identically.

This means that both Thevenin and Norton equivalent circuits should produce the same voltage across the load terminals with no load resistor attached. With the Thevenin equivalent, the open-circuited voltage would be equal to the Thevenin source voltage (no circuit current present to drop voltage across the series resistor), which is 11.2 volts in this case. With the Norton equivalent circuit, all 14 amps from the Norton current source would have to flow through the 0.8 Ω Norton resistance, producing the exact same voltage, 11.2 volts (E=IR). Thus, we can say that the Thevenin voltage is equal to the Norton current times the Norton resistance:

So, if we wanted to convert a Norton equivalent circuit to a Thevenin equivalent circuit, we could use the same resistance and calculate the Thevenin voltage with Ohm’s Law.

Conversely, both Thevenin and Norton equivalent circuits should generate the same amount of current through a short circuit across the load terminals. With the Norton equivalent, the short-circuit current would be exactly equal to the Norton source current, which is 14 amps in this case. With the Thevenin equivalent, all 11.2 volts would be applied across the 0.8 Ω Thevenin resistance, producing the exact same current through the short, 14 amps (I=E/R). Thus, we can say that the Norton current is equal to the Thevenin voltage divided by the Thevenin resistance:

This equivalence between Thevenin and Norton circuits can be a useful tool in itself, as we shall see in the next section.

• REVIEW:
• Thevenin and Norton resistances are equal.
• Thevenin voltage is equal to Norton current times Norton resistance.
• Norton current is equal to Thevenin voltage divided by Thevenin resistance.

Source Transformation

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.

The source transformation of a circuit is the transformation of a power source from a voltage source to a current source, or a current source to a voltage source.

In other words, we transform the power source from either voltage to current, or current to voltage. 

The direction of the arrow of the current source follows the positive sign of the voltage source.

EXAMPLE

In this circuit, we should identify the voltage at the 4Ω resistor. We will start transforming at the right side of the circuit using simple ohms law.

V= 2A (2Ω)

V= 4 V

The resulting circuit would be:

Combine the series impedance’s.

Then, transform it again.

I=V/R

I= 4/ 2-j1

I= 1.788∟26.56 A

Combine the parallel impedances.

Z= j3(2-j1)/ j3+(2-j1)

Z= 2.37 ∟18.43 Ω

Then transform again so that we can subtract the voltage source and finally obtain V.

V= (1.788∟26.56) (2.25+j0.75)

V= 2.998 + j2.998

V= V1-V2

V= (2.998 +j2.998)-(10)

V= 7.6168 ∟156.82 V

We can now use voltage division to obtain V.

V=(7.616∟156.82)(4)/4+ (2.25+j0.75)

V= 4.84 ∟149.97 V

Superposition Theorem

The Superposition Theorem states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone.

The principle of Superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However to apply the superposition principle, we must keep two things in mind:

1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current sources by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.

2. Dependent sources are left intact because they are controlled by circuit variables.

Steps to Apply Superposition:

1. Turn off all independent sources except one source. Find the output voltage r current due to that active source using the techniques.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to independent sources.

Mesh Analysis

Another method of analyzing circuits and the counterpart of nodal analysis is the mesh analysis. Also known as loop analysis or the mesh-current method. A mesh is a loop which does not contain any other loops within it. The current through a mesh is known as mesh current.

The steps in determining mesh currents:

• Assign the meshes found in the circuit as mesh currents i1, i2, … , in.

• Apply KVL to each of the assigned meshes to form equations. Use Ohm’s law (V=IR) to express the voltages in terms of the mesh currents.

• Solve the formed equations from step 2 to get the value of the unknown mesh currents. 

Example circuit:

Nodal Analysis

Nodal analysis is one of the many methods in solving and finding a specific value of a parameter in electronic circuit analysis. The aim of using nodal analysis is to determine the voltage in each node that’s relative to the reference node, which is the ground GND where voltage is equal to 0. This means that all the other nodes present in the circuit are referred to as the non-reference nodes; the ones that has voltage you are trying to solve for. Depends, of course, if you do need the voltage present in them or not.

Steps on how to determine node voltages:

 

• Determine the nodes of the circuit, and then select a node as the reference node (ground GND). Then assign the non-reference nodes to voltages V1, V2, Vx, or whatever you feel comfortable with.

• Apply KCL (Kirchhoff’s Current Law) to each of the non-reference nodes. Use Ohm’s law (V=IR) to express the branch currents in terms of node voltages. I use the shortcut method, though the same principles are still applied.

• Solve the resulting simultaneous equations formed from the non-reference nodes to obtain the unknown node voltages.

• Always remember that the number of equations formed should be equal to the number of unknowns. 

Example circuit:

 

Impedance and Admittance

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.

In some ways, you could say that the impedance acts the same as the resistance in a circuit, but not completely.

The admittance Y is the reciprocal of impedance, measured in siemens.

 

Resistors, capacitors, and inductors are the three elements that cause impedance. Though you have to convert the capacitance C and the inductance L to get their equivalent impedances Z, by applying these formulas: