equation, reactances. Thus, Current = Voltage , Or Xei = (Xc) 1 + (Xc)2.
Capacitive reactance The total reactance of capacitors connected in parallel is found in the same way total resistance is computed in a parallel circuit:
is similar to Ohm's law and the equation for current in an inductive circuit. The greater the frequency, the less the reactance. Hence, the capacitive reactance, where: f = frequency in c.p.s.
C = capacity in farads.
2 т = 6.28.
Problem:
A series circuit is assumed in which the impressed voltage is 110 volts at 60 c.p.s., and the capacitance of a condenser is 80 uf. Find the capacitive reactance and the current flow.
Solution:
To find capacitive reactance, the equation X. = 27 fc is used. First, the capacitance, 30 f., is changed to farads by dividing 80 by 1,000,000, since 1 million microfarads is equal to 1 farad. This quotient equals .000080 farad. This is substituted in the equation and X: = 6.28 × 60 × .000080 X. = 33.2 ohms reactance.
Find the current flow:
1 = 110 33.2 I = 3.31 amperes.
(X): = (Хc)1 Phase of Current and Voltage in Reactive Circuits When current and voltage pass through zero and reach maximum value at the same time, the current and voltage are said to be in phase (A of figure 8-187). If the current and voltage pass through zero and reach the maximum values at different times, the current and voltage are said to be out of phase. In a circuit containing only inductance, the current reaches a maximum value later than the voltage, lagging the voltage by 90°, or one-fourth cycle (B of figure 8-187). In a circuit containing only capacitance, the current reaches its maximum value ahead of the voltage and the current leads the voltage by 90°, or one-fourth cycle (C of figure 8-187). The amount the current lags or leads the voltage in a circuit depends on the relative amounts of resistance, inductance, and capacitance in the circuit.
OHM'S LAW FOR A. C. CIRCUITS The rules and equations for d.c. circuits apply to a.c. circuits only when the circuits contain resistance alone, as in the case of lamps and heating elements. In order to use effective values of voltage and current in a.c. circuits, the effect of inductance and capacitance with resistance must be considered.
The combined effect of resistance, inductive reactance, and capacitive reactance makes up the total opposition to current flow in an a.c. circuit.
This total opposition is called impedance and is represented by the letter "Z." The unit for the measurement of impedance is the ohm.
Capacitive Reactances in Series and in Parallel When capacitors are connected in series, the total reactance is equal to the sum of the individual Series A. C. Circuits If an a.c. circuit consists of resistance only, the value of the impedance is the same as the 352
A for the EMF EMF EMF Current and voltage in phase Effect of inductance Effect of capacitance B FIGURE 8-187. Phase of current and voltage.
resistance, and Ohm's law for an a.c. circuit, 1 - is exacily the same as for a de. sirenit. In figure 8-188 a series circuit containing a lamp with Il ohms resistance connected across a source is illustrated. To find how much current will flow if 110 volts d.c. are applied and how much current will flow if 110 volts a.c. are applied, the following examples are solved:
I = R E I - → (where Z = R) = 115 110V = 10 amperes d.c. = 10 amperes a.c.
When a.c. circuits contain resistance and either inductance or capacitance, the impedance, L, is not the same as the resistance, R. The impedance of a circuit is the circuit's total opposition to the flow of current. In an a.c. circuit, this opposition consists of resistance and reactance, either inductive or capacitive, or elements of both.
Resistance and reactance cannot be added directly, but they can be considered as two forces acting at right angles to each other. Thus, the relation between resistance, reactance, and impedance may be illustrated by a right triangle, as shown in figure 8-189.
Since these quantities may be related to the sides of a right triangle, the formula for finding the impedance, or total opposition to current flow in an a.c. circuit, can be found by using the law of right triangles. This theorem, called the Pythagorean theorem, applies to any right triangle. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Thus, the value of any side of a right triangle can be found if the other two sides are known. If an
a.c. circuit contains resistance and inductance, as shown in figure 8-190, the relation between the sides can be stated as:
22 = R + XL.
Reactance Impedance 110V - Xc Z R Resistance FIGURE 8-188. Applying d.c. and a.c. to a circuit. FIGURE 8-189. Impedance triangle.
Then the current flow, 110V AC 62 1 = Z E 60 cycles 0.021 henrys 10 FIGURE 8-190. A circuit containing resistance and inductance.
The square root of both sides of the equation gives Z = VR + X2.
This formula can be used to determine the impedance when the values of inductive reactance and resistance are known. It can be modified to solve for impedance in circuits containing capacitive reactance and resistance by substituting Xc in the formula in place of Xr. In circuits containing resistance with both inductive and capacitive reactance, the reactances can be combined, but because their effects in the circuit are exactly opposite, they are combined by subtraction:
X = X1 - Xc or X = Xc - Xz (the smaller is always subtracted from the larger).
In figure 8-190, a series circuit consisting of resistance and inductance connected in series is connected to a source of 110 volts at 60 cycles per second. The resistive element is a lamp with 6 ohms resistance, and the inductive element is a coil with an inductance of 0.021 benry. What is the value of the impedance and the current through the lamp and the coil?
Solution:
First, the inductive reactance of the coil is computed:
I = 11 amperes current.
The voltage drop across the resistance (En) is ER = IXR ER - 11 X 6 = 66 volts.
The voltage drop across the inductance (Ext) is Ex: = IX XI Exz = 11 × 8 = 88 volts.
The sum of the two voltages is greater than the impressed voltage. This results from the fact that the two voltages are out of phase and, as such, represent the maximum voltage. If the voltage in the circuit is measured by a voltmeter, it will be approximately 110 volts, the impressed voltage.
This can be proved by the equation E = V(ER)? + (ExL)2 E = V662 + 882 E = V4356 + 7744 E = V12100 E = 110 volts.
In figure 8-191, a series circuit is illustrated in which a capacitor of 200 uf. is connected in series with a 10-ohm lamp. What is the value of the impedance, the current flow, and the voltage drop across the lamp?
XI = 2=XfXL Xx = 6.28 × 60 × 0.021 Xz = 8 ohms inductive reactance. 110V AC 1022 Next, the total impedance is computed: 60 cycles 200 ut ;
Z VRE + XE Z 62 ÷ 82 Z 36 + 64 Z · 100 FIGURE 8-191. A circuit containing resistance Z = 10 ohms impedance. and capacitance.
Solution: The voltage drop across the capacitor (Exc) is First the capacitance is changed from uf. to Exe = 1 X Xc farads. Since 1 million microfarads equal 1 farad, then Exc = 6.7 × 13 Exe = 86.1 volts.
200 200 uf. = 1,000,000 = .000200 farads. The sum of these two voltages does not equal the applied voltage, since the current leads the voltage. To find the applied voltage, the formula Xc = 2r fC Er = V (ER)' + (Exc)' is used.
Er = V 672 + 86.12 Xc = 6.28 × 60 × .000200 farads Er = V 4489 + 7413 Er = V11902 Xo = .07536 Er = 110 volts.
When the circuit contains resistance, inductance, Xo = 13 ohms capacitive reactance. and capacitance, the equation To find the impedance, Z = VR + (Xi - Xc)2 Z = 1 VR° + Ac is used to find the impedance.
= V 102 + 132 Example:
Z = V100 + 169 What is the impedance of a series circuit = V269 (figure 8-192), consisting of a capacitor with a Z = 16.4 ohms capacitive reactance. reactance of 7 ohms, an inductor with a reactance of 10 ohms, and a resistor with a resistance of To find the current, 4 ohms?
E Solution:
= Z Z = VR + (XI - Xc)2 110 Z = V42 + (10 - 7)2 = 16.1 Z = V43 + 32 = 6.7 amperes. Z = V25 The voltage drop across the lamp (ER) is Z = 5 ohms.
ER = 6.7 × 10 Assuming that the reactance of the capacitor is ER = 67 volts. 10 ohms and the reactance of the inductor is 7 ohms, then Xc is greater than Xr. Thus, 2 = VR+ (XI - Xc) 4:2 110V AC 2 = 143+ (7 - 10)2 70 60 cycles I = 143 + (-3)2 108 Z = V16 + 9 FIGURE 8-192. A circuit containing resistance, Z = V25 inductance, and capacitance. Z = 5 ohms.
into the formula given:
1 5 henrys 1 ≤ 10000 = 2 × 3.14 × 60 × 0.000002 1 10,000 or
0.00075360 - 7.536 = 1,3270 capacitive reactance.
FIGURE 8-193. A.C. parallel circuit containing inductance and resistance. To find the impedance, the impedance formula used in a series a.c. circuit must be modified to fit the parallel circuit:
Parallel A. C. Circuits RXc The methods used in solving parallel a.c. circuit problems are basically the same as those used for series a.c. circuits. Out-of-phase voltages and 10,000 × 1327 currents can be added by using the law of right = V (10,000) + (1327)% triangles, but in solving circuit problems, the currents through the branches are added, since the = 0.13150 (approx.).
voltage drops across the various branches are the To find the current through the capacitance:
same and are equal to the applied voltage. In figure 8-193, a parallel a.c. circuit containing an 10 - Ка inductance and a resistance is shown schematically.
is 0.0584 ampere, and the current flowing through The current fowing through the inductance, I1, = 1327 110 the resistance is 0.11 ampere. What is the total current in the circuit? = 0.0829 ampere.
To find the current flowing through the Solution: resistance:
= V (0.0584)% + (0.11)2 In = R = V0.0155 110 = 0.1245 ampere. = 10,000 = 0.011 ampere.
Since inductive reactance causes voltage to lead the current, the total current, which contains a component of inductive current, lags the applied To find the total current in the circuit:
voltage. If the current and voltages are plotted, I = VIR + Iở the angle between the two, called the phase angle, Ir = V (0.011) + (.0829)2 illustrates the amount the current lags the voltage.
In figure 8-194, a 110-volt generator is con- = 0.0836 ampere (approx.).
nected to a load consisting of a 2-uf. capacitance and a 10,000-ohm resistance in parallel. What is the value of the impedance and total current flow?
Solution: • 2 uf. ≤ 10,000 г 110 First, find the capacitive reactance of the Volts circuit:
Xo = 2= fC FIGURE 8-194. A parallel a.c. circuit containing Changing 2uf. to farads and entering the values capacitance and resistance.
Resonance It has been shown that both inductive reactance (X1 = 2r fL) and capacitive reactance (Xo = 2+ fC are functions of an alternating current frequency.
Decreasing the frequency decreases the ohmic value of the inductive reactance, but a decrease in frequency increases the capacitive reactance. At some particular frequency, known as the resonant frequency, the reactive effects of a capacitor and an inductor will be equal. Since these effects are the opposite of one another, they will cancel, leaving only the ohmie value of the resistance to oppose current flow in a circuit. If the value of resistance is small or consists only of the resistance in the conductors, the value of current flow can become very high.
In a circuit where the inductor and capacitor are in series, and the frequency is the resonant frequency, or frequency of resonance, the circuit is said to be "in resonance" and is referred to as a series resonant circuit. The symbol for resonant frequency is Fn.
If, at the frequency of resonance, the inductive reactance is equal to the capacitive reactance, then XL = Х, or 1 2T fL = 2T fC Dividing both sides by 2 fL, Fr? = (2x) LC Extracting the square root of both sides gives Fn = 2+ VLČ' Where F. is the resonant frequency in cycles per second, C is the capacitance in farads, and L is the inductance in henrys. With this formula the frequency at which a capacitor and inductor will be resonant can be determined.
To find the inductive reactance of a circuit use Xx =2 (7) tL The impedance formula used in a series ac circuit must be modified to fit a parallel circuit.
RZL Z= · R2=XL2 FIGURE 8-195. A parallel resonant circuit.
To find the parallel networks of inductanee and capacitive reactors use AL NO X= To find the parallel networks with resistance capacitive and inductance use:
R X1 Xc Z= XxXo+ (RXL-RXo) 2 Since at the resonant frequency X1 cancels Xc, the current can become very large, depending on the amount of resistance. In such cases, the voltage drop across the inductor or capacitor will often be higher than the applied voltage.
In a parallel resonant circuit (figure 8-195), the reactances are equal and equal currents will flow through the coil and the capacitor.
Since the inductive reactance causes the current through the coil to lag the voltage by 90°, and the capacitive reactance causes the current through the capacitor to lead the voltage by 90°, the two currents are 180° out of phase. The cancelling effect of such currents would mean that no current would flow from the generator and the parallel combination of the inductor and the capacitor would appear as an infinite impedance. In practice, no such circuit is possible, since some value of resistance is always present, and the parallel circuit, sometimes called a tank circuit, acts as a very high impedance. It is also called an antiresonant circuit, since its effect in a circuit is opposite to that of a series-resonant circuit, in which the impedance is very low.
Power in A. C. Circuits In a d.c. circuit, power is obtained by the equation, P = EI, (watts equal volts times amperes). Thus, if 1 ampere of current flows in a circuit at a pressure of 200 volts, the power is 200 watts. The product of the volts and the amperes is the true power in the circuit.
In an a.c. circuit, a voltmeter indicates the effective voltage and an ammeter indicates the 357
Reactive power Apparent power Volts X Amperes True Power Watts FIGURE 8-196. Power relations in a.c. circuit.
effective current. The product of these two readings is called the apparent power. Only when the a.c.
circuit is made up of pure resistance is the apparent power equal to the true power (figure 8-196).
When there is capacitance or inductance in the circuit, the current and voltage are not exactly in phase, and the true power is less than the apparent power. The true power is obtained by a wattmeter reading. The ratio of the true power to the apparent power is called the power factor and is usually expressed in percent. In equation form, the relationship is:
Power Factor (PF) = 100 X Watts (True Power) Volts X Amperes (Apparent Power) ' Problem:
A 220-volt a.c. motor takes 50 amperes from the line, but a wattmeter in the line shows that only 9,350 watts are taken by the motor, What is the apparent power and the power factor?
Solution:
Apparent power = Volts X Amperes.
Apparent power = 220 × 50 = 11,000 watts or volt-amperes.
PF = Watts (True Power) × 100 VA (Apparent Power) PF = 9,350 × 100 11,000 PF = 85, or 85%.
TRANSFORMERS A transformer changes electrical energy of a given voltage into electrical energy at a different voltage level. It consists of two coils which are not electrically connected, but which are arranged in such a way that the magnetic field surrounding one coil cuts through the other coil. When an alternating voltage is applied to (across) one coil, the varying magnetic field set up around that coil creates an alternating voltage in the other coil by mutual induction. A transformer can also be used with pulsating d.c., but a pure d.c. voltage cannot be used, since only a varying voltage creates the varying magnetic field which is the basis of the mutual induction process.
A transformer consists of three basic parts, as shown in figure 8-197. These are an iron core which provides a circuit of low reluctance for magnetic lines of force, a primary winding which receives the electrical energy from the source of applied voltage, and a secondary winding which receives electrical energy by induction from the primary coil.
The primary and secondary of this closed-core transformer are wound on a closed core to obtain maximum inductive effect between the two coils.
There are two classes of transformers: (1) Voltage transformers used for stepping up or stepping down voltages, and (2) current transformers used in instrument circuits.
In voltage transformers the primary coils are connected in parallel across the supply voltage, as shown in A of figure 8-198. The primary windings of current transformers are connected in series in the primary circuit (B of figure 8-198). Of the two types, the voltage transformer is the more common.
There are many types of voltage transformers.
Most of these are either step-up or step-down transformers. The factor which determines whether a transformer is a step-up or step-down type is the "turns" ratio. The turns ratio is the ratio of the number of turns in the primary winding to the number of turns in the secondary winding. For example, the turns ratio of the step-down transformer shown in A of figure 8-199 is 5 to 1, since there are five times as many turns in the primary as in the secondary. The step-up Primary coil Iron core Secondary coil FIGURE 8-197. An iron-core transformer.
AC power supply AC pover supply To load Load Meter FIcURE 8-198. Voltage and current transformers.
transformer shown in B of figure 8-199 has a 1-to-4 turns ratio.
The ratio of the transformer input voltage to the output voltage is the same as the turns ratio if the transformer is 100 percent efficient. Thus, when 10 volts are applied to the primary of the transformer shown in A of figure 8-199, two volts are induced in the secondary. If 10 volts are applied to the primary of the transformer in B of figure 8-199, the output voltage across the terminals of the secondary will be 40 volts.
No transformer can be constructed that is 100 percent efficient, although iron-core transformers can approach this figure. This is because all the magnetic lines of force set up in the primary do not cut across the turns of the secondary coil. A certain amount of the magnetic flux, called leakage flux, leaks out of the magnetic circuit. The measure of how well the flux of the primary is coupled into the secondary is called the "coefficient of coupling."
For example, if it is assumed that the primary of a transformer develops 10,000 lines of force and only 9,000 cut across the secondary, the coefficient of coupling would be .9 or, stated another way, the transformer would be 90 percent efficient.
When an a.c. voltage is connected across the primary terminals of a transformer, an alternating current will flow and self-induce a voltage in the primary coil which is opposite and nearly equal to the applied voltage. The difference between these two voltages allows just enough current in the primary to magnetize its core. This is called the exciting, or magnetizing, current. The magnetic field caused by this exciting current cuts across the secondary coil and induces a voltage by mutual induction. If a load is connected across the secondary coil, the load current flowing through the secondary coil will produce a magnetic field which will tend to neutralize the magnetic field produced by the primary current. This will reduce the self-induced (opposition) voltage in the primary coil and allow more primary current to flow. The primary current increases as the secondary load current increases, and decreases as the secondary load current decreases. When the secondary load is removed, the primary current is again reduced to the small exciting current sufficient only to magnetize the iron core of the transformer.
If a transformer steps up the voltage, it will step down the current by the same ratio. This should be evident if the power formula is considered, for the power (I X E) of the output (secondary) electrical energy is the same as the input (primary) power minus that energy loss in the transforming process. Thus, if 10 volts and 4 amps (40 watts of power) are used in the primary to produce a magnetic field, there will be 40 watts of power developed in the secondary (disregarding any loss). If the transformer has a step-up ratio of 4 to 1, the voltage across the secondary will be 40 volts and the current will be 1 amp. The voltage is 4 times greater and the current is one-fourth the primary circuit value, but the power (I X E value) is the same.
2 turns secondary 10 turos primary 2 turs primary 8 turns secondary B FIGURE 8-199. A step-down and a step-up transformer.
Iron Care Red Red-Yellow High-voltage winding Red Black Yellow PRIMARY Yellow-Blue 5-volt FIGURE 8-200. Power supply transformer. Yellow winding When the turns ratio and the input voltage are known, the output voltage can be determined as follows:
Where E is the voltage of the primary, E, is the output voltage of the secondary, and Ni and N2 are the number of turns of the primary and secondary, respectively.
Transposing the equation to find the output voltage gives:
The most commonly used types of voltage transformers are as follows:
(1) Power transformers are used to step up or step down voltages and current in many types of power supplies. They range in size from the small power transformer shown in figure 8-200 used in a radio receiver to the large transformers used to step down highpower line voltage to the 110-120 volt level used in homes.
In figure 8-201, the schematic symbol for an iron-core transformer is shown. In this case the secondary is made up of three separate windings. Each winding supplies a different circuit with a specific voltage, which saves the weight, space, and expense of three separate transformers. Each secondary has a midpoint connection, called a "center tap," which provides a selection of half the voltage across the whole winding.
The leads from the various windings are color-coded by the manufacturer, as labeled in figure 8-201. This is a standard color code, but other codes or numbers may be used.
Black Green Green-Yellow 6 volt Green winding
SECONDARY
WINDINGS FIGURE 8-201. Schematic symbol for an iron-core power transformer.
(2) Audio transformers resemble power transformers. They have only one secondary and are designed to operate over the range of audio frequencies (20 to 20,000 c.p.s.).
(3) RF transformers are designed to operate in equipment that functions in the radio range ot frequencies. The symbol for the RF transformer is the same as for an RF choke coil.
It has an air core as shown in figure 8-202.
(4) Autotransformers are normally used in power circuits; however, they may be designed for other uses. Two different symbols for autotransformers used in power or audio circuits are shown in figure 8-203.
If used in an RF communication or navigation circuit (B of figure 8-203), it is the same, except there is no symbol for an iron core. The autotransformer uses part of a winding as a primary; and, depending on whether it is step-up or step-down, it uses all or part of the same winding as the secondary. For example, the autotransformer shown in A of figure 8-203 could use the following possible choices for primary and secondary terminals.
FIGURE 8-202. An air-core transformer.
Current Transformers Current transformers are used in a.c. power supply systems to sense generator line current and to provide a current, proportional to the line current, for circuit protection and control devices.
The current transformer is a ring-type transformer using a current-carrying power lead as a primary (either the power lead or the ground lead of the a.c. generator). The current in the primary induces a current in the secondary by magnetic induction.
The sides of all current transformers are marked "H1" and "H2" on the unit base. The transformers must be installed with the "HI" side toward the generator in the circuit in order to have proper polarity. The secondary of the transformer should never be left open while the system is being operated; to do so could cause dangerously high voltages, and could overheat the transformer.
Therefore, the transformer output connections should always be connected with a jumper when the transformer is not being used but is left in the system.
Transformer Losses In addition to the power loss caused by imperfect coupling, transformers are subject to "copper" and "iron" losses. Copper loss is caused by the resistance of the conductor comprising the turns of the coil. The iron losses are of two types called hysteresis loss and eddy current loss. Hysteresis loss is the electrical energy required to magnetize the transformer core, first in one direction and then in the other, in step with the applied alternating voltage. Eddy current loss is caused by electric currents (eddy currents) induced in the transformer core by the varying magnetic fields. To reduce eddy current losses, cores are made of laminations coated with an insulation, which reduces the circulation of in- Power in Transformers Since a transformer does not add any electricity to the circuit but merely changes or transforms the electricity that already exists in the circuit from one voltage to another, the total amount of energy in a circuit must remain the same. If it were possible to construct a perfect transformer, there would be no loss of power in it; power would be transferred undiminished from one voltage to another.
Since power is the product of volts times amperes, an increase in voltage by the transformer must result in a decrease in current and vice versa.
There cannot be more power in the secondary side of a transformer than there is in the primary. The product of amperes times volts remains the same.
The transmission of power over long distances is accomplished by using transformers. At the power source the voltage is stepped up in order to reduce the line loss during transmission. At the point of utilization, the voltage is stepped down, since it is not feasible to use high voltage to operate motors, lights, or other electrical appliances.
Connecting Transformers in A. C. Circuits Before studying the various means of connecting transformers in a.c. circuits, the differences between single-phase and three-phase circuits must be clearly understood. In a single-phase circuit the voltage is generated by one alternator coil. This -3 single-phase voltage may be taken from a singlephase alternator, or from one phase of a threephase alternator, as explained later in the study of a.c. generators.
In a three-phase circuit three voltages are Input generated by an alternator with three coils so spaced within the alternator that the three volt- Output ages generated are equal but reach their maximum values at different times. In each phase of a 400- B cycle, three-phase generator, a cycle is generated every 7400 second.
FIGURE 8-203. Autotransformers. In its rotation, the magnetic pole passes one 361
Phase A 3 Phase motor Phase B Phase C FIGURE 8-204. Three-phase generator using a three conductors.
coil and generates a maximum voltage; one-third cycle (1200 second) later, this same pole passes another coil and generates a maximum voltage in it; and the next one-third cycle later, it passes still another coil and generates a maximum voltage in it. This causes the maximum voltages generated in the three coils always to be one-third cycle (1200 second) apart.
The early three-phase generators were connected to their loads with six wires and all six leads in the circuit carried the current. Later, experiments proved that the generator would furnish as much power with the coils connected so that only three wires were needed for all three phases as shown in figure 8-204. The use of three wires is standard for the transmission of threephase power today. The return current from any one alternator coil always flows back through the other two wires in the three-phase circuit.
Three-phase motors and other three-phase loads are connected with their coils or load elements arranged so that three transmission lines are required for delivery of power. Transformers that are used for stepping the voltage up or down in a three-phase circuit are electrically connected so 60V 90V 120V 30V FIGURE 8-206. Tapped transformer secondary.
that power is delivered to the primary and taken from the secondary by the standard three-wire system.
However, single-phase transformers and singlephase lights and motors may be connected across any one phase of a three-phase circuit, as shown in figure 8-205. When single-phase loads are connected to three-phase circuits, the loads are distributed equally among the three phases in order to balance the loads on the three generator coils.
Another use of the transformer is the singlephase transformer with several taps in the secondary. With this type of transformer, the voltage can be lowered to provide several working voltages, as shown in figure 8-206.
A center tapped transformer, powering a motor requiring 220 volts along with four lights requiring 110 volts, is shown in figure 8-207. The motor is connected across the entire transformer output, and the lights are connected from the center tap to one end of the transformer. With this connection only half of the secondary output is used.
This type of transformer connection is used FICURE 8-205. Step-down transformer using FIGURE 8-207. Step-down transformer using two-wire system. a three-wire system.
3 Phase C AC Voltage Primaries Secondaries B FIGURE 8-208. Wye-to-wye connection.
extensively on aircraft because of the combinations of voltages that can be taken from one transformer. Various voltages can be picked off the secondary winding of the transformer by inserting taps (during manufacture) at various points along the secondary windings.
The various amounts of voltage are obtained by connecting to any two taps or to one tap and either end.
Transformers for three-phase circuits can be connected in any one of several combinations of the wye (y) and delta (4) connections. The connection used depends on the requirements for the transformer.
When the wye connection is used in three-phase B 3 Phase AC Voltage Primaries Secondaries transformers, a fourth or neutral wire may be used. The neutral wire connects single-phase equipment to the transformer. Voltages (115v) between any one of the three-phase lines and the neutral wire can be used for power for devices such as lights or single-phase motors.
In combination, all four wires can furnish power at 208 volts, three-phase, for operating threephase equipment, such as three-phase motors or rectifiers. When only three-phase equipment is used, the ground wire may be omitted. This leaves a three-phase, three-wire system as illustrated in figure 8-208.
Figure 8-209 shows the primary and secondary with a delta connection. With this type of connection the transformer has the same voltage output as the line voltage. Between any two phases the voltage is 240 volts. In this type of connection, wires A, B, and C can furnish 240-volt, three-phase power for the operation of three-phase equipment.
The type of connection used for the primary coils may or may not be the same as the type of connection used for the secondary coils. For example, the primary may be a delta connection and the secondary a wye connection. This is called a delta-wye-connected transformer. Other combinations are delta-delta, wye-delta, and wyewye.
Troubleshooting Transformers There are occasions when a transformer must be checked for opens or shorts, and it is often necessary to determine that a transformer is a step-up or step-down transformer.
An open winding in a transformer can be located by connecting an ohmmeter as shown in figure
8-210. Connected as shown, the ohmmeter would read infinity. If there were no open in the coil, the ohmmeter would indicate the resistance of wire in the coil. Both primary and secondary can be checked in the same manner.
open FIGURE 8-210. Checking for an open in a FIGURE 8-209. Delta-to-delta connection. transformer winding.