have approximately constant capacitance, may K (Dielectric Material constant) then be further divided, according to the type of dielectric used, into the following classes: paper, Air 1.0 oil, mica, and electrolytic capacitors. Ceramic Resin_- 2.5 capacitors are also used in some circuits.
Asbestos paper_. 2.7 When connecting electrolytic capacitors in a Hard rubber. 2.8 circuit, the proper polarity must be observed.
Isolantite_ _ Dry paper-- 3.5 3.5 "ground," which means that this terminal connects Paper capacitors may have one terminal marked Common glass _ 4.2 to the outside foil. Polarity does not ordinarily Quartz___ 4.5 have to be observed in connecting paper, oil, mica, Mica_ 4.5-7.5 or ceramic capacitors.
Porcelain_ 5.5 Flint glass. 7.0 Paper Capacitors Crown glass. 7.9 The plates of paper capacitors are strips of metal FIGURE 8-178. Dielectric constants.
alternating current is impressed on the circuit (figure 8-179), the charge on the plates constantly changes. This means that electricity must flow first from Y clockwise around to X, then from X counterclockwise around to Y, then from Y clockwise around to X, and so on. Although no current flows through the insulator between the plates of the capacitor, it constantly flows in the remainder of the circuit between X and Y. In a circuit in which there is only capacitance, current leads the impressed voltage as contrasted with a circuit in which there is inductance, where the current lags the voltage.
The unit of measurement of capacitance is the farad, for which the symbol is the letter "f." The farad is too large for practical use, and the units generally used are the microfarad (uf.), one millionth of a farad, and the micromicrofarad (uuf.), one millionth of a microfarad.
Types of Capacitors Capacitors may be divided into two groups:
fixed and variable. The fixed capacitors, which foil separated by waxed paper (figure 8-180). The capacitance of paper capacitors ranges from about 200 uf. to several uf. The strips of foil and paper are rolled together to form a cylindrical cartridge, which is then sealed in wax to keep out moisture and to prevent corrosion and leakage. Two metal leads are soldered to the plates, one extending from each end of the cylinder. The assembly is enclosed either in a cardboard cover or in a hard, molded plastic covering Bathtub-type capacitors consist of papercapacitor cartridges hermetically sealed in metal containers. The container often serves as a common terminal for several enclosed capacitors, but when not a terminal, the cover serves as a shield against electrical interference (figure 8-181).
Oil Capacitors In radio and radar transmitters, voltages high enough to cause arcing, or breakdown, of paper dielectrics are often employed. Consequently, in these applications capacitors that use oil or oilimpregnated paper for the dielectric material are preferred. Capacitors of this type are considerably more expensive than ordinary paper capacitors, and their use is generally restricted to radio and radar transmitting equipment (figure 8-182).
Mica Capacitors The fixed mica capacitor is made of metal foil AC Generator plates that are separated by sheets of mica, which form the dielectric. The whole assembly is covered in molded plastic, which keeps out moisture. Mica is an excellent dielectric and will withstand higher voltages than paper without allowing arcing FIGURE 8-179. Capacitor in an a.c. circuit. between the plates. Common values of mica 348
Negative Lead Waxed paper Tinfoil Lead 5 mfd 600 wvdc Tinfoil FIGURE 8-180. Paper capacitors.
capacitors range from approximately 50 uf, to about 0.02 uf. Mica capacitors are shown in figure 8-183.
Electrolytic Capacitors For capacitances greater than a few microfarads, the plate areas of paper or mica capacitors must become very large; thus, electrolytic capacitors are usually employed instead. These units provide large capacitance in small physical sizes. Their values range from 1 to about 1,500 microfarads.
Unlike the other types, electrolytic capacitors are generally polarized, and should be subjected to direct voltage, or pulsating direct voltage only;
however, a special type of electrolytic capacitor is made for use in motors.
The electrolytic capacitor is widely used in electronic circuits and consists of two metal plates separated by an electrolyte. The electrolyte in contact with the negative terminal, either in paste or liquid form, comprises the negative electrode.
The dielectric is an exceedingly thin film of oxide deposited on the positive electrode of the capacitor.
The positive electrode, which is an aluminum sheet, is folded to achieve maximum area.
The capacitor is subjected to a forming process during manufacture, in which current is passed through it. The flow of current results in the deposit of the thin coating of oxide on the aluminum plate.
The close spacing of the negative and positive electrodes gives rise to the comparatively high capacitance value, but allows greater possibility of voltage breakdown and leakage of electrons from one electrode to the other.
Two kinds of electrolytic capacitors are in use:
(1) Wet-electrolytic and (2) dry-electrolytic capacitors. In the former, the electrolyte is a liquid and the container must be leakproof. This type should always be mounted in a vertical position.
The electrolyte of the dry-electrolytic unit is a paste contained in a separator made of an absorbent material such as gauze or paper. The separator not only holds the electrolyte in place but also prevents short-circuiting the plates.
Dry-electrolytic capacitors are made in both cylindrical and rectangular-block form and may be contained either within cardboard or metal covers. Since the electrolyte cannot spill, the dry capacitor may be mounted in any convenient position. Electrolytic capacitors are shown in figure 8-184.
Capacitors in Parallel and in Series Capacitors may be combined in parallel or series to give equivalent values, which may be either the FIGURE 8-181. Bathtub-case paper capacitor. FIGURE 8-182. Oil capacitor.
FIGURE 8-183. Mica capacitors. A Parallel sum of the individual values (in parallel) or a value less than that of the smallest capacitance (in series). Figure 8-185 shows the parallel and series connections.
Two units used in the measurement of capacitance are the farad and the coulomb. As previously defined, the farad is the amount of capacitance present in a capacitor when one coulomb of electrical energy is stored on the plates and one volt is applied across the capacitor. One coulomb is the electrical charge of 6.28 billion billion 1 1 electrons. From this it can be seen that + a Q (in coulombs) C (in farads) = E (in volts) In A of figure 8-185 the voltage, E, is the same for all the capacitors. The total charge, Q« is the sum of all the individual charges, Qu Q2, and Q3.
B Series FIGURE 8-185. Capacitors in parallel and in series.
Using the basic equation for the capacitor, C== The total charge is Q, = CE, where C is the total capacitance. Since the total charge on capacitors in parallel is the sum of the individual capacitor charges, Using both equations for total charge develops the equation GIE = GIE + CrE + CE.
Dividing both sides of this equation by & gives Gi = CI + Cz + C3.
This formula is used to determine the total capacitance of any number of capacitors in parallel.
In the series arrangement, (B of figure 8-185), FIGURE 8-184. Electrolytic capacitors. the current is the same in all parts of the circuit.
Each capacitor develops a voltage during charge, and the sum of the voltages of all the capacitors Dielectric K (volts per .001 inch) Dielectric Strength must equal the applied voltage, E. By the capacitor equation, the applied voltage, E, is equal Air 1.0 80 to the total charge divided by the total capacitance, or Paper (1) Paraffined 2.2 1,200
(2) Beeswaxed 3.1 1,800 B = Glass 4.2 200 Castor oil 4.7 380 The total charge, Q« is equal to the charge on Bakelite 6.0 500 any one of the capacitors because the same Mica 6.0 2,000 current flows in all for the same length of time, and Fiber 6.5 50 because the charge equals current multiplied by time in seconds (Q:=1 xT). Therefore, FIGURE 8-186. Strength of some dielectric Q: = 01 = 02 = 03 , materials.
and, since in a circuit with capacitors in series Ei = Es + Er + E3 , where E1, E2, and E, are the voltages of the three capacitors. Then Dividing both sides of the equation by Q, gives = + = + Cs The reciprocal of the total capacitance of any number of capacitors in series is equal to the sum of the reciprocals of the individual values.
Parallel capacitors combine by a rule similar to that used to combine resistors in series. Series capacitors combine by a rule similar to that for combining parallel resistors.
In the series arrangement of two capacitors, Ci and C2 the total capacitance is given by the G + Ci Voltage Rating of Capacitors In selecting or substituting a capacitor for use in a particular circuit, the following must be considered: (1) The value of capacitance desired and (2) the amount of voltage to which the capacitor is to be subjected. If the voltage applied across the plates is too great, the dielectric will break down and arcing will occur between the plates. The capacitor is then short-circuited, and the possible flow of direct current through it can cause damage to other parts of the equipment.
Capacitors have a voltage rating that should not be exceeded The working voltage of the capacitor is the maximum voltage that can be steadily applied without danger of arc-over. The working voltage depends on (1) the type of material used as the dielectric and (2) the thickness of the dielectric.
The voltage rating of the capacitor is a factor in determining the capacitance because capacitance decreases as the thickness of the dielectric increases. A high-voltage capacitor that has a thick dielectric must have a larger plate area in order to have the same capacitance as a similar low-voltage capacitor having a thin dielectric. The strength of some commonly used dielectric materials is listed in figure 8-186. The voltage rating also depends on frequency because the losses, and the resultant heating effect, increase as the frequency increases.
A capacitor that can be safely charged to 500 volts d.c. cannot be safely subjected to a.c. or pulsating d.c. whose effective values are 500 volts.
An alternating voltage of 500 volts (r.m.s.) has a peak voltage of 707 volts, and a capacitor to which it is applied should have a working voltage of at least 750 volts. The capacitor should be selected so that its working voltage is at least 50 percent greater than the highest voltage to be applied to it.
Capacitance Reactance Capacitance, like inductance, offers opposition to the flow of current. This opposition is called capacitive reactance and is measured in ohms. The symbol for capacitive reactance is X.. The 351
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.