CHAPTER XLVII ALTERNATING CURRENT DIAGRAMS

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Whenever an alternating pressure is impressed on a circuit, part of it is spent in overcoming the resistance, and the rest goes to balance the reverse pressure due to self-induction.

The total pressure applied to the circuit is known as the impressed pressure, as distinguished from that portion of it called the active pressure which is used to overcome the resistance, and that portion called the self-induction pressure used to balance the reverse pressure of self-induction.

The intensity of the reverse pressure induced in a circuit due to self-induction is proportional to the rate of change in the current strength.

Thus a current, changing at the rate of one ampere per second, in flowing through a coil having a coefficient of self-induction of one henry, will induce a reverse pressure of one volt.

Ques. Describe how the rate of change in current strength varies, and how this affects the reverse pressure.

Ans. The alternating current varies from zero to maximum strength in one-quarter period, that is, in one-quarter revolution of the generating loop or 90° as represented by the sine curve in fig. 1,307. Now, during, say, the first 10 degrees of rotation (from 0 to A), the current jumps from zero value to A', or 4 amperes, according to the scale; during some intermediate 10 degrees of the quarter revolution, as from B to C, the current increases from B' to C' or 2½ amperes, and during another 10 degrees as from D to E, at the end of one-quarter revolution where the sine curve reaches its amplitude, it rises and falls ½ ampere. It is thus seen that the rate of change varies from a maximum when the current is least, to zero when the current is at its maximum. Accordingly, the reverse pressure of self-induction being proportional to the rate of change in the current strength, is greatest when the current is at zero value, and zero when the current is at its maximum.

Fig. 1,307.—Sine curve showing that the rate of change in the strength of an alternating current is greatest when the current is least, and zero when the current is at a maximum. This is evident from the diagram, since during say the first 10° as OA, the current increases 4 amperes; during BC, 2½ amperes; during DE it rises and falls ½ ampere. The reverse pressure of self-induction being proportional to the rate of change of the current, is a maximum when the current is zero, and zero when the current is a maximum, giving a phase difference of 90° between reverse pressure of self-induction and current.

This relation is shown by curves in fig. 1,308, and it should be noted that the reverse pressure and current are 90° apart in phase. For this reason many alternating current problems may be solved graphically by the use of right angle triangles, the sides, drawn to some arbitrary scale, to represent the quantities involved, such as resistance, reactance, impedance, etc.

Properties of Right Angle Triangles.—In order to understand the graphical method of solving alternating current problems, it is necessary to know why certain relations exist between the sides of a right angle triangle. For instance, in every right angle triangle:

The square of the hypothenuse is equal to the sum of the squares on the other two sides.

That is, condensing this statement into the form of an equation:

hypothenuse2=base2+altitude2 (1)

the horizontal side being called the base and the vertical side, the altitude.

This may be called the equation of the right angle triangle.

Fig. 1,308.—Sine curves showing phase relation between current and reverse pressure of self-induction. This reverse pressure, being proportional to the rate of change in the current strength, is greatest when the current is at zero value, and zero when the current is maximum, and in phase is 90° behind the current.

Fig. 1,309.In a right angle triangle the square on the hypothenuse is equal to the sum of the squares on the other two sides. That is: hypothenuse2=base2+altitude2. Draw AB, 4 inches long, and BC, 3 inches long and at right angles to AB. Join AC, which will be found to be 5 inches long. From the diagram, it must be clear that the square on AC=sum of squares on AB and BC; that is, 52=42+32. Further, 42=52-32; 32=52-42; 5=v(42+32); 4=v(52-32); 3=v(52-42).

Ques. Why is the square of the hypothenuse of a right angle triangle equal to the sum of the squares of the other two sides?

Ans. This may be explained with the aid of fig. 1,309. Draw a line AB, 4 inches in length and erect a perpendicular BC, 3 inches in height; connect A and C, giving the right angle triangle ABC. It will be found that AC the hypothenuse of this triangle is 5 inches long. If squares be constructed on all three sides of the triangle, the square on the hypothenuse will have an area of 25 sq. ins.; the square on the base, 16 sq. ins., and the square on the altitude, 9 sq. ins. Then from the figure 52=42+32, that is 25=16+9.

Repeating equation (1), it is evident from the figure that

hypothenuse2 ? ? base2+altitude2 ?
? = ?
?
52 ? ? 42+32 ?

that is,

25=16+9.

In the right angle triangle, the following relations also hold:

base2 = hypothenuse2 - altitude2 (2)
(42 = 52 - 32)
altitude2 = hypothenuse2 - base2 (3)
(32 = 52 - 42)

In working impedance problems, it is not the square of any of the quantities which the sides of the triangle are used to represent that is required, but the quantities themselves, that is, the sides. Hence extracting the square root in equations (1), (2) and (3), the following are obtained:

hypothenuse = v(base2+altitude2) (4)
(5 = v(42+32))
base = v(hypothenuse2-altitude2) (5)
(4 = v(52-32))
altitude = v(hypothenuse2-base2) (6)
(3 = v(52-42))

Representation of Forces by Lines.—A single force may be represented in a drawing by a straight line, 1, the point of application of the force being indicated by an extremity of the line, 2, the intensity of the force by the length of the line, and 3, the direction of the force by the direction of the line, an arrow head being placed at an extremity defining the direction.

Thus in fig. 1,310, the force necessary to balance the thrust on the steam piston may be represented by the straight line f whose length measured on any convenient scale represents the intensity of the force, and whose direction represents the direction of the force.

Fig. 1,310.—Diagram illustrating the representation of forces by straight lines. If 80 lbs. of steam be applied to a piston of 4 square inches area, the total pressure acting on the piston is 4×80=320 lbs. This may be balanced by an equal and opposite force. To represent the latter by a line, select any convenient scale whose divisions represent any convenient number of pounds—1, 3, 5 or, as here taken, 25 lbs. If the scale selected be divided into inches with ¼-inch divisions, then each ¼ inch represents a force of 25 lbs.; or, as usually stated, 1"=100 lbs. Strictly speaking 1" is equivalent to 100 lbs. Draw the line f=3.2 ins., then its length represents the magnitude of the force or 320 lbs., that is, 3.2×100=320 lbs.

Composition of Forces.—This is the operation of finding a single force whose effect is the same as the combined effect of two or more given forces. The required force is called the resultant of the given forces.

The composition of forces may be illustrated by the effect of the wind and tide on a sailboat as in fig. 1,311. Supposing the boat be acted upon by the wind so that in a given time, say half an hour, it would be moved in the direction and a distance represented by the line AB, and that in the same time the tide would carry it from A to C. Now, lay down AB, to any convenient scale, representing the effect of the wind, and AC that of the tide, and draw BD equal and parallel to AC, and CD equal and parallel to AB, then the diagonal AD will represent the direction and distance the boat will move under the combined effect of wind and tide.

Fig. 1,311.—Parallelogram of forces for boat acted upon by both wind and tide.

Ques. In fig. 1,311 what is the line AD called?

Ans. The resultant, that is, it represents the actual movement of the boat resulting from the combined forces of wind and tide.

Ques. What are the forces, AB and AD in fig. 1,311, represented by the sides of the parallelogram, and which act upon a body to produce the resultant, called?

Ans. The components.

Fig. 1,312.—Parallelogram of forces; method of obtaining the resultant of two components acting at right angles.

EXAMPLE.—Two forces, one of 3 lbs. and one of 4 lbs. act at a point a in a body and at right angles, what is the resultant?

Take any convenient scale, say 1 in.=1 lb., and lay off (fig. 1312.) AB=4 ins.=4 lbs.; also, AC (at right angles to AB)=3 ins.=3 lbs. Draw CD and BD parallel to AB and AC respectively, and join AD. The line AD is the resultant of the components AB and AC, and when measured on the same scale from which AB and AC were drawn will be found to be 5 inches long, which represents 5 lbs. acting in the direction AD.

Circuits containing Resistance and Inductance.—In circuits of this kind where the impressed pressure encounters both resistance and inductance, it may be looked upon as split up into two components, as already explained, one of which is necessary to overcome the resistance, and the other, the inductance. That is, the impressed pressure is split up into

  • 1. Active pressure, to overcome resistance;
  • 2. Self-induction pressure to overcome inductance.

Fig. 1,313.—Diagram illustrating the active, and self-induction pressures, or the two components of the impressed pressure in circuits containing resistance and inductance. The active pressure is the volts required to overcome the resistance of the circuit. In the figure only the portion from A to C is considered as having resistance (the rest being negligibly small) except at R, a resistance equivalent to that of the inductive coil is inserted next to the non-inductive coil, so Pa will give the total "ohmic drop" or active pressure, that is, the pressure necessary to force any equivalent direct current from A to C. This active pressure Pa or component of the impressed pressure is in phase with the current. The other component or self-induction pressure Pi that is the reactance drop necessary to overcome the reverse pressure of self-induction and is at right angles to the current and 90° ahead of the current in phase. It is registered by a voltmeter between B and C, less the pressure due to ohmic resistance of the inductive coil. The impressed pressure Pim then or total pressure required to force electricity around the circuit not including the resistance R, (which is removed from the circuit when the reading of the impressed pressure is taken), is equal to the square root of the sum of the squares of the two components, that is, Pim = v(Pa}2+P_{i2).

The active pressure is in phase with the current.

The self induction pressure is at right angles to the current and 90 degrees ahead of the current in phase.

Ques. Why is the active pressure in phase with the current?

Ans. The pressure used in overcoming resistance is from Ohm's law, E=RI. Hence, when the current is zero, E is zero, and when the current is a maximum E is a maximum. Hence, that component of the impressed pressure necessary to overcome the resistance must be in phase with the current.

Ques. Why is this?

Ans. Since the reverse pressure of self induction is 90° behind the current, the component of the impressed pressure necessary to overcome the reverse pressure of self induction, being opposite to this, will be represented as being 90° ahead of the current.

The distinction between the reverse pressure of self-induction, that is, the induced pressure, and the pressure necessary to overcome self-induction should be carefully noted. They are two equal and opposite forces, that is, two balancing forces just as is shown in fig. 1,310. Here, in analogy, the thrust of the piston may represent the induced pressure and the equal and opposite force indicated by the arrow f, the component of the impressed pressure necessary to balance the induced pressure.

Fig. 1,314.—Graphical method of obtaining the impressed pressure in circuits containing resistance and inductance, having given the ohmic drop, and reactance drop due to inductance. With any convenient scale lay off AB=ohmic drop and erect the perpendicular BC = reactance drop (using same scale). Join AC, whose length (measured with same scale) will give the impressed pressure. Constructing a parallelogram with dotted lines AD and CD, it is evident that AC is the resultant of the two components AB and BC, or its equal AD.

The Active Pressure or "Ohmic Drop."—The component of the impressed pressure necessary to overcome resistance, is from Ohm's law:

active pressure=ohmic resistance×virtual current

that is

Ea=RoIv (1)

this is the "ohmic drop" and may be represented by a line AB, fig. 1,314 drawn to any convenient scale, as for instance, 1 in.=10 volts.

The Self-induction Pressure or "Reactance Drop."—The component of the impressed pressure necessary to overcome the induced pressure, is from Ohm's law:

inductance pressure=inductance reactance×virtual current;

that is,

Ei=XiIv (2)

Fig. 1,315.—Diagram for impressed pressure on circuit containing 5 volts ohmic drop and 15 volts reactance drop.

Now the reactance Xi, that is the spurious resistance, is obtained from the formula

Xi=2pfL (3)

as explained on page 1,038, and in order to obtain the volts necessary to overcome this spurious resistance, that is, the "reactance drop" as it is called, the value of Xi in equation (3) must be substituted in equation (2), giving

Ei=2pfLI (4)

writing simply I for the virtual pressure.

Since the pressure impressed on a circuit is considered as made up of two components, one in phase with the current and one at right angles to the current, the component Ei or "reactance drop" as given in equation (4) maybe represented by the line BC in fig. 1,314, at right angles to AB, and of a length BC, measured with the same scale as was measured AB, to correspond to the value indicated by equation (4).

EXAMPLE.—In an alternating circuit, having an ohmic drop of 5 volts, and a reactance drop of 15 volts, what is the impressed pressure?

With a scale of say, ¼ inch=one volt, lay off, in fig. 1,315, AB=5 volts=1¼ in., and, at right angles to it, BC=15 volts=15/4 or 3¾ ins. Join AC; this measures 3.95 inches, which is equivalent to 3.95×4=15.8 volts, the impressed pressure. By using good paper, such as bristol board, a 6H pencil, engineers' scale and triangles or square, such problems are solved with precision. By calculation impressed pressure=v(52+152)=15.8 volts. Note that the diagram is drawn with the side BC horizontal instead of AB—simply to save space.

Fig. 1,316.—Diagram of circuit containing 5 volts ohmic drop, and 15 volts reactance drop.

Fig. 1,317.—Diagram for obtaining reactance drop in circuit containing 5 volts ohmic drop, and 15.8 volts impressed pressure.

EXAMPLE.—In an alternating circuit, having an ohmic drop of 5 volts and an impressed pressure of 15.8 volts, what is the reactance drop?

In fig. 1,317, draw a horizontal line of indefinite length and at any point B erect a perpendicular AB=5 volts. With A as center and radius of length equivalent to 15.8 volts, describe an arc cutting the horizontal line at C. This gives BC, the reactance drop required, which by measurement is 15 volts.

Fig. 1,318.—Diagram for obtaining ohmic drop in the circuit fig. 1,316 when impressed pressure and reactance drop are given. Lay off BC to scale=reactance drop; draw AB at right angle and of indefinite length; with C as center and radius of length=impressed pressure, describe an arc cutting ohmic drop line at A, then AB=ohmic drop=5 volts by measurement.

Fig. 1,319.—Graphical method of finding angle of lag when the ohmic drop and reactance drop are given. The angle of lag f, is that angle included between the impressed pressure and the ohmic drop lines, that is, between AC and AB.

EXAMPLE.—An alternating current of 10 amperes having a frequency of 60, is impressed on a circuit containing a resistance of 5 ohms and an inductance of 15 milli-henrys. What is the impressed pressure?

The active pressure or ohmic drop is 5×10=50 volts.

Fig. 1,320.—Diagram of circuit containing 5 ohms resistance, 15 milli-henrys inductance, with 3 ampere 60 frequency current.

The inductance reactance or Xi=is 2×3.1416×60×.015=5.66 ohms. Substituting this and the current value 10 amperes in the formula for inductance pressure or reactance drop (equation 2 on page 1,077) gives Ei=5.65×10=56.5 volts.

Fig. 1,321.—Diagram for impressed pressure on circuit containing 5 ohms resistance and inductance of 15 milli-henrys, the current being 10 amperes with frequency of 60.

In fig. 1,321, lay off AB=50 volts, and BC=56.6 volts. Using a scale of 20 volts to the inch gives AB=2.5 ins., and BC=2.83 ins. Joining AC gives the impressed voltage, which by measurement is 75.4 volts.

In some problems it is required to find the impedance of a circuit in which the ohmic and spurious resistances are given. This is done in a manner similar to finding the impressed pressure.

Fig. 1,322.—Graphical method of obtaining the impedance in circuits containing resistance and inductance, having given the resistance and reactance, that is, the ohmic resistance and spurious resistance. With any convenient scale lay off AB=resistance, and erect the perpendicular BC=reactance (using the same scale); join AC, whose length (measured with the same scale) will give the impedance.

Ohmic resistance and spurious resistance or inductance reactance both tend to reduce an alternating current. Their combined action or impedance is equal to the square root of the sum of their squares, that is,

impedance=v(resistance2+reactance2)

This relation is represented graphically by the side of a right angle triangle as in fig. 1322, in which the hypothenuse corresponds to the impedance, and the sides to the resistance and reactance.

EXAMPLE.—In a certain circuit the resistance is 4 ohms, and the reactance 3 ohms. What is the impedance?

In fig. 1,323, lay off, on any scale AB=4 ohms and erect the perpendicular BC=3 ohms. Join AC, which gives the impedance, and which is, measured with the same scale, 5 ohms.

EXAMPLE.—A coil of wire has a resistance of 20 ohms and an inductance of 15 milli-henrys. What is its impedance for a current having a frequency of 100?

Fig. 1,323.—Diagram for obtaining the impedance of a circuit containing 4 ohms resistance and 3 ohms reactance.

The ohmic value of the inductance, that is, the reactance is

2pfL=2×3.1416×100×.015=9.42 ohms.

In fig. 1,324, lay off, on any scale, AB=20 ohms, and the perpendicular BC to length=9.42 ohms. Join AC, which gives the impedance, which is, measured on the same scale, 22.1 ohms.

EXAMPLE.—What is the angle of lag in a circuit having a resistance of 4 ohms and a reactance of 3 ohms?

Construct the impedance diagram in the usual way as in fig. 1,325, then the angle included between the impedance and resistance lines (denoted by f) is the angle of lag, that is, the angle BAC. By measurement with a protractor it is 37 degrees. By calculation the tangent of the angle of lag or

BC 3
tan f =
=
or .75
AB 4

From the table on page 451, the angle is approximately 37°.

Fig. 1,324.—Diagram for impedance of circuit containing 20 ohms resistance, and inductance of 15 milli-henrys, when the frequency is 100.

Fig. 1,325.—Diagram showing angle of lag for current containing 4 ohms resistance and 3 ohms reactance.

Circuits containing Resistance and Capacity.—The effect of capacity in an alternating current circuit is to cause the current to lead the pressure, since the reaction of a condenser, instead of tending to prolong the current, tends to drive it back.

Careful distinction should be made between capacity in series with a circuit and capacity in parallel with a branch of a circuit. The discussion here refers to capacity in series, which means that the circuit is not continuous but the ends are joined to a condenser, as shown at the right in fig. 1,326, so that no current can flow except into and out of the condenser.

Fig. 1,326,—Circuit diagram illustrating the distinction between capacity in series and capacity in parallel. The condition for capacity in series is that the circuit must be discontinuous as at M; for capacity in parallel the main circuit must be continuous; this means that the capacity must be inserted in a branch of the main circuit as at A. In the figure the capacity S is connected in series with respect to the branch, that is, the branch is discontinuous, but it is in parallel with respect to the main circuit, when the latter is continuous, that is, when the switch W is closed. If W be opened, the main circuit becomes discontinuous and S is changed from in parallel to in series connection.

Ques. In circuits containing resistance and capacity upon what does the amount of lead depend?

Ans. Upon the relative values of the resistance and the capacity reactance.

Ques. Describe the action of a condenser when current is applied.

Ans. When the current begins to flow into a condenser, that is, when the flow is maximum, the back pressure set up by the condenser (called the condenser pressure) is zero, and when the flow finally becomes zero, the condenser pressure is a maximum.

Fig. 1,327.—Current and pressure curves showing that the condenser pressure is 90° ahead of the current. A current flowing into a condenser encounters a gradually increasing pressure which opposes it, beginning from zero pressure when the current enters at maximum flow and increasing to the same value as the current pressure, at which time the current ceases to flow. Hence, since the current varies from zero to maximum in one quarter period, or 90°, the phase difference between current and condenser pressure is 90°. The condenser pressure reaching a positive maximum when the current starts from zero on the positive wave, is 90° ahead of the current.

Ques. What does this indicate?

Ans. It shows that the phase difference between the wave representing the condenser pressure and the current is 90°, as illustrated in fig. 1,327.

Ques. Is the condenser pressure ahead or behind the current and why?

Ans. It is ahead of the current. The condenser pressure, when the condenser is discharged being zero, the current enters at maximum velocity as at A in fig. 1,327, and gradually decreases to zero as the condenser pressure rises to maximum at B, this change taking place in one-quarter period. Thus the condenser pressure, which opposes the current, being at a maximum when the current begins its cycle is 90° ahead of the current, as is more clearly seen in the last quarter of the cycle (fig. 1,327).

Fig. 1,328.—Current and pressure curves, showing phase relation between the current, condenser pressure, and impressed or capacity pressure necessary to overcome the condenser pressure. The capacity pressure, since it must overcome the condenser pressure, is equal and opposite to the condenser pressure, that is, the phase difference is 180°. The condenser pressure being 90° ahead of the current, the impressed pressure is 90° behind the current.

Ques. What is the phase relation between the condenser pressure and the pressure applied to the condenser to overcome the condenser pressure?

Ans. The pressure applied to the condenser to overcome the condenser pressure, or as it is called, the capacity pressure, must be opposite to the condenser pressure, or 90° behind the current.

In circuits containing resistance and capacity, the total pressure impressed on the circuit, or impressed pressure, as it is called, is made up of two components:

1. The active pressure, or pressure necessary to overcome the resistance;

The active pressure is in phase with the current.

2. The capacity pressure, or pressure necessary to overcome the condenser pressure,

The capacity pressure is 90 degrees behind the current.

Fig. 1,329.—Graphical method of obtaining the impressed pressure in circuits containing resistance and capacity, having given the ohmic drop and reactance drop due to capacity. With any convenient scale, lay off AB=ohmic drop, and at right angles to AB draw BC=reactance drop (using the same scale). Join AC, whose length (measured with the same scale) will give the impressed pressure. The mathematical expressions for the three quantities are given inside the triangle, and explained in the text.

Problems involving resistance and capacity are solved similarly to those including resistance and inductance.

The Active Pressure or "Ohmic Drop."—This, as before explained is represented, in fig. 1,329, by a line AB, which in magnitude equals, by Ohm's law, the product of the resistance multiplied by the current, that is,

Ea=RoIv (1)

Fig. 1,330.—Diagram of circuit containing a resistance of 30 ohms and capacity of 125 microfarads. The calculation for impressed pressure, ohmic drop, and reactance drop for a current of 8 amperes at frequency 60 is given in the example on page 1,089, the diagram for impressed pressure being given in fig. 1,331.

The Capacity Pressure or "Reactance Drop."—This component of the impressed pressure, is, applying Ohm's law,

capacity pressure=capacity reactance×virtual current.

Ec=XcIv (2)

That is, the expression for capacity reactance Xc, that is, for the value of capacity in ohms is, as explained on page 1,048,

1
Xc =
(3)
2pfC

Substituting this value of Xc in equation (2) and writing I for virtual current.

I
Ec =
(4)
2pfC

CAUTION—The reader should distinguish between the 1 (one) in (3) and the letter I in (4); both look alike.

Since the capacity pressure is 90° behind the current, it is represented in fig. 1,329, by a line BC, drawn downward, at right angles to AB, and of a length corresponding to the capacity pressure, that is, to the reactance drop.

The Impressed Pressure.—Having determined the ohmic and reactance drops and represented them in the diagram, fig. 1,329, by lines AB and BC respectively, a line AC joining A and C, will then be the resultant of the two component pressures, that is, it will represent the impressed pressure or total pressure applied to the circuit.

In the diagram it should be noted that the active pressure is called the ohmic drop, and the capacity pressure, the reactance drop.

EXAMPLE.—A circuit as shown in fig. 1,330 contains a resistance of 30 ohms, and a capacity of 125 microfarads. If an alternating current of 8 amperes with frequency 60 be flowing in the circuit, what is the ohmic drop, the reactance drop, and the impressed pressure?

The ohmic drop or active pressure is, substituting in formula (1) on page 1,087,

Ea=30×8=240 volts

which is the reading of voltmeter A in fig. 1,330.

Fig. 1,331.—Diagram for obtaining the impressed pressure of the circuit shown in fig. 1,330.

The reactance drop or

I 8
Ec =
=
= 170 volts
2pfC 2×3.1416×60×.000125

in substituting, note that the capacity C of 125 microfarads is reduced to .000125 farad.

Using a scale of say 1 inch=80 volts, lay off in fig. 1,331, AB equal to the ohmic drop of 240 volts; on this scale AB=3 inches. Lay off at right angles, BC=reactance drop=170 volts=2.125 inches. Join AC, which gives the impressed voltage, (that is the reading of voltmeter I in fig. 1,330,) which measures 294 volts.

By calculation, impressed pressure=v(2402+1702)=294 volts.

EXAMPLE.—In the circuit shown in fig. 1,330, what is the angle of lead?

The tangent of the angle of lead is given by the quotient of the reactance divided by the resistance of the circuit. That is,

reactance reactance drop
tan f =
=
resistance resistance drop
Ec I
tan f =
=
÷ Ea (1)
Ea 2pfC

The tangent is given a negative sign because lead is opposed to lag and because the positive value is assigned to lag. Substituting in (1)

170 2.125"
tan f =
or
= -.71
240 3"

the angle corresponding is approximately 35¼° (see table page 451).

Figs. 1,332 and 1,333.—Diagrams for circuits containing inductance and capacity. Since inductance and capacity act 180° apart, their reactances, or their ohmic drops may be represented by oppositely directed lines. These may be drawn above and below a reference line, as in fig. 1,332, and their algebraic sum taken, or both may be drawn on the same side of the reference line and their difference in lengths, as CD, fig. 1,333, measured. Recourse to a diagram for obtaining the resultant reactance in circuits containing inductance and capacity is unnecessary as it is simply a matter of taking the difference of two quantities.

Circuits Containing Inductance and Capacity.—The effect of capacity in a circuit is exactly the opposite of inductance, that is, one tends to neutralize the other. The method of representing each graphically has been shown in the preceding figures. Since they act oppositely, that is 180° apart, the reactance due to each may be calculated and the values thus found, represented by oppositely directed vertical lines: the inductance resistance upward from a reference line, and the capacity resistance downward from the same reference line. The difference then is the resultant impedance. This method is shown in fig. 1,332, but it is more conveniently done as in fig. 1,333.

Fig. 1,334.—Diagram of circuit containing 30 milli-henrys inductance and 125 microfarads capacity, with current of 20 amperes, 100 frequency.

EXAMPLE.—In a circuit, as in fig. 1,334, containing an inductance of 30 milli-henrys and a capacity of 125 microfarads, how many volts must be impressed on the circuit to produce a current of 20 amperes having a frequency of 100.

The inductance reactance is

Xi=2pfL=2×3.1416×100×.03=18.85 ohms.

Substituting this and the current value of 20 amperes in the formula for inductance pressure

Ei=RiI=18.85×20=377 volts.

Reducing 125 microfarads to .000125 farad, and substituting in the formula for capacity pressure

I 20
Ec =
=
= 255 volts.
2pfC 2×3.1416×100×.000125

A diagram is unnecessary in obtaining the impressed pressure since it is simply the difference between inductance pressure and capacity pressure (the circuit being assumed to have no resistance), that is

impressed pressure=Ei - Ec=377 - 255=122 volts.

EXAMPLE.—A circuit in which a current of 20 amperes is flowing at a frequency of 100, has an inductance reactance of 18.25 ohms, and a capacity of 125 microfarads. What is the impedance?

The reactance due to capacity is

1 1
Xc =
=
= 12.76 ohms.
2pfC 2×3.1416×100×.000125

The impedance of the circuit then is the difference between the two reactances, that is impedance=inductance reactance-capacity reactance, or

Z=Xi - Xc=18.25 - 12.76=5.49 ohms.

Fig. 1,335.—Impedance diagram for circuit (of above example) containing inductance and capacity. With any convenient scale, erect a perpendicular AB=18.25 ohms, and CD = 12.76 ohms. Continue CD by dotted line to D' so that CD'=AB, then DD'=AB - CD = inductance reactance-capacity reactance, which is equal to the impedance. Expressed by letters Z=Xi-Xc=DD', which by measurement=5.49 ohms.

Circuits Containing Resistance, Inductance, and Capacity.—When the three quantities resistance, inductance, and capacity, are present in a circuit, the combined effect is easily understood by remembering that inductance and capacity always act oppositely, that is, they tend to neutralize each other. Hence, in problems involving the three quantities, the resultant of inductance and capacity is first obtained, which, together with the resistance, is used in determining the final effect.

Capacity introduced into a circuit containing inductance reduces the latter and if enough be introduced, inductance will be neutralized, giving a resonant circuit which will act as though only resistance were present.

Fig. 1,336.—Impedance diagram for circuit containing resistance, inductance and capacity. The symbols correspond to those used in equation (1) below. In constructing the diagram from the given values, lay off AB=resistance; at B, draw a line at right angles, on which lay off above the resistance line, BC=inductive reactance, and below, BD=capacity reactance, then the resultant reactance=BC - BD=BD'. Join A and D', then AD'=impedance.

Ques. What is the expression for impedance of a circuit containing resistance, inductance and capacity?

Ans. It is equal to the square root of the sum of the resistance squared plus the square of inductance reactance minus capacity reactance.

This is expressed plainer in the form of an equation as follows:

impedance=v(resistance2+(inductance reactance-capacity reactance)2)

or, using symbols,

Z=v(R2+(Xi - Xc)2) (1)

Ques. If the capacity reactance be larger than the inductance reactance, how does this affect the sign of (Xi-Xc)2?

Ans. The sign of the resultant reactance of inductance and capacity will be negative if capacity be the greater, but since in the formula the reactance is squared, the sign will be positive.

Fig. 1,337.—Impedance diagram of a circuit containing 25 ohms resistance, 30 ohms inductance, and 40 ohms capacity. The resultant reactance being due to excess of capacity, the impedance line AC' falls below the horizontal line AB, indicating that the current leads pressure.

EXAMPLE.—What is the impedance in a circuit having 25 ohms resistance, 30 ohms inductance reactance, and 40 ohms capacity reactance?

To solve this problem graphically, draw the line AB, in fig. 1,337, equal to 25 ohms resistance, using any convenient scale.

At B draw upward at right angles BC=30 ohms; draw from C downward CC'=40 ohms. This gives-BC' (= BC - CC') showing the capacity reactance to be 10 ohms in excess of the inductance reactance. Such a circuit is equivalent to one having no inductance but the same resistance and 10 ohms capacity reactance.

The diagram is completed in the usual way by joining AC giving the required impedance, which by measurement is 26.9 ohms.

By calculation, Z=v(252+(30 - 40)2)=v(252+(-10)2)=26.9.

Form of Impedance Equation without Ohmic Values.—

Using the expressions 2pfL for inductance reactance and 1/(2pfC) for capacity reactance, and substituting in equation (1) on page 1,093 gives the following:

Z=v(R2+(2pfL - 1/(2pfC))2) (2)

which is the proper form of equation (1) to use in solving problems in which the ohmic values of inductance and capacity must be calculated.

Fig. 1,338.—EXAMPLE: A resistance of 20 ohms and an inductance of .02 henry are connected in parallel as in the diagram. What is the impedance, and how many volts are required for 50 amperes, when the frequency is 78.6? SOLUTION: The time constants are not alike, hence the geometric sum of the reciprocals must be taken as the reciprocal of the required impedance. That is, the combined conductivity will be the hypothenuse of the right triangle, of which the ohmic conductivity and the reactive conductivity are the two sides, respectively. Accordingly: 1/R=1/20=.05, and 1/(2pfL)=1/10=.1, from which, 1/R=v((1/R1)2+(1/(2pfL))2)=.111. Whence Z=1/.111=9 ohms.

Fig. 1,339.—Diagram of circuit containing 23 ohms resistance, 41 milli-henrys inductance, and 51 microfarads capacity, with current supplied at a frequency of 150.

EXAMPLE.—A current has a frequency of 150. It passes through a circuit, as in fig. 1,339, of 23 ohms resistance, of 41 milli-henrys inductance, and of 51 microfarads capacity. What is the impedance?

The inductance reactance or

Xi=2pfL=2×3.1416×150×.041=38.64 ohms

(note that 41 milli-henrys are reduced to .041 henry before substituting in the above equation).

The capacity reactance, or

1 1
Xc =
=
= 20.8 ohms
2pfC 2×3.1416×150×.000051

(note that 51 microfarads are reduced to .000051 farad before substituting in the above equation).

Substituting the values as calculated for 2pfL and 1/(2pfC) in equation (2)

Z=v(232+(38.64 - 20.8)2)=29.1 ohms.

To solve the problem graphically, lay off in fig. 1,340, the line AB equal to 23 ohms resistance, using any convenient scale. Draw upward and at right angles to AB the line BC=38.64 ohms inductance reactance, and from C lay off downward CC'=20.8 ohms capacity reactance. The resultant reactance is BC' and being above the horizontal line AB shows that inductance reactance is in excess of capacity reactance by the amount BC'. Join AC' which gives the impedance sought, and which by measurement is 29.1 ohms.

In order to obtain the impressed pressure in circuits containing resistance, inductance and reactance, an equation similar to (2) on page 1,095 is used which is made up from the following:

Eo = RI (3)
Ei = 2pfLI (4)
I
Ec =
(5)
2pfC

Fig. 1,340.—Impedance diagram for the circuit shown in fig. 1,339. Note that the resultant reactance being due to excess of inductance, the impedance line AC' falls above the horizontal line AB. This indicates that the current lags behind the pressure.

When all three quantities, resistance, inductance, and capacity are present, the equation is as follows:

impressed pressure = v(ohmic drop2+(inductive drop-capacity drop)2)
Eim = v(Eo2+(Ei-Ec)2) (6)

Substituting in this last equation (6), the values given in (3), (4) and (5)

Eim = v(R2I2+(2pfLI - (I/(2pfC)))2)
= Iv(R2+(2pfL - (1/(2pfC)))2) (7)

Fig. 1,341.—Diagram of circuit containing 25 ohms resistance, .15 henry inductance, and 125 microfarads capacity, with current of 8 amperes at 60 frequency.

Ques. What does the quantity under the square root sign in equation (7) represent?

Ans. It is the impedance of a circuit possessing resistance, inductance, and capacity.

Ques. Why?

Ans. Because it is that quantity which multiplied by the current gives the pressure, which is in accordance with Ohm's law.

Fig. 1,342.—Diagram for finding the pressure necessary to be impressed on the circuit shown in fig. 1,341, to produce a current of 8 amperes.

EXAMPLE.—An alternator is connected to a circuit having, as in fig. 1,341, 25 ohms resistance, an inductance of .15 henry, and a capacity of 125 microfarads. What pressure must be impressed on the circuit to allow 8 amperes to flow at a frequency of 60?

The ohmic drop is

Eo=RI=25×8=200 volts.

The inductance drop is

Ei=2pfLI=2×3.1416×60×.15×8=452 volts

The capacity drop is

I 8
Ec =
=
= 170 volts.
2pfC 2×3.1416×60×.000125

Substituting the values thus found,

impressed pressure = v(Eo2+(Ei - Ec)2)
= v(2002+(452 - 170)2)
= v(2002+2822)
= v(119524)
= 345.7 volts.
                                                                                                                                                                                                                                                                                                           

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