~ALTERNATING CURRENT DIAGRAMS~
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.
[Illustration: 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.
[Illustration: 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.]
[Illustration: 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 = √(4² + 3²);
4 = √(5² - 3²); 3 = √(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_ = √(_base²_ + _altitude²_) (4)
(5 = √(4² + 3²))
_base_ = √(_hypothenuse²_ - _altitude²_) (5)
(4 = √(5² - 3²))
_altitude_ = √(_hypothenuse²_ - _base²_) (6)
(3 = √(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.
[Illustration: 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.
[Illustration: 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_.
[Illustration: 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.
[Illustration: 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 = √(Pₐ² + Pᵢ²).]
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.
[Illustration: 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)
[Illustration: 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 = 2π_f_L (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 = 2π_f_LI (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 = √(5² + 15²) = 15.8 volts. Note that the diagram is
drawn with the side BC horizontal instead of AB--simply to save
space.
[Illustration: FIG. 1,316.--Diagram of circuit containing 5 volts ohmic
drop, and 15 volts reactance drop.]
[Illustration: 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.
[Illustration: 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.]
[Illustration: FIG. 1,319.--Graphical method of finding angle
of lag when the ohmic drop and reactance drop are given. The
angle of lag φ, 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.
[Illustration: 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.
[Illustration: 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.
[Illustration: 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_ = √(_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?
[Illustration: 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
2π_f_L = 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 φ) 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 φ = -- = - or .75
AB 4
From the table on page 451, the angle is approximately 37°.
[Illustration: FIG. 1,324.--Diagram for impedance of circuit containing
20 ohms resistance, and inductance of 15 milli-henrys, when the
frequency is 100.]
[Illustration: 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.
[Illustration: 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.
[Illustration: 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).
[Illustration: 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.
[Illustration: 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)
[Illustration: 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)
2π_f_C
Substituting this value of Xc in equation (2) and writing I for
virtual current.
~I
Ec = ------- (4)
2π_f_C~
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.
[Illustration: 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
2π_f_C 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 = √(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 φ_ = ------------ = -----------------
_resistance_ _resistance drop_
Ec I
tan φ = -- = ------ ÷ Eₐ (1)
Eₐ 2π_f_C
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 φ = --- or ------ = -.71
240 3"
the angle corresponding is approximately 35¼° (see table page
451).
[Illustration: 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.
[Illustration: 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 = 2π_f_L = 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.
2π_f_C 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.
2π_f_C 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.
[Illustration: 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.
[Illustration: 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_ =
√(_resistance²_ + (_inductance reactance_ - _capacity reactance_)²)
or, using symbols,
~Z = √(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.
[Illustration: 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 = √(25² + (30 - 40)²) = √(25² + (-10)²) = 26.9.
~Form of Impedance Equation without Ohmic Values.--~
Using the expressions 2π_f_L for inductance reactance and 1 / (2π_f_C)
for capacity reactance, and substituting in equation (1) on page 1,093
gives the following:
Z = √(R² + (2π_f_L - 1 / (2π_f_C))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.
[Illustration: 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 / (2π_f_L) = 1 / 10 = .1, from which,
1 / R = √((1 / R₁)² + (1 / (2π_f_L))²) = .111. Whence Z = 1 / .111 = 9
ohms.]
[Illustration: 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 = 2π_f_L = 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
2π_f_C 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 2π_f_L and
1 / (2π_f_C) in equation (2)
Z = √(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 = 2π_f_LI (4)
I
Ec = ------ (5)
2π_f_C
[Illustration: 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 = √(ohmic drop² + (inductive drop - capacity drop)²)_
Eᵢₘ = √(Eₒ² + (Eᵢ - Ec)²) (6)
Substituting in this last equation (6), the values given in (3), (4)
and (5)
Eᵢₘ = √(R²I² + (2π_f_LI - (I / (2π_f_C)))²)
= I √(R² + (2π_f_L - (1 / (2π_f_C)))²) (7)
[Illustration 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.
[Illustration: 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 = 2π_f_LI = 2 × 3.1416 × 60 × .15 × 8 = 452 volts
The capacity drop is
I 8
Ec = ------ = ------------------------- = 170 volts.
2π_f_C 2 × 3.1416 × 60 × .000125
Substituting the values thus found,
impressed pressure = √(Eₒ² + (Eᵢ - Ec)²)
= √(200² + (452 - 170)²)
= √(200² + 282²)
= √(119524)
= 345.7 volts.