CHAPTER XLVII
~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.
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.
~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.
~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.
~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_.
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.
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.
~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)
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.
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.
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.
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.
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.
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?
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°.
~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.
~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.
~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).
~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.
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)
~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.
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).
~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.
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.
~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.
~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.
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.
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
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)
~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.
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.