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 = √(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.
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 2πfL for inductance reactance and 1 / (2πfC) for capacity reactance, and substituting in equation (1) on page 1,093 gives the following:
Z = √(R2 + (2πfL - 1 / (2πfC))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 / (2πfL) = 1 / 10 = .1, from which, 1 / R = √((1 / R1)2 + (1 / (2πfL))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 = 2πfL = 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 | ||
| 2πfC | 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 2πfL and 1 / (2πfC) 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 | = | 2πfLI | (4) |
| I | |||
| Ec | = | (5) | |
| 2πfC |
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 + (2πfLI - (I / (2πfC)))2) | |
| = | I√(R2 + (2πfL - (1 / (2πfC)))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 = 2πfLI = 2 × 3.1416 × 60 × .15 × 8 = 452 volts
The capacity drop is
| I | 8 | |||||
| Ec | = | = | = | 170 volts. | ||
| 2πfC | 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. |