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