The impedance of an inductive circuit which does not contain capacity is equal to the square root of the sum of the squares of the resistance and reactance, that is

impedance = √(resistance2 + reactance2) (1)

Fig. 1,297.—Diagram showing alternating circuit containing resistance, inductance, and capacity. Formula for calculating the impedance of this circuit is Z = √(R2 + (Xi - Xc)2) in which, Z = impedance; R = resistance; Xi = inductance reactance; Xc = capacity reactance. Example: What is the impedance when R = 4, Xi = 94.2, and Xc = 72.4? Substituting Z = √(42 + (94.2 - 72.4)2) = 22.2 ohms. Where the ohmic values of inductance and capacity are given as in this example, the calculation of impedance is very simple, but when inductance and capacity are given in milli-henrys and microfarads respectively, it is necessary to first calculate their ohmic values as in figs. 1,295 and 1,296.

EXAMPLE I.—If an alternating pressure of 100 volts be impressed on a coil of wire having a resistance of 6 ohms and inductance of 8 ohms, what is the impedance of the circuit and how many amperes will flow through the coil? In the example here given, 6 ohms is the resistance and 8 ohms the reactance. Substituting these in equation (1)

Impedance = √(62 + 82) = √(100) = 10 ohms.

The current in amperes which will flow through the coil is, by Ohm's law using impedance in the same way as resistance.

volts100 volts
current=
=
=10 amperes.
impedance10 ohms

The reactance is not always given but instead in some problems the frequency of the current and inductance of the circuit. An expression to fit such cases is obtained by substituting 2πfL for the reactance as follows: (using symbols for impedance and resistance)