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.
Fig. 1,316.—Diagram of circuit containing 5 volts ohmic drop, and 15 volts reactance drop.
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.