may be taken to represent that voltage. Similarly the line OA gives the drop due to the ohmic resistance, and the line OB shows the voltage required to overcome the reactance. That is, the impressed E.M.F. OD is resolved into two components at right angles to each other, one of which, IR, overcomes the resistance, and the other, pLI, overcomes the reactance. The reactance is, in fact, an opposing E.M.F., having an effective value, OF, equal but directly opposite to pLI, and lagging 90° behind the current, the phase of the latter being represented by the line OA.

These relations, which are very important, may be stated as follows: With a given current I the effect of resistance is equivalent to a counter E.M.F. equal to IR, and represented by the line OG. With the same current, the reactance is actually an opposing E.M.F. having an effective value equal to pLI, and indicated by

OF. The combined effect of the two is equivalent to a counter E.M.F. equal to OH, which is their resultant. Hence OD, the impressed E.M.F., must exactly balance OH; that is, it is equal and opposite as shown. The phase of the impressed E.M.F. is represented by the line OD, with respect to which the current phase OA has an angle of lag 4, and OF, the reactance E.M.F., has a further lag of 90° behind the current.

Precisely similar diagrams and reasoning apply to the capacity reactance and angle of lead in Fig. 89 and to the combined inductance and capacity reactances in Fig. 90.

Composition and Resolution of E.M.F. and Current. In the manner explained above, two or more alternating E.M.F.'s may be combined to form a resultant, or one E.M.F. may be resolved

into two or more components; and the same is true of alternating currents. For example, two alternators are running in series, the value and phase of the E.M.F. of one being represented by OM in Fig. 93, and the value and relative phase of the other's E.M.F. being represented by ON. The combined effect is the same as that of one E.M.F. having the value and phase OP, which is the resultant of the two. If OM and ON represent the phase and values of two alternating currents, the phase and amount of the resultant current are given by OP. This method applies to any

P

N

Figs. 93 and 94. Components and Resultants of E.M.F. and Current.

number of E.M.F's or currents having any phase angles, the various components being combined successively in pairs. Instead of constructing a complete parallelogram, it is sufficient to lay off OT to represent one E.M.F., and TV to represent the phase and value of another E.M.F., then OV is their joint phase and amount. In other words, their resultant is the vector sum OV of the two vectors OT and TV, representing their respective values and phases.

Impedance of Parallel Circuits. With two or more simple resistances in parallel, the joint conductance is the sum of the several conductances; hence the joint resistance is the reciprocal of the sum of the reciprocals, that is:

In the above, R is the joint resistance of the resistances 1, 1⁄2, etc., in parallel. If no inductance or capacity is present, this is true of alternating circuits, and always applies to steady direct currents. When there is inductance or capacity or both in one or more of several alternating circuits in parallel, it becomes necessary to consider the phases of the currents in the different branches.

Two inductive coils A and B in parallel are assumed to be supplied with an alternating voltage by the conductors C and D in Fig. 95. To deduce a method for finding their joint impedance, let us suppose that OW in Fig. 96 represents the phase and value of the current in the coil A, and OY the phase and value of the current in B, then OX gives the phase and value of the resultant

Figs. 95 and 96. Impedance of Parallel Circuits.

current, which actually flows on the supply wires C and D. If the effective potential difference between C and D is one volt, and 1

the impedance of the coil A is J, then its current OW is I'

1

and

if the impedance of B is J, its current OY is The resultant Ji

1

J'

current OX must be in which is the joint impedance of the two. The reciprocal of impedance is called admittance; therefore, the joint admittance of two or more alternating circuits in parallel is the resultant of their several admittances. The equivalent impedance is found by taking the reciprocal of the joint admittance. Many combinations of resistance, inductance, and capacity in series and in parallel are calculated out in the work on Alternating Currents," by Professors D. C. and J. P. Jackson, pp. 151-220. The following examples are sufficient to show the method:

Examples. - A non-inductive resistance R1 of 10 ohms is in parallel with an inductive resistance of 10 ohms and .01 henry as indicated in Fig. 97. What is the joint impedance and angle of lag for an alternating current having a fre

Figs. 97, 98, and 99. Inductive and Non-inductive Circuits in Parallel.

quency of 127,? The impedance of the lower branch may be determined by constructing the triangle shown in Fig. 98, in which the base = R2 = 10 and the perpendicular = 2πfL = 800 L = 8. Hence, by calculation or by measurement

we find that the impedance, J = 12.8 and the angle of lag 2=38°40′. In an

=

1

other diagram (Fig. 99) the conductance =.100 is laid off horizontally to

R1

=

1

represent the upper branch, and the admittance = .078 is laid off at an angle

J2

2=38°40′ to represent the lower branch. Completing the parallelogram, it is

=

1

found that the joint admittance = .168, so that the joint impedance /= 5.95

In Fig. 100 a coil R1 = 10 ohms, and L = .01042 henry is in parallel with a circuit containing a resistance R2 = 10 ohms and a condenser having a capacity K=150 microfarad, what is the joint impedance and angle of lag for an alternating current having a frequency of 1274? The impedance of the coil is found to be J1 = 13 ohms, and the angle of lag 139°501 in Fig. 101, and the impedance of the condenser circuit is = 13 ohms, and the angle of lead 2 =

1

= 39°50 in Fig. 102. The corresponding admittances, each of which is 1 ÷ 13 = .077, are laid off in Fig. 103 with the same angles of lag and lead. The joint admittance

=

.117, so that the joint impedance /= 8.47 ohms, and the resultant angle of lag = 0, since the lag of one circuit balances the lead of the other. At 100 volts the main conductors F and G would supply 100 ÷ 8,47 = 11.7 amperes. The power factor is 100 per cent (cos += 1) so that the true power would be 100 × 11.7 = 1170 watts. The current in each branch would be 100 ÷ 137.7 amperes, one lagging 39°50′, and the other leading 39°50′, with

respect to the impressed E.M.F. The fact that the actual current supplied (= 11.7 amperes) is not equal to the sum of those in the two branches (2 × 7.7 = 15.4 amperes) shows that the discharging of one partly acts to charge the other, which is always the case when parallel circuits are not in the same phase.

Other combinations of resistance, inductance, and capacity may be treated in a similar manner.

Self-Inductance of Lines and Circuits. It is possible to calculate the inductance of coils of wire, but difficult to do so, and in most cases it is determined by comparison with standard inductances. The inductance of aërial lines is easily calculated, since they are generally parallel, and the medium has a fixed permeability practically equal to one. Wires laid upon a wall of wood, or other non-conducting, non-magnetic substance, or otherwise placed out of proximity to such a substance, have practically the same inductance as aërial lines.

The following formulas may be used to determine the selfinductance of two parallel aërial wires forming part of the same circuit, and composed of copper or other non-magnetic material.

L

I per centimeter = (.5 + 2 log, 10

(51)

In this expression L is the self-inductance in henrys per centimeter of each wire, A is the interaxial distance between the two wires, and is the radius of each. The dimensions A and may r be expressed in terms of any unit, provided it is the same for both. Since the Napierian logarithms in the above equation are 2.302585 times the common logarithms, and as it is more convenient to use the diameter d of the wire instead of its radius, we have without appreciable error:

For each of two parallel iron wires we have the following expressions, in which the only change is the first constant :

2 A

L Z per mile = (12070 + 740 log 10 (58)

d