a speed of 1,000 feet per minute, the motors having 70 per cent efficiency.

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Solution: Perpendicular rise of car is 1,000 feet X.05=50 feet. Weight of car in pounds is 8 X 2,000 16,000 pounds. Product of lift and weight is 50 X 16,000 800,000 foot pounds. Force required to propel car is 30 X 8240 pounds. Product of force and distance is 240 X 1,000=240,000 foot pounds. Sum of the two products is 800,000 + 240,000 1,040,000 total foot pounds.

Product of 33,000 by efficiency is 33,000 X .70= 23,100. Electrical H. P. is the quotient of 1,040,000 23,10045.021 H. P. Ans.

CHAPTER XI.

ALTERNATING CURRENTS.

By far the greater part of calculations in the domain of alternating currents lie in the realm of trigonometry and the intricacies of the calculus. On this account it is hoped that the following presentation of some of the simpler formulæ may prove welcome to the craft.

A current flowing alternately in opposite directions may be considered as increasing from zero to a certain amount flowing in, say, the positive direction, then diminishing to zero and increasing to an equal amount flowing in the negative direction and again decreasing to a zero value. This action is repeated indefinitely. The sequence of a positive and negative current as just described is called a cycle.

The frequency of an alternating current is the number of cycles passed through in one second. An alternation is half a cycle. That is to say, an alternation may be taken as either the positive or the negative wave of the current.

The frequency may be expressed not only in cycles per second but in alternations per minute.

Since one cycle equals two alternations we can interchange these expressions as follows:

Rule 80. A. Having given the cycles per second, to find the alternations per minute multiply the cycles per second by 120. B. Having given the alternations per minute, to find the cycles per second divide the alternations per minute by 120.

EXAMPLES.

If a current has 60 cycles per second, how many alternations are there per minute?

Solution: 60 X 1207,200 alternations.

A current has 15,000 alternations per minute; how many cycles per second are there?

Solution: 15,000 ÷ 120 125 cycles per second.

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A bipolar dynamo having an armature with but a single coil wound upon it (like an ordinary magneto generator) gives one complete cycle of current for every revolution of the armature. That is to say, its frequency equals the number of revolutions per second. A four-pole generator will have a frequency equal to twice the revolutions per second, etc.

Rule 81. To find the frequency of any alternator, divide the revolutions per minute by 60 and multiply the quotient by the number of pairs of poles in the field.

EXAMPLE.

Find the frequency of a 16-pole alternator running at 937.5 revolutions per minute.

Solution: 937.5÷60=15.625 rev. per second. 15.625 X 8 frequency of 125 cycles per second.

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Electrical measuring instruments used on alternating currents do not indicate the maximum volts or amperes of such circuits, but the effective values are what they show. These effective values are the same as those of a continuous current performing the same work.

Rule 82. The maximum volts or amperes of an alternating current may be found by multiplying the average volts or amperes by 1.11. Reciprocally, the average values can be found by taking .707 times the maximum values.

Note that these figures are strictly true only for an exactly sinusoidal current.

EXAMPLE.

Find the maximum pressure of an alternating

current of 55 volts.

Solution: 55 X 1.11 61.05 volts. Ans.

SELF-INDUCTION.

In an alternating current circuit the flow of a current under a given voltage is determined not only by the resistance of the conductor in ohms but also by the self-induction of the circuit. Suppose a current to start at zero and increase to 10 amperes in a coil of 1,000 turns of wire. This magnetizing force, growing from zero to 10,000 ampere-turns, surrounds the coil with lines of force whose action

upon the current in the coil is such as to resist its increase. Conversely, when the current is decreasing from 10 amperes to zero, the lines of force change their direction and tend to prolong the flow of current. This opposing effect which acts on a varying or an alternating current is caller the counter E. M. F. or E. M. F. of self-induction and is measured in volts.

Rule 83. The E. M. F. of self-induction of a given coil is found by multiplying together 12.5664, the total number of turns in the coil, the number of turns per centimeter length of the coil, the sectional area of the core of the coil in square centimeters, the permeability of the magnetic circuit and the current; divide the resulting product by 1,000,000,000, multiplied by the time taken by the current to reach its maximum value.

EXAMPLE.

Find the volts of counter E. M. F. in a coil of 300 turns wound uniformly on a ring made of soft iron wire, the ring having a mean circumference of 60 centimeters and an effective sectional area of 25 square centimeters; its permeability to be taken as 200, and a current of 5 amperes in the coil requires .02 second to reach its maximum value.

Solution: Product of 12.5664 × 300 × 5 × 25 X 200 X 5 is 471,240,000. Product of 10' X .02 is 20,000,000. Dividing the former product by the latter gives 23.562 volts, Answer.

The coefficient of self-induction or, as it is more