Optic projection : principles, installation and use of the magic lantern, projection microscope, reflecting lantern, moving picture machine, fully illustrated with plates and with over 400 text-figures (1914)

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CH. XIII] DIRECT CURRENT UNITS 477 § 657. The Ohm. — This is the unit of resistance to the flow of an electric current. It is represented by the resistance, at zero centigrade, of a column of mercury 106.3 centimeters long, of uniform cross sectional area, and weighing 14.4521 grams. Such a column of mercury will have a cross sectional area of one square mm. Ohmage is a term analogous with voltage and amperage. It is used to designate the amount of resistance in ohms of an electric circuit. A conductor may have little resistance, as copper, etc., or it may have great resistance like German silver. Naturally then copper wire is used largely for electric circuits, and German silver wire for making resistors or rheostats (§ 7240.}. § 658. The Watt. — This is the unit of activity and is the rate at which work can be done by a current of one ampere under a pressure of one volt. One watt means the doing of work at the rate of io7 ergs per second, or one joule per second. This is approximately equal to the lifting of i kilogram, io centimeters every second. § 659. Kilowatt. — A kilowatt is 1,000 watts. This term is more common than watt. It is equal to 1.34 horse power. § 660. The watts which any direct current represents are obtained by multiplying the quantity of current flowing by the pressure — that is, the amperes by the volts. Thus, if there were an amperage of one and a voltage of one, there would be an activity of one watt. If the voltage were io and the amperage 100, or the voltage 100 and the amperage io, there would be an activity of 1,000 watts, or one kilowatt. § 661. Kilowatt-hour. — This is the unit of electrical energy or work, which is in commercial use and which is used as a basis for making the charges to consumers. A kilowatt-hour is the work represented by one kilowatt when acting for one hour. In order to find the amount of work done by an electric current it is necessary not only to know the rate at which the work is being done but also the time during which this rate is continued. Thus,