12. How many horse-power are there in a water-fall 212 ft. high over which flows 800 cu. ft. of water per second? Express this power in kilowatts.
13. What horse-power must be applied to a dynamo having an efficiency of go per cent. if it is to light 20 arc lamps in series, each taking 10 amperes at 60 volts?
(3) The Heat Effect of Electric Currents
292. The Production of Heat by an Electric Current.—When no chemical or mechanical work is done by an electric current its energy is employed in overcoming the resistance of the conducting circuit and is transformed into heat. This effect has many practical applications and some disadvantages. Many devices employ the heating effect of electric currents, (a) the electric furnace, (b) electric lights, (c) heating coils for street cars, (d) devices about the home, as flat irons, toasters, etc. Sometimes the heat produced by an electric current in the wires of a device such as a transformer is so large in amount that especial means of cooling are employed. Unusually heavy currents have been known to melt the conducting wires of circuits and electrical devices. Hence all circuits for electric power as well as many others that ordinarily carry small currents are protected by fuses. An electric fuse is a short piece of wire that will melt and break the circuit if the current exceeds a determined value. The fuse wire is usually enclosed in an incombustible holder. Fuse wire is frequently made of lead or of an alloy of lead and other easily fusible metals. (See Figs. 271 and 272.)
Fig. 271.—A type of enclosed fuse.
Fig. 272.—A link fuse (above); plug fuses (below).
293. Heat Developed in a Conductor.—A rule for computing the amount of heat produced in an electric circuit by a given current has been accurately determined by experiment. It has been found that 1 calorie of heat (Art. 142), is produced by an expenditure of 4.2 joules of electrical (or other) energy. In other words, 1 joule will produce 1/4.2 or 0.24 calorie. Now the number of joules of electrical energy in an electric circuit is expressed by the following formula:
Joules = volts × amperes × seconds, or since 1 joule = 0.24 calorie,
Calories = volts × amperes × seconds × 0.24 or
H = EI × t × 0.24 (1)
By Ohm's law, I = E/R or E = I × R, substituting in equation (1) IR for its equal E we have