Now, if instead of interposing four inches of air, or the human body, between the mains of our 110-volt dynamo, we connected an incandescent lamp across the mains, how much electricity would flow from the generator? An incandescent lamp consists of a vacuum bulb of glass, in which is mounted a slender thread of carbonized fibre, or fine tungsten wire. To complete a circuit, the current must flow through this wire or filament. In flowing through it, the electric current turns the wire or filament white hot--incandescent--and thus turns electricity back into light, with a small loss in heat. In an ordinary 16 candlepower carbon lamp, the resistance of this filament is 220 ohms. Therefore the amount of current that a 110-volt generator can force through that filament is 110/220, or 1/2 ampere. [Illustration: Armature and field coils of a direct current dynamo] One hundred lamps would provide 100 paths of 220 ohms resistance each to carry current, and the amount required to light 100 such lamps would be 100 × 1/2 or 50 amperes. Every electrical device--a lamp, a stove, an iron, a motor, etc.,--must, by regulations of the Fire Underwriters' Board be plainly marked with the voltage of the current for which it is designed and the amount of current it will consume. This is usually done by indicating its capacity in watts, which as we have seen, means volts times amperes, and from this one can figure ohms, by the above formulas. _A Short Circuit_ We said a few paragraphs back that under no conditions must two bare wires leading from electric mains be permitted to touch each other, without some form of resistance being interposed in the form of lamps, or other devices. Let us see what would happen if two such bare wires did touch each other. Our dynamo as we discover by reading its plate, is rated to deliver 50 amperes, let us say, at 110 volts pressure. Modern dynamos are rated liberally, and can stand 100% overload for short periods of time, without dangerous overheating. Let us say that the mains conveying current from the armature to the switchboard are five feet long, and of No. 2 B. & S. gauge copper wire, a size which will carry 50 amperes without heating appreciably. The resistance of this 10 feet of No. 2 copper wire, is, as we find by consulting a wire table, .001560 ohms. If we touch the ends of these two five-foot wires together, we instantly open a clear path for the flow of electric current, limited only by the carrying capacity of the wire and the back pressure of .001560 ohms resistance. Using Ohm's Law, C equals E divided by R, we find that C (amperes) equals 110/.001560 or _70,515 amperes_! [Illustration: A direct current dynamo]