(110 volts)/(330 + 220 ohms) = .2 ampere
Suppose now we should undertake to use the same flasher and the same lamp on a 220-volt current. This might push more current through than the small wire could carry. It might melt, or its insulation might burn off before a made contact with b; if not the lamp would certainly burn out after the contact. If we undertook to operate with this flasher several 32-candle-power lamps instead of one upon the 110-volt circuit, the result would be the same, for in that case the resistance would be reduced and, therefore, a greater current would pass than the wire could carry without undue heating.
Fig. 86
The boys were at first troubled to see how increasing the number of lamps in a circuit would decrease the resistance in that circuit. [Fig. 86] was drawn to explain the matter. The lamps l, l, l, etc., are connected in parallel. Each lamp makes an independent connection from one feed wire to the other. The flasher a acts as a switch to close the circuit for the whole.
Now if we think of these wires as pipes to conduct water we would say that water flows from D to E through ten pipes more readily than through one. It would meet with only one tenth as much resistance. The result would be the same, if we should substitute for the ten pipes one pipe ten times as large in cross section. So it is with wires which are conducting electricity. Introduce two in parallel, and you allow twice as much current to pass by reducing the resistance to one half. Ten parallel conductors reduce the resistance to one tenth and allow ten times as much current to pass.
Fig. 87