Our spark coil has a vibrator which acts precisely like the hammer of the bell to make and break the primary circuit and thus make rapid changes in the magnetic field produced by the primary coil. The primary coil of the spark coil is many times larger than the coil of the bell, that is, it contains many more turns of wire. It has much more iron in the core. We use upon it five cells instead of the two cells upon the bell. The result of all this is that we have a much more powerful magnetic field than that in the bell and many more watts of energy from which to induce a secondary current. Now the number of turns employed in the secondary circuit of our spark coil is very great, stepping its voltage up to thousands where the bell induced hundreds. [Illustration: Fig. 121] Suppose we now repeat our experiment in which we tried to light the gasolene in the watch crystal, using now the spark coil of the boat instead of our small "home-made" coil. In Fig. 121, B is the battery of five dry cells. _S_ is a switch. _V_ is the vibrator, which, like the hammer of an electric bell, makes and breaks the primary circuit. Of course the coil has a core of iron, although that is not here represented, and, of course, the coil has many hundred turns instead of the few here represented, and of course also it is built up of many layers instead of one as here represented. The secondary has very many more turns than the primary, but those in which the primary current passes are common to both circuits. There is also a condenser--not here represented, and not to be described in this book. The result of all this is that the secondary circuit has a voltage of between 5000 and 10,000, and a spark jumps across the gap at _c_ between one sixteenth and one eighth of an inch long. This spark is hot enough to light the gasolene which I have put in the watch crystal at _c_. [Illustration: Fig. 122] Let us return to the bell for a few minutes. I have here a miniature lamp which requires 10 volts and .1 ampere, that is, 1 watt, which I will connect at _S_ (Fig. 122). When now I close the primary circuit with two cells at _P_ you notice that the lamp lights up, but faintly. It is not receiving .1 ampere. Remember we have only .75 watt at our disposal and this lamp requires 1 watt. Hence it is getting only three quarters enough energy. We connect in a third cell and now it lights up to full brilliancy. The resistance of this lamp must be about 100 ohms. (10 volts)/(100 ohms) = .1 ampere The resistance of the four boys might have been 60,000 ohms, and the voltage of the secondary circuit might in that case have been, say, 150. (150 volts)/(60,000 ohms) = .0052 ampere How does it happen that the secondary current had a pressure of 150 volts on the boys but cannot supply even the 10 volts required by the lamp? Perhaps we can be brought to appreciate the answer to that question best by asking ourselves some others quite like it.