Any one of the three answers may be right. It all depends on the characteristic of the tube as we are operating it, and that depends not only upon the type and design of tube but also upon what voltages we are using in our batteries. Suppose the variations in the voltage of the grid are as represented in Fig. 55, and that the characteristic of the tube is as shown in the same figure. Then obviously the first answer is correct. You can see for yourself that when the grid becomes positive the current in the plate circuit can’t increase much anyway. For the other half of the cycle, that is, while the grid is negative, the current in the plate is very much decreased. The decrease in one half-cycle is larger than the increase during the other half-cycle, so that on the average the current is less when the 131signal is coming in. The dotted line shows the average current.
Suppose that we take the same tube and use a B-battery of lower voltage. The characteristic will have the same shape but there will not be as much current unless the grid helps, so that the characteristic will be like that of Fig. 56. This characteristic crosses the axis of zero volts at a smaller number of mil-amperes than does the other because the B-batteries can’t pull as hard as they did in the other case.
You can see the result. When the grid becomes positive it helps and increases the plate current. When it becomes negative it opposes and decreases the plate current. But the increase just balances the decrease, so that on the average the current is unchanged, as shown by the dotted line.
On the other hand, if we use a still smaller voltage of B-battery we get a characteristic which shows a still smaller current when the grid is at zero potential. 132For this case, as shown in Fig. 57, the plate current is larger on the average when there is an incoming signal.
If we want to know whether or not there is any incoming signal we will not use the tube in the second condition, that of Fig. 56, because it won’t tell us anything. On the other hand why use the tube under the first conditions where we need a large plate battery? If we can get the same result, that is an indication when the other station is signalling, by using a small battery let’s do it that way for batteries cost money. For that reason we shall confine ourselves to the study of what takes place under the conditions of Fig. 57.
We now know that when a signal is being sent by the distant station the current in the plate circuit of our audion at the receiving station is greater, on the average. We are ready to see what effect this has on the telephone receiver. And to do this requires a little study of how the telephone receiver works and why.
I shall not stop now to tell you much about the telephone receiver for it deserves a whole letter all to itself. You know that a magnet attracts iron. Suppose you wind a coil of insulated wire around a bar magnet or put the magnet inside such a coil as in Fig. 58. Send a stream of electrons through the turns of the coil–a steady stream such as comes from the battery shown in 133the figure. The strength of the magnet is altered. For one direction of the electron stream through the coil the magnet is stronger. For the opposite direction of current the magnet will be weaker.