Influence Machine

Whenever this occurs there is a loss of energy, and therefore the whole work employed in turning the machine is not converted into electrification in an available form, but part is spent in producing the heat and noise of electric sparks.

I have therefore thought it desirable to shew how an electrical machine may be constructed which is not subject to this loss of efficiency. I do not propose it as a useful form of machine, but as an example of the method by which the contrivance called in heat-engines a regenerator may be applied to an electrical machine to prevent loss of work.

In the figure let A, B, C, A′ , B ′ , C ′ represent hollow fixed conduc- tors, so arranged that the carrier P passes in succession within each of them. Of these A, A′ and B, B ′ nearly surround the carrier when it is at the middle point of its passage, but C, C ′ do not cover it so much. We shall suppose A, B, C to be connected with a Leyden jar of great capacity at potential V, and A′ , B ′ , C ′ to be connected with an- other jar at potential −V ′ . P is one of the carriers moving Fig. 42. in a circle from A to C ′ , &c. and touching in its course certain springs, of which a and a′ are connected with A and A′ respectively, and e, e′ are connected with the earth.

Let us suppose that when the carrier P is in the middle of A the coefficient of induction between P and A is −A. The capacity of P in this position is greater than A, since it is not completely surrounded by the receiver A. Let it be A + a. Then if the potential of P is U, and that of A, V, the charge on P will be (A + a)U − AV.

Now let P be in contact with the spring a when in the middle of the re- ceiver A, then the potential of P is V, the same as that of A, and its charge is therefore aV.

If P now leaves the spring a it carries with it the charge aV. As P leaves A its potential diminishes, and it diminishes still more when it comes within the influence of C ′ , which is negatively electrified. If when P comes within C its coefficient of induction on C is −C ′ , and its capacity is C ′ + c ′ , then, if U is the potential of P the charge on P is (C ′ + c ′ )U + C ′ V ′ = aV. If C ′ V ′ = aV, then at this point U the potential of P will be reduced to zero. Let P at this point come in contact with the spring e′ which is connected with the earth. Since the potential of P is equal to that of the spring there will be no spark at contact.

This conductor C ′ , by which the carrier is enabled to be connected to earth without a spark, answers to the contrivance called a regenerator in heat- engines. We shall therefore call it a Regenerator.

Now let P move on, still in contact with the earth-spring e′ , till it comes into the middle of the inductor B, the potential of which is V. If −B is the coefficient of induction between P and B at this point, then, since U = 0 the charge on P will be −BV.

When P moves away from the earth-spring it carries this charge with it. As it moves out of the positive inductor B towards the negative receiver A′ its potential will be increasingly negative. At the middle of A′ , if it retained its charge, its potential would be − A′ V ′ + BV , A′ + a ′

and if BV is greater than a′ V ′ its numerical value will be greater than that of V ′ . Hence there is some point before P reaches the middle of A′ where its potential is −V ′ .

At this point let it come in contact with the negative receiver-spring a′ . There will be no spark since the two bodies are at the same potential. Let P move on to the middle of A′ , still in contact with the spring, and therefore at the same potential with A′ . During this motion it communicates a negative charge to A′ .

At the middle of A′ it leaves the spring and carries away a charge −a′ V ′ towards the positive regenerator C, where its potential is reduced to zero and it touches the earth-spring e. It then slides along the earth-spring into the negative inductor B ′ , during which motion it acquires a positive charge B ′ V ′ which it finally communicates to the positive receiver A, and the cycle of operations is repeated.

During this cycle the positive receiver has lost a charge aV and gained a charge B ′ V ′ . Hence the total gain of positive electricity is B ′ V ′ − aV.

Similarly the total gain of negative electricity is BV − a′ V ′ . By making the inductors so as to be as close to the surface of the carrier as is consistent with insulation, B and B ′ may be made large, and by making the receivers so as nearly to surround the carrier when it is within them, a and a′ may be made very small, and then the charges of both the Leyden jars will be increased in every revolution.

The conditions to be fulfilled by the regenerators are C ′ V ′ = aV, and CV = a′ V ′ .

Since a and a′ are small the regenerators do not require to be either large or very close to the carriers.