E/R = C or Electromotive force/Resistance = Current.
Or if we like to use the initials of volts, ampères, and ohms, instead of the general terms, E, R, and C, we may write V/R = A, or Volts/Ohms = Ampères.
From this it appears that 1 volt will send a current of 1 ampère through a total resistance of 1 ohm, since 1 divided by 1 equals 1. So also 1 volt can send a current of 4 ampères through a resistance of ¼ of an ohm, since 1 divided by ¼ is equal to 4. We can therefore always double the current by halving the resistance; or we may obtain the same result by doubling the E.M.F., allowing the resistance to remain the same. In performing this with batteries we must bear in mind that the metals, carbon, and liquids in a battery do themselves set up resistance. This resistance is known as "internal resistance," and must always be reckoned in these calculations. We can halve the internal resistance by doubling the size of the negative plate, or what amounts to the same thing by connecting two similar cells "in parallel;" that is to say, with both their zincs together, to form a positive plate of double size, and both carbons or coppers together to form a single negative of twice the dimensions of that in one cell. Any number
of cells thus coupled together "in parallel" have their resistances reduced just in proportion as their number is increased; hence 8 cells, each having a resistance of 1 ohm if coupled together in parallel would have a joint resistance of ⅛ ohm only. The E.M.F. would remain the same, since this does not depend on the size of the plate (see [§ 38]). The arrangement of cells in parallel is shown at [Fig. 16], where three Leclanché cells are illustrated thus coupled. The following little table gives an idea of the E.M.F. in volts, and the internal
resistance in ohms, of the cells mostly used in electric bell work.
Fig. 16.
TABLE SHOWING E.M.F. AND R. OF BATTERIES.