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HANDBOOK OF PROJECTION FOR
through the grid, according to how the dial switch is adjusted. We will therefore have a total of from 25+25=50, to 25+40=65 amperes at the arc with this combination. Were we to connect the same two rheostats in series on D. C. the resultant current would be from 10 to 12+ amperes. It is figured as follows. The Powers is a 25-ampere 110 volt instrument, hence it has, roughly, (110 — 50)-^25=2^ ohms resistance. The grid, when working at 25 amperes, must have the same resistance, hence there will be a total of 2^+2^ ohms resistance when they are opposed to the voltage in series when the grid is delivering its minimum amperage. The resistance of the arc will be approximately two ohms, hence ll(H-(2^+2^+2) will equal the amperage. This amounts to about 16 amperes. If the grid were set on the
Figure 138.
40-ampere contact, we would then have (110 — 50)-*-40= practically 1.5 ohms, which added to the resistance of the Powers makes 2^2+1^=4 ohms. We would then have lNH-(2l/2-\1^+2)=18+ amperes delivered. If the current were A. C. we would then have the same thing, except that instead of subtracting 50 from 110 we would subtract the resistance of the alternating current arc, which might be taken at about 30.
Our readers should understand, however, that these figures are approximate only. It is impossible to figure the matter accurately, for the reason that the arc resistance varies with the length of the arc, also the rheostatic resistance varies with (a) temperature of the coils or grids; (b) with their age; also merely because a rheostat is stamped "110 volt 25 ampere" it does not follow that it has exactly the resistance this would indicate. Then, too, the supply voltage may not be just what you think it is, hence it follows that the results