Our equation C = E / R must then obviously become
C (10 ampères) = E (200 volts) / Total Resistance (20 ohms).
The resistance of the arc itself being the same as before, viz. 4.5 ohms, it is obviously necessary to put an extra fixed resistance equal to 15.5 ohms in series with it in order to make up the total of 20 ohms.
Now leave the arc unattended until the resistance of 4.5 ohms has again become 5 ohms; the only effect is that our current, instead of remaining at 10 ampères, has become 200 / 20.5 or 9.8 nearly, a difference which is imperceptible.
This is not all, for it is an elementary rule in electrical science that the total E.M.F. of any circuit distributes itself along that circuit in proportion to the distribution of resistance.
In other words, our original E.M.F. of 200 volts will so distribute itself as to reserve, so to speak, an E.M.F. of 45 volts for the arc, while the resistance of this remains at 4.5 ohms, but directly this resistance increases, the E.M.F. at the arc lamp terminals automatically rises, and therefore the actual diminution in current is even less than the figures above quoted.
Should the arc tend to 'break' or go out, the resistance across it automatically becomes infinite and the whole 200 volts is at that moment available to prevent the occurrence.
Under these conditions, therefore, the operator can safely leave the arc for many minutes at a time. In carrying out experimental work I have often left the lantern, walked up to the screen, discussed results with a friend, and walked back, and the arc has shown no signs of misbehaviour whatever.
