International projectionist (Oct 1931-Sept 1933)

8 INTERNATIONAL PROJECTIONIST December 1931 at this point. We place the meter across the resistance and get only 5 volts. Apparently some voltage has been lost between the battery and the resistance. Taking the voltage across the resistance, which is 5, and dividing by 5, the number of ohms, again gives us 1 amp. as the current. To illustrate a different angle, we draw the circuit shown in Figure 2. R has been moved next to the battery, where it is connected by a wire having practically no resistance. The loop of wire extending from R to X and back to the battery is 200 feet long and has 5 ohms resistance, just as the two wires had in Figure 1 after R had been moved 100 feet from the battery. An ammeter will show a flow of 1 amp. A voltmeter across the iattery shows 10 volts. The meter is 710W placed across R, and the reading is -5 volts as before. One lead from the meter is then touched to point 2 at the lower end of the hattery, and with the other lead placed ■on 3 at the far end of the resistance, a reading of 5 volts is obtained. Inspection ■of the drawing will show that we are measuring the voltage across the wire that connects the battery and the resistance, that is, around the 200-foot loop. It takes 5 volts of the battery potential to. force the current through the loop of wire. We know that the resistance of the wire is 5 ohms. Dividing thfe voltage drop in the wire by the resistance of the wire gives us 1, which is the number of amperes flowing. As the current remains constant so long as no change is made in the circuit and the battery is not discharged, then our answer in amps, must be 1, regardless of how the calculation is made, and right here we must watch our step. Notice the italics in the previous paragraph. To apply ohms-law we must be very careful not to get the various parts ■of the circuit mixed up. Had we taken the voltage of the battery and divided it by the resistance of the long loop of wire, we should have had a wrong answer. Mistakes such as this are very easy to make when dealing with circuits that are more or less complicated, but this is no fault of ohms-law. The law always holds good, and when it seems as though it will not work in some cases, it is because we do not use it properly. The voltage of the battery (Fig. 2), is also the voltage across that part of the circuit starting at 1, through the resistance, R, out on the long stretch of wire to X, and back again to point 2 at the ^ 3 DISTRIBUTION of more than 450 baskets of food to poor families just before Christmas is the proud record of Local Union 160 of Cleveland. This job was handled in its entirety by the Local: the neediest poor families were canvassed and listed, the food was bought and the baskets packed and distributed by the Local membership. Also, the cost of every bit of the food as well as the job of canvassing, packing and distribution was borne by the Local. Every member of the Local participated in the manual labor involved in this great task. Before work, after work, and often long into the night, men who had put in a full working day in projection rooms cheerfully "went to it" and contributed their bit to the success of the plan. There may be better means than this for building community goodwill for a labor organization, but if there are, we have near heard of them. J. J. F. other end of the battery. In applying ohms-law, when we consider the voltage across the entire circuit we must also consider the resistance of exactly the same circuit. The result will then be correct. Before passing on from Figure 2, let us make a different application of ohmslaw. It is desirable to know what the resistance of the wire is from R to the point X. It would probably be 2.5 ohms, but this is not positively known. The voltmeter terminals are applied at 3 and at X, and the voltage is found to be 2.5. 2.5 divided by 1 gives 2.5, the number of ohms in that part of the wire. To measure the voltage in this part of the wire it is necessary to have a wire on the meter about 100 feet long to reach X. In the present problem this makes no difference, but there are some cases where that length of wire would give an erroneous reading. We will discuss that phase of the problem in connection with other circuits later on. Assume that we have two vacuum tubes wired in parallel, the normal filament voltage being 4.5 and the current through each tube 1.6 amps., or a total of 3.2 amps, for both tubes. A rheostat is in series with the tubes so as to cut the voltage from 6 to that required by the filament. The source of current is a 12volt storage battery. A difference of 6 volts exists between the battery voltage R UK ^5 Figure 2 Figure 3 and the voltage we need across the circuit in the amplifier, so a resistance is used to consume the excess voltage. We desire to find out what value of resistance it will take to produce the 6-volt drop. The unknown quantity is the resistance in ohms; the two known values are the voltage to be dropped, which is 6, and the current that is to flow through the additional resistance, 3.2 amps. Dividing 6 by 3.2 gives 1.875, the number of ohms required. In practice, a resistance of 1.8 or 1.9 ohms would be used. The rheostat will take care of the difference caused by the fixed resistance being of a slightly different value. Figure 3 is a diagram of the circuit. Actually there would be a switcli and a number of wires connected to other circuits in the amplifier, but these do not affect the filament current and we need not take them into consideration. R is the fixed resistance, RH is the rheostat, and the two resistances, T, represent the filaments of the vacuum tubes. While we have the circuit in Figure 3 under discussion, let us see what occurs if one of the tubes should burn out. As it stands now, we do not know the resistance of the filaments nor the resistance of that portion of the rheostat which is in use. We know that the current through the rheostat must be 3.2 amps., and we know the rheostat must cause a drop of 1.5 volts, from 6 to that required by the tubes, which is 4.5 volts. 1.5 divided by 3.2 gives a figure very close to .469 ohms, the resistance being used in the rheostat. The filament resistance of one tube is found in the same manner — 4.5 divided by 1.6, or 2.8 ohms. Possibly it isn't clear why we divide by 3.2 in one instance while in the other we divide by 1.6. The current through one tube will be 1.6 amps, when the voltage impressed on the filament is 4.5, consequently we must divide the voltage by 1.6. The rheostat is adjusted until the current through both tubes is 3.2 amps. This same current is also flowing through the rheostat, which causes a drop in the rheostat of 1.5 volts, so 1.5 is divided by 3.2 to find the number of ohms. Having found the resistance of all the apparatus in the circuit we add them all together: — 1.9 ohms in the fixed resistance, .47 ohms in the rheostat (.47 is close enough to the actual value, .469 ohms) ; and 2.8 ohms in the filament of the tube. The sum of all three resistances is 5.17 ohms. The resistance of the wires is negligible. Dividing the battery voltage, 12, by the total resistance in the circuit, 5.17, gives us a current of 2.S amps, through one tube. This much current through a tube designed to carry only 1.6 amps, will very rapidly ruin it. Next month our article will delve deeply into testing, and we will find ohms-law creeping into the picture again and again as we go along.