International projectionist (Jan-Dec 1935)

REVIEW NUMBER May 1935 law. 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, which, 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 12-volt storage battery. A difference of 6 volts exists between the battery voltage 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 switch 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. Figure 3 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 2.3 amps, through one tube. This much current through a tube designed to carry only 1.6 amps, will very rapidly ruin it. IN FIGURE 4 is shown a simple twowire circuit. It is surprising how many things can happen in such a circuit and the different methods that can be used in testing it. The use to which this circuit is put often has a large bearing on the test and in interpreting the results thereof. Assume that the drawing represents a line carrying 110 volts and that the end at A is connected to the source of supply. The wires at B go to the device that constitutes the load, but for the present we will assume that it has been disconnected by opening a switch. Near the end B are two fuses. Both wires should be continuous from A to B, that is, if nothing is wrong. For the first test we place the leads from a test lamp across the terminals of the fuse block at the end nearest B. If the lamp lights, it is an indication that current is flowing through the lamp and that it probably is coming from A. This might sound queer, but the fact that the lamp lights when connected to the end of the line at B is not positive proof that the line is continuous. The line could be broken, as at X in Figure 5, and if there is a circuit of some kind around the break, as shown by the dotted line, the lamp will light. It is true that such a condition is not likely to occur in a line such as we are discussing. It is not at all unlikely to happen in an amplifier or its associated apparatus, however, with the usual result of noisy operation or possibly one or more tubes working at a wrong voltage, etc. The unlikely condition is the one that stumps the man who does not consider all possibilities. As a rule, it can be taken for granted that the current is coming from A when the test lamp lights. If any doubt exists (and it must be established if this is so), simply opening both sides of the circuit at A will establish the fact. If this causes the lamp to go out, it shows that A is the source of power; if the lamp remains lighted, the current must be coming from some other place. Figure 6 shows how such a condition might occur. We have a line from A to B as before, but it is broken at X-X and two wires are brought down to C where they are connected to a different power supply. Suppose that the lamp does not light. The next step will be to see that the test circuit is 0. K. Place the test lamp across a circuit that is known to be in good condition. This can be done by testing across the fuses that connect to the projection room lights, the amplifier circuit, or somewhere in the circuit between the motor generator and the arc. If the projection room lights work, if the amplifier is lighted, or if the arc is lighted we know that these circuits are 0. K. Should the test lamp light when put across any of these, it shows that our test circuit was in good shape and that the current was not getting to point B in this line. If it is shown that the test circuit is at fault, we try another lamp, one that has been working somewhere, or we can take the lamp out of the test circuit and try it in a socket in which a light has been burning. If the lamp was in good condition, the trouble is either a broken wire, or the lamp was not screwed down in the socket. Now that the test circuit is O. K., we again test at B (Fig. 6). Suppose that we still get no indication of current. The test terminals are then placed across the other end of the fuse block, the end ! farthest from B, and we find that the , lamp lights here, which shows that the trouble is either in the fuses or in some part of the fuse block. One wire from the test lamp is then touched to C (Fig. 4) , and the other wire to D. If the lamp does not light, it shows that the lower fuse is probably blown. Ninety-nine times out of a 100 it will be the fuse; but we must remember that it is possible for the trouble to be in the fuse receptacle. If the lamp lights when this test is made, it shows that B A E D Figure 4 X Figure 5 B