Radio Broadcast (Nov. 1925-Apr 1926)

APRIL, 1926 THE USE OF THE FILAMENT RESISTANCE 683 Or,where the resistance of r1 is the same as, r2orr3 then R = N + I" Where N = Number of resistance units FIG. 4 A series-parallel circuit. The total resistance of the circuit is determined by first calculating the parallel circuit resistance of rl, r2, and r3 and adding that total to r4 remaining constant, then double the current will flow. This gives rise to the equation 1 = ^ where I is the current in amperes, E is the pressure in volts and R the resistance in ohms. From this equation it is possible by transposing, to find any one value where the other two are p known. That is to say E = IxR and R = — • A circuit comprising these three factors is shown in Fig. i . If a resistance is paralleled by another of the same value, then two paths are provided for the flow of current so the total resistance to this flow is cut in half. If the resistances are added to. each other, that is, connected in series, then the current flow is retarded because the total circuit resistance has been increased. To determine the total resistance of a circuit where resistances are in parallel the formula R = 1 i J_ I JL ls employed, as shown in Fig. 2. R.1 + R2 + R.3 Where the resistances are in series, the total resistance is equal to the sum of all the resistances or where the resistance per unit is the same, then the total is equal to the value of one unit multiplied by the number of units employed. Expressed algebraically R = r, + r2 + r3, etc. or R = r,XN where R = total resistance r = resistance per unit and N = number of units. This is illustrated in Fig. 3. ft is possible to combine resistances in a circuit so that a series-parallel arrangement is produced. This is the case where it is desired to know the total resistance of a circuit comprising several tubes in parallel with a single rheostat in series with the tubes and battery. To calculate this total resistance, it is first necessary to find the resistance of all the tubes in parallel. Then when this value is known it is added to the value of the resistance of the rheostat. This is illustrated in Fig. 4. In the matter of determining the resistance of the tube filament Ohm's Law is employed first and then where it is desired to know the total resistance of a circuit, where such an arrangement exists as in Fig. 4 then the formula for resistances in series is employed. Take, for example, a five-volt tube. Its filament should be energized by the battery so that .25 ampere of current flows in the circuit when the voltage at the tube terminals is 5. With these two known factors it is possible to determine the resistance of the circuit. Since the resistance of the battery and wire for the circuit is negligible, therefore the resistance computed will be purely tube resistance. Applying the tube, E = rated voltage of tube, I = rated filament current in amperes. Then R = — =20 25 ohms. WHY RESISTANCES ARE NEEDED "MOW if six volts is applied to a filament. * ^ the current will be correspondingly greater than with five volts. In order to keep the current at that point stipulated by the tube manufacturer, it is necessary to decrease this voltage by adding resistance to the circuit. Until a short while ago the one means for regulating the current flow and voltage in a filament circuit was by means of the rheostat but lately there has been placed on the market the filament ballast, otherwise termed filament regulator. That is, they automatically decrease the battery voltage to the correct point for application to the filament terminals. It is a known fact that the battery voltage remains quite constant over the major portion of its discharge life but at the end takes a decided and sudden drop. A curve illustrating this is shown in Fig. 5. It is because of this voltage life-retaining property of the battery that filament ballasts have proved satisfactory for use as filament controls. For those who desire simplicity of control, the filament ballast will commend itself. Writers of radio articles have differed widely for years concerning the correct value of rheostat to be used in a filament circuit. The best possible advice, and the easiest to follow is that our old friend Ohm's Law be used. Where two factors or values of this equation are known, the third can be determined by the application of the formula. Let's look over a typical filament circuit consisting of tube, rheostat and battery, such as that in Fig. 6 B. The filament Rof the tube is considered as a resistance and, therefore, its value may be rated in ohms. The battery E is the source of the energy which lights the filament and has a certain voltage, usually six. The rheostat r, has a variable external resistance whose total resistance we do not know, but wish to ascertain. If there were no rheostat r, in the circuit and the voltage of the battery were 5, then the total resistance of the circuit would be 20 ohms. If the battery voltage were raised to 6 then .3 amperes of current would flow in the circuit instead of the rated .25 amperes. Now by introducing a resistance in the form of the rheostat r, not only is the current reduced but the voltage at the filament terminals is accordingly diminished. By applying the E 6 formula R = j then R = — = 24 ohms which is the total resistance of the circuit. How much B -(¥) Where r = Rheostat resistance E = Battery voltage Ei= Filament terminal voltage I = Filamentamperes Example : r =(1) " (1) = 4 0hms FIG. 6 Determining the resistance value of the unit 'r' in the circuit above involves the use of the formula as shown. The battery voltage, the tube resistance and the current are usually known ; from these values it is possible to calculate the unknown resistance is necessary in the rheostat? The answer may be found by subtracting the circuit resistance at 5 volts from the circuit resistance at 6 volts i. e. 24—20 = 4 ohms as shown in Fig. 6A. From this we see that with 4 ohms in the circuit where a fully charged 6-volt storage battery is employed, .25 amperes of current will flow. Theoretically, as the charge in the battery decreases, the voltage decreases; therefore, to keep the circuit characteristics at their rated level, it is necessary to cut out part of the external resistance to compensate for the corresponding drop in battery voltage. Now the main rub comes in the advocacy of rheostats larger than 4 ohms where only one tube is to be controlled by it. Of course, if one rheostat controls more than one tube, the proper resistance value may be calculated since usually the tube filaments are in parallel and as such the total resistance of these filaments is figured from the formula R = r,+~^M — ~^ etc. To use a rheostat of 4 ohms means that when the movable arm touches the first turn of the wire BATTERY LIFE & VOLTAGE DROP 7.0 6.5 6.0 C 5.5 5.0 4.5 ! ! ! 1 L Dl9 > ! ! ! 1 1 0 10 15 20 60 65 70 75 80 Ohm's Law R = where R resistance of 25 30 35 40 45 50 55 DISCHARGE LIFE IN AMPERE-HOURS FIG. 5 When a fully charged battery is first used, its voltage is slightly above six volts. The major portion of its life, however, is at a fairly constant voltage level, the gradual drop being from 6 to 5.7 volts. When this low point has been reached, the battery is considered discharged and is in need of recharge. As the 5.7 point is reached the rheostat in a filament circuit is of greatest use because as the rheostat arm is advanced the voltage at the tube terminals is maintained at its highest point