Radio Broadcast (May 1928-Apr 1929)

206 RADIO BROADCAST AUGUST, 1928 No* 4 Radio Broadcast's Home Study Sheets August, 1928 Ohm's Law THE most fundamental rule in all electrical work is known as Ohm's Law — from its discoverer, a German experimenter. Previous to its discovery experimenters had only vague notions regarding the amount of current that passed through a circuit under given conditions of voltage and resistance. This law states that in any electrical circuit, the current in amperes equals voltage in volts divided by resistance in ohms. In electrical language, this means that I (intensity of current) equals E (electrical pressure or voltage) divided by R (resistance) ' RADIO BROADCAST LABORATORY — k H E 5.4-2 \E R"(260-90)» 10~3 j „ = .02x 103 1 R = 20 j FIG. I This law may be stated in three ways, viz., 1.1 = E/R 2. E = I x R 3. R = E/I To get a working knowledge of this fundamental law, we need the following: LIST OF APPARATUS 1. A source of current, say a storage battery or several dry cells connected in series. 2. Two resistances, about 20 ohms each. One-may be a rheostat such as was used in Experiment 2 and the other a similar rheostat or a fixed resistor. The Yaxley 20-ohm De Luxe resistor for example. 3. A voltmeter that will read up to six volts. A Model 301 Weston was used in the Laboratory. 4. A milliammeter reading about 300 milliamperes. The one used in the Laboratory was a Jewell Pattern 54. 5. Hook-up wire. PROCEDURE The baseboard set-up described in Experiment 2 may be used for this experiment by connecting the second resistance, i. e. the 20-ohm rheostat or the fixed resistance, R2, and the voltmeter across the filament terminals of the socket. The connections are shown in Fig. 3, that is, minus A to 3, plus A to the meter and the other meter terminal to 4. The schematic diagram of the set-up is given in Fig. 2. Turn the rheostat arm slowly and note the voltmeter and milliammeter readings. If either reads backwards, reverse the connections to it. Note down as in column 1 and 2 in table 1 the voltage and the current as the rheostat is varied. As a final reading, short circuit the rheostat and allow the full battery voltage to be applied across R». Read the current in the circuit. Then remove R2 but leave the voltmeter across the filament terminals to which R2 was attached. Read the current flowing now. This is the current taken by the meter. Using the third way of expressing Ohm's law, viz., that resistance is equal to the voltage divided by the current, calculate the resistance of R2 and of the voltmeter. Plot on cross section paper as in Fig. 1 the current against the voltage, using the vertical scale for the voltage. DISCUSSION The experiment we have just performed is what is known as the voltmeter-ammeter method of measuring a resistance. All that is needed to determine an unknown resistance is to measure the current through it when a known voltage is across it. The total voltage of the battery has not changed during this experiment, but the voltage across the resistor, R2, has varied with each setting of the rheostat. What has happened to the remainder of the battery voltage? Clearly it has been cut down by the rheostat, and if we place the voltmeter across this variable resistance as the experiment is repeated, we shall perceive that the total voltage, that is, the voltage lost across resistor R2 and that across; the rheostat, adds up to the terminal voltage of the battery. Since we know the voltage across the rheostat, that is, the battery voltage minus the voltage measured across resistance R2, and the current through it (the current in a series circuit such as this is the same in all parts of the circuit) we may use the second method of stating Ohm's law to determine the resistance of the rheostat at each reading in our table and fill in the values. When the current is plotted against the voltage, a straight line results. The slope of this line, that is, the vertical units divided by the horizontal units, is the resistance of the circuit. If this line did not turn out to be straight, we should have to assume that the resistance of something in the circuit changed with the current through it. This is true of the vacuum tube filament. It has a temperature coefficient, that is, its resistance changes with increase in temperature. This matter of temperature coefficient is discussed in the Signal Corps' Principles Underlying Radio Communication on page 37 in Morecroft's Principles of Radio Communication on page 25, in all physics textbooks and in books on electricity. An "IR drop" is the technical expression for the voltage appearing ■VWWVWWV— HltflflH FIG. 2 across a resistance when a current flows through it; it is sometimes merely referred to as a voltage drop. It may be calculated, as may be seen from its name, by multiplying the current by the resistance. Similarly the value of an unknown voltage may be determined by observing how much current it can force through a known resistance. Thus a meter for measuring voltage, known as a voltmeter, is simply a sensitive current measuring device calibrated in volts rather than in amperes. Throughout an experiment of this kind, and the calculations which go with it, the proper units must be used to make the formula bring the correct answer. The rule uses amperes, volts, and ohms. In this experiment we have used a meter which measures thousandths of amperes, or milliamperes. In order to use the various ways of stating Ohm's Law, we must convert these milliamperes into amperes and then proceed. For example, in the Laboratory the current in the circuit was 94 milliamperes when the voltage across R2 was 2 volts. In this case one cannot divide 2 by 94 and expect to get an answer in ohms; instead we must realize that 94 milliamperes are 0.094 amperes and divide accordingly. Thus Ohm's Law, which can be stated in any one of three ways, is useful in determining any one of three fundamental electrical quantities, voltage, current, and resistance. PROBLEMS 1. Throughout this experiment there have been two "IR" drops in the circuit. Where are they? What should their sums be? 2. Calculate all of the values of current and voltage drops that would exist if R2 had a resistance of 10, 40, 400 ohms and Ri were equal to 0, 1,2 ohms. 3. What current would be taken from the battery if a receiver using five i-ampere tubes were operated from it? 4. If the storage battery has a useful life, on one charge, of 100 ampere hours, how long can you operate a five tube (201 A) receiver at three hours per day without recharging the battery? TABLE 1 600 ER2 I ^i = R2 Eb Er2 2 94 21.3 4 2 3 140 21.3 4 1 4 190 21.0 6 1 5 236 21.0 6 1 6 290 20.7 6 0 R2 open, I = 10 = Im meter current ER2 = voltage across R2 I = current in milliamperes R2 = resistance of R2 Ri = resistance of rheostat = Rm = resistance of meter ERi = voltage across rheostat Eb = voltage of battery ERi Eb — Er2 FIG. 3