Showmen's Trade Review (Oct-Dec 1944)

42 SHOWMEN'S TRADE REVIEW October 7. 1944 PROJECTIOI\ {Continued jrom Page 41) determining the total resistance of a circuit when the individual resistances of the components of the circuit are known. We know irom our Ohm's Law that the voltage of our circuit divided by the resistance of our circuit will indicate to us the amount of current that we are to expect to flow in that circuit. So in order to liave an example that will explain our theory we will select a few lamps and solve for the several values of the circuit. Now let us picture what we may call the DUMBBELL EQUATION This dumbbell of ours will come in mighty handy because at times people forget the exact relations of these formulae. The line under the "E" and the " W" mean : that which is above should be divided by one of those that is below ; the line between the "I" and the "R" and between the "E" and the "I" mean that if we multiply the values represented by the letters, we will produce the value of the letter above the horizontal line. Now let's take the first problem and see how it fits into our plan. The wattage of the lamp is fifty. The line voltage is 100. We need to find the resistance and the current of the circuit. So 100 is below the line on the right hand side of the dumbbell and SO is above the line. The other letter below the line is "I" so if we divide 50 by 100 we will get an answer of 0.5, or onehalf an ampere. Now let us transfer this value of current to the left hand side of the dumbbell. We will then have 100 in place of "E" and 0.5 in place of "I." The equation now demands that we divide the 100 by the 0.5 and when we have done this our answer will be 200 and this will be the value of the resistance for the single lamp. Now it is easy to see that if we perform according to our plan as outlined by the dumbbell equations we can multiply the values below the line and produce the value above the line and if we divide the value above the line by either of the values below the line we will produce the value below the line by which we did not divide. Now try the second problem and see how you make out. Now take a look at the third problem. This seems quite different. We can easily see that if any current is to flow through one lamp it must also flow through the other. This is what is termed a "series" circuit. However let us look at it this way. We found the resistance of one lamp to be equal to 200 ohms in our first problem. And we know that all of these lamps are of the same size and wattage. Since the current must now pass through the two lamps before it can light either, we know that the current is now going to be forced to pass through twice as much resistance as it did in the first problem. Therefore the total resistance of the "Series" circuit is bound to be 400 ohms. In other words whenever we have devices connected in series with each other the separate resistances when added together will give us the total resistance of the circuit. So by adding the two together we may now place this value in the "R" position of the left hand of the dumbbell equation and proceed as before and find out the missing value. This will be 100/400 equals 0.25. So now we will have one-quarter of an ampere. Now shift our values of voltage and current to the left hand side of the equation and perform. The answer will now be 25. Now we can see that when resistances are connected in series, the total resistance of the circuit increases and the current decreases and also the wattage consumed by the circuit decreases. There is another fact worth figuring out. In the same third problem we know that a total current of one-quarter of an ampere will flow. The value of the current will be the same for both of the lamps. Therefore the conclusion is obvious and is : That in all parts of a series circuit the current is the same. Since we have a value of 0.25 amperes and we have a resistance in each lamp of 200 ohms and since in our dumbbell equation on the left hand side we see that I times R or IR will equal the voltage we find that the voltage across each lamp will equal 50. So here is another conclusion : The voltages across a SECTION of a series circuit will be equal to the current of the circuit times the resistance of THAT SECTION of the circuit. Then we can say that across the first lamp we lost a voltage of 50 — the voltage drop occasioned by the first lamp. Now we have 50 volts with which to light the second lamp. Now if we add the voltage drops across each lamp we will find that the sum equals our line voltage. Therefore we can say that the sum of all of the voltage drops of a series circuit is equal to the line voltage or the applied voltage. If we replace one of the lamps with a variable wire wound resistor and slowly vary the value of the resistor we will find that our lamp will increase and decrease its brilliancy as we move the variable arm of the control. This permits a greater and / You who realize how important it is to have projectors that can stand the gaff of continuous, day-in, day-out use — consider the war record of DeVRY before you buy. Learn how precision projectors such as these have assured men of the U. S. 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