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November 1957 Popular Electronics[Table of Contents] People old and young enjoy waxing nostalgic about and learning some of the history of early electronics. Popular Electronics was published from October 1954 through April 1985. All copyrights are hereby acknowledged. See all articles from Popular Electronics. 
As with on my Airplane and Rockets hobby website, a big part of my motivation for scanning and posting these vintage electronics magazine articles has been twofold. The first reason is to provide access to historical documents for educational reasons. The second reason is to have the names of people and places published in text format (everything OCRed) so that someone doing a Web search for himself, a relative, or a friend, might run across it here. I receive emails occasionally from readers who are thrilled to find those names in an old article.
By Herb S. Brier, W9EGQ
In October we discussed some of the fundamental properties of current, voltage, and resistance, and how the relationship between them is expressed concisely by Ohm's law: E = IR, I = E/R, R = E/I. We also talked about the formulas for calculating the power in electrical circuits, P = EI and P = I^{2}R, and how to use them answering simple questions involving Ohm's law and power calculation that may found in the Novice examination. Now, we'll go over the information necessary to understand and answer questions that might appear in the General/Technician/Conditional exam.
In the General Class examination, there are usually a couple of questions based on the diagram in Fig. 1, such as: "What is thee bias voltage on the tube?" And: "What is the resistance of R2?" It is important to read each question separately and then study the diagram to determine which of the data given is pertinent to that question. Once this is done, the problem is half solved.
For example, there is nothing difficult about determining the resistance of R2. From Ohm's law, R = E/I; therefore, you simply divide the voltage across R2 by the current flowing through it. Both of these quantities are given. Remembering to change milliamperes to amperes, the solution will be:
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Referring to Fig. 2(B), the first step in calculating the resistances in the voltage divider is to decide on how much "bleeder current" to allow to flow through it. Normally, about 10% of the current capacity of the power supply is allocated for the purpose. Ten percent of 200 ma. is 20 ma. When added to the 160 ma. drawn by the transmitter, this makes the total keydown current drain on the supply 180 ma.
Starting at the bottom of the divider, the only current that flows through R1 is the bleeder current, while the voltage across it is obviously 100 volts. Substituting these values in the appropriate Ohm's law formula gives:
Under noload conditions, the voltage at the divider taps will be higher than when they are under load. The main advantage of a voltage divider over series dropping resistors is that voltage variation between the load and noload conditions is minimized.
Resistances in Parallel. Figure 3 shows three resistances in parallel. The formula for calculating the effective resistance of such a combination is:
If you are a whiz at algebra, you know exactly how to solve this formula and need not worry that you may be asked about it in the examination. If not, you will probably say: "Fine, but what does it mean?" Fortunately, it's easy to understand with a bit of explanation.
Fig. 3. How to calculate the effective resistance when three resistances are connected in parallel.
Referring to Fig. 3 and taking each resistance in turn, we know from Ohm's law that the currents through them are: I1 = E/R1, I2 = E/R2, and I3 = E/R3. Furthermore, the total current is the sum of the individual currents:
I_{total} = I1 + I2 + I3, or I_{total}
= E/R1 + E/R2 + E/R3.
Reverting to the fundamental Ohm's
law again, the effective resistance of the three resistances
in parallel is:
Of course, we can make E any value we wish, as long as it is the same in each part of the equation. For convenience, E is usually made equal to one volt in parallelresistance calculations, making the last equation the same as the first one given here.
The effective resistance of 2, 4, and 5 ohm resistors in parallel is 1.05 (to two decimal places). I suggest you verify this answer for practice and then make up a few practice problems of your own. In checking your answers, remember that the effective resistance is always less than that of any of the individual resistances.
Posted June 27, 2012