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What's Your EQ?
September 1961 Radio-Electronics

September 1961 Radio-Electronics

September 1961 Radio-Electronics Cover - RF Cafe[Table of Contents]

Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Electronics, published 1930-1988. All copyrights hereby acknowledged.

These "What's Your EQ?"  (electronics quotient) challenges that appeared in Radio-Electronics magazine are sort of the electronics enthusiast  version of The Old Farmer's Almanac "Old and New Mathematical Puzzles." Whereas the OFA puzzles could be on any subject - sometimes including electricity - the RE puzzles are typically based on circuit analysis and troubleshooting issues. Two of this month's problems are new versions of old themes. Any time you see a schematic with series and parallel combinations of components (usually of the same kind), the first thing to do is attempt to redraw it in a more familiar configuration. The Black Box problems often have more than one solution; in this instance, at least three "correct" answers were submitted. The "Over the River" puzzle is reminiscent of an OFA type puzzle. Have fun.

What's Your EQ?

What's Your EQ?, September 1961 Radio-Electronics - RF CafeHere we go again. Sharpen up your pencils, put on that thinking cap and settle down to some interesting puzzles.

If you've got one of your own that will stump our readers, especially if it's on an engineering level, send it to us. We pay $10 and up for each one accepted. Write to EQ editor, Radio-Electronics, 154 W. 14 St., New York 11, N. Y. Answers to July puzzles are on page 89. Answers to this month's puzzles appear next month.

Series-Parallel Capacitors - RF CafeSeries-Parallel Capacitors

Five paper capacitors are connected as shown. The capacitances are given in microfarads. Now, what is the capacitance between points A and B?

- T. A. Varhelyi

 

Over the River

Here is a puzzle that should increase aspirin sales:

You have a cable containing 100 wires running under a river. None of the wires are color-coded. The problem is to identify (using numbered labels or cable markers) the corresponding wires on both sides of the river. Only one round trip across the river is permitted and the only piece of equipment that may be used is a continuity checker (not an ohmmeter).

- Ronald J. Draus

 

Black Box No. 3 - RF CafeBlack Box No. 3

In the figure, note that at the left, one cell identical to the two cells at the right is used to power the black box. Further, note that the current drain of the black box in both cases is the same.

Question: What's in the black box?

- George R. Wisner

 

Automated Voting

Automated Voting - RF CafeTo solve the usual office problem of whether to have music or not, one boss installed a double-pole switch on his own desk and single-pole switches on the desks of each of his three men, John, Dick and Harry, as shown in the figure. When anyone wants music, he votes for it by closing the switch at his desk. Closing the boss' switch counts as two votes, while closing any of the other three switches counts as one vote. When there are three or more votes for music, the radio comes on.

Wire up the circuit. No additional parts are needed, and the switches need not be altered. Before deciding you have done the job right, check to make sure the radio will come on for any combination of three or more votes, and never for any combination of less than three votes.

- Darrell L. Geiger

 

----- Solutions Below -----


Quizzes from vintage electronics magazines such as Popular Electronics, Electronics-World, QST, Radio-Electronics, and Radio News were published over the years - some really simple and others not so simple. Robert P. Balin created most of the quizzes for Popular Electronics. This is a listing of all I have posted thus far.

RF Cafe Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Solutions

Series-Parallel Capacitors Solution - RF CafeSeries-Parallel Capacitors

This is a typical sample of the present-day right-angled drawing. (It can make any problem unnecessarily complicated.) Drawn in bridge form as shown here, it is easy to see that the 5-μf capacitor has no effect on the circuit, and that we have two parallel branches of two 2-μf capacitors in series with each other. Capacitance of each branch: 1-μf; total capacitance: 2-μf.

 

Over the River Solution - RF CafeOver the River

The solution shown is for 10 wires but it applies to any even number of wires greater than 4. - Start by tying wires into pairs except for four wires. Tie three of these together and leave one free. Label the pairs A-A B-B C-C; then label the set of three D-D-D. The remaining wire is then labeled 1. Move over to the other side. Find the free wire and label it 1. Find the pairs and label them 2-3, 4-5, 6-7; then label the set of three 8-9-10. Tie 1-2-10 together. Then tie the following wires into pairs 3-4, 5-6, 7-8; leave 9 free. Move back to the first side. Open all the wires that are tied together. Find the free wire and label it 9. Find the wires that are connected to No. 1. Label the D-wire 10 and the C-wire 2, then the other C-wire must be 3. Label it. Find the wire connected to No. 3 and label it 4. Since it is an A-wire, label the other A-wire 5. Find the wire connected to No. 5 and label it 6. Since it is a B-wire, label the other B-wire 7. Find the wire connected to No. 7 and label it 8. This completes wire identification.

 

Automated Voting Solution - RF CafeAutomated Voting

Now everyone gets to vote. Three employees can outvote the boss - when they want to listen to the radio, that is. Wonder how this system works when Dick wants a raise?

 

Black Box Mixup

Almost too simple. The black box merely contains a nice, neat short circuit. The cells, of course, both have equal internal resistances through which the current, I, flows. When two cells are in series, the resistance as well as the voltage doubles, leaving current I essentially constant.

It turns out that readers came up with other solutions to the Black Box problem. Their methods appeared in later issues as follows:

Black Box No. 3 Again (November 1961 issue)

That there may always be more than one solution to a problem is pointed out in a letter from two members of the Bell Telephone Laboratories staff:

"Since the illustration shows a two-terminal system we think we have a novel solution. The Black Box could consist of one field-effect varistor which exhibits the property of a constant current under conditions of variable voltage, provided E is greater than the pinchoff voltage and 2E is less than the breakdown voltage of the device.

Further information on the field-effect-varistor may be found in the January issue of the Proceedings of the IRE, pages 44 to 56." -R. P. Massey and S. V. Natale

[The original Black Box No.3 contained simply a dead short, so that the current was limited only by the internal resistance of the cells. Thus current remained the same whether one cell or two cells in series were connected to its input. -Editor]

Black Box No.5 (January 1962 issue)

(September)

Two ingenious solutions to Black Box No. 3 (September, 1961) have been submitted. If you remember, the puzzle was a black box with two terminals. First one dry cell was connected to it, then two dry cells. In each case, the current was the same. What was in the black box was actually a nice, solid short circuit. Therefore the current depended only on the internal resistance of the cells, which of course in-creased in proportion to the voltage (number of cells in series), keeping the current the same.

Reader Ray R. Stark suggests that the external cells might have been 2-volt storage cells, and that two dry cells could be contained in the black box. Then whether one wet cell at 2 volts was applied, or two wet cells at 4 volts, the difference in both cases would have been 1 volt, and the current the same, though in reverse direction.

Readers W. Kaune and J. Molberg of Seattle have another solution. In theirs, there is a resistor, R1, in parallel with a relay and a second resistor, R2. R2 and the relay coil together have exactly the same resistance as R1. Normally closed contacts of the relay connect the circuit through R1, and the relay is supposed to operate anywhere between I/2 and I. With one dry cell applied, the current divides between the two paths. With two dry cells, 2I starts to flow. This immediately actuates the relay, opening the circuit to R1. R2 and the relay now pass the same amount of current as one cell put through the two parallel channels.

 

 

Posted September 5

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