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What's Your EQ?
November 1962 Radio-Electronics

November 1962 Radio-Electronics

November 1962 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.

Back in 1962 when these "What's Your EQ?" (EQ=Electronics Quotient, a la IQ) puzzlers appeared in Radio-Electronics magazine, very few people had any exposure to printed circuit boards. Most electronics assemblies were still using point-to-point connections, with component leads soldered to terminal lugs on other components (can capacitors, potentiometers, switches, lamps, fuse holders, etc.) or to terminal strips. "A Tracking Problem?," while being a cinch to us nowadays, was not such an intuitive task at the time. You will get it in mere seconds, but a technician or hobbyist in 1962 might have scratched his bean a little. "Capacitance Problem" is a run-of-the-mill first semester circuit analysis problem. Hint: Think in terms of "Q" (i.e., charge). "Distribution Problem" is a tad bit more work, but you can do it.

What's Your EQ?

"What's Your EQ?, November 1962 Radio-Electronics - RF CafeThree puzzlers for the student, theoretician and practical man. They may look simple, but double-check your answers before you say you've solved them. If you've got an interesting or unusual answer send it to us. We are especially interested in service stinkers or engineering stumpers on actual electronic equipment. We are getting so many letters we can't answer individual ones, but we'll print the more interesting solutions (the ones the original authors never thought of). We will pay $10 and up for each one accepted. Write EQ Editor, Radio-Electronics, 154 West 14th St., New York, N. Y.

Capacitance Problem - RF CafeCapacitance Problem

Two capacitors, a 20-μf and a 5-μf unit, are each charged to 200 volts, then connected in series so their voltages add up to 400 (see diagram). Switch S is then closed, putting a 500,000-ohm load across the series combination.

What will be the end condition of this circuit, and how long will it take to reach it? (For the purposes of this problem, an R-C circuit can be considered to reach its final condition in five time constants. )

- Walther Richter

 

A Tracking Problem? - RF CafeA Tracking Problem?

An early worker with printed-circuit boards was given the board shown in the diagram with three components to be mounted as shown, and a set of terminals as indicated. The trick is to layout leads to all of the components, in straightforward fashion, without jumping through to the other side of the board, or using any of the other tricks sometimes seen in printed circuits.

- Rene E. Pittet, Jr.

 

Distribution Problem - RF CafeDistribution Problem  

Here is a simple series-parallel circuit which might be found in practical work. Enough values are given to make it possible to discover the others. Do you think you can work it?

- Cpl. David B. Schulz

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

Answers to What's Your Eq?

Answers to puzzles on page 47

A Capacitance Problem

The resistance sees the series combination of the 20-μf and the 5-μf capacitor - in other words, 4 μf. The time constant is therefore 4 μf times 0.5 megohm, or 2 seconds. In 10 seconds, the voltage across the resistance - between points A and B - is practically zero. But how about the voltages across the individual capacitors? A 4-μf capacitance, charged to 400 volts, holds a charge of 1,600 microcoulombs. This is the charge that has flowed through the resistor during the discharge process. But the 20-μf capacitance held a charge of 200 x 20 = 4,000 microcoulombs at the beginning. After 1,600 have been taken out, it still holds 2,500. The voltage across its terminals is therefore 120, with the same polarity as at the start. The 5-μf capacitor originally held 5 x 200 = 1,000 microcoulombs; if 1,600 were displaced during the discharge, it first discharged to zero and then charged to the opposite polarity with 600 microcoulombs. This makes the voltage across it also 120, but with a polarity opposite to the original.

 

A Tracking Problem? Solution - RF CafeA Tracking Problem?

This puzzle was proposed (and solved) back in the days of Sam Lloyd, long before printed circuitry - or even radio - was though of. The solution:

 

Distribution Problem

E1 = R1 (I1 + I2)

= 26 (0.8 + I2)

= 20.8 + 26I2

E2 = 40I2

E1 + E2 = 100

20.8 + 26I2 + 40I2 = 100

20.8 + 66I2 = 100

66I2 = 79.2

I2 = 1.2 amps

From here on the problem is easy, and can be solved with Ohm's law:

E = 100 volts

E1 = 52 volts

E2 + E3 = 48 volts

I1 = 0.8 amps

I2 = 1.2 amps

R1 = 26 ohms

R2 = 60 ohms

R3 = 40 ohms

 
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About RF Cafe

Kirt Blattenberger - RF Cafe Webmaster

Copyright: 1996 - 2025
Webmaster: Kirt Blattenberger,
   BSEE - KB3UON - AMA92498

RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The World Wide Web (Internet) was largely an unknown entity at the time and bandwidth was a scarce commodity. Dial-up modems blazed along at 14.4 kbps while tying up your telephone line, and a lady's voice announced "You've Got Mail" when a new message arrived...

Copyright  1996 - 2026

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