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What's Your EQ?
May 1962 Radio-Electronics

May 1962 Radio-Electronics

May 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.

Time to put on the thinking cap again for three more "What's Your EQ?" circuit challenges, compliments of Radio-Electronics magazine in May 1962. The first is a classic "black box" type problem which, from reading its description, involves some sort of resonant circuit. that's all I'll say on that. The next, called "An Easy One?" should, by the way it is drawn, be a clue that it might be easier to solve if you re-draw it to make a familiar-looking circuit. Hint: Summons the spirit of Sir Charles Wheatstone. Just the name of the last one, "Iterative Network," is enough to induce a cold sweat. As with most of these "What's Your EQ?" problems, successful completion of a first year college circuits course is plenty to get through them. A few are better attempted by people with hands-on experience troubleshooting circuits, but don't let that scare you off.

What's Your EQ?

It's stumper time again. Here are three little beauties that will give you a run for the money. They may look simple, but double-check your answers before you say you've solved them. For those that get stuck, or think that it just can't be done, see the answers next month. If you've got an interesting or unusual answer send it to us. 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). Also, we're in the market for puzzlers and will pay $10 and up for each one accepted. Write to EQ Editor, Radio-Electronics, 154 West 14th St., New York, N. Y.

Variable-Current Black Box

This particular Black Box has two terminals. When 50 volts dc is applied, 2 amperes flows. If 100 volts at 60 cycles is the supply source, the current is 12 amperes and the power 1,200 watts. With 151 volts at 400 cycles, the current is 10 amperes.

Draw the circuit and give the values of the components in the Black Box.

- Frank A. Lopez

 

What's Your EQ?, May 1962 Radio-Electronics - RF CafeAn Easy One?

Some readers have complained that many of the problems require calculus - or at least quadratics - to figure out. Here is one that can be done without calculation. Observe the circuit pattern and determine the voltage across the 1-ohm resistor.

- Albert C. W. Saunders

 

Iterative Network - RF CafeIterative Network

Engineers will recognize this problem as the very simplest type of iterative network, but it will be interesting and useful to beginners. Good training for transmission lines, too!

The circuit shown above consists of 1-ohm resistors. The lattice extends to infinity. What is the total resistance of R, looking into points A and B?

- Mordehai Arditti.


Quizzes from vintage electronics magazines such as Popular Electronics, Electronics-World, QST, 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

 

May Solutions

 

Variable-Current Black Box - RF CafeVariable-Current Black Box

The circuit is as shown. To solve, first calculate the dc, which must be passing through a resistor of 25 ohms. Since the voltage-current product at 60 cycles equals the power, a series-resonant circuit is suggested. The current at 400 cycles permits the reactance to be calculated. Then the inductance, capacitance and resistance in the ac circuit can be found.

 

Wheatstone Bridge circuit - RF CafeAn Easy One

If we layout the resistors in the pattern shown in the sketch, it becomes an obvious balanced bridge. So the drop across the 1-ohm resistor is zero.

 

parallel combination of R and 1 ohm - RF CafeIterative Network

We can make this circuit (Fig. 1) more easily understandable by breaking it just after the first section and labeling the new inputs C and D. Since the lattice extends to infinity, we also see resistance R at points C and D. Now we can draw an equivalent circuit as in Fig. 2. This is a simple series-parallel combination which we can solve.

R= 1 + 1 + R/(R + 1)

The last term is the parallel combination of R and 1 ohm. Multiplying both sides of the equation (by R + 1 to get rid of the fraction, we get:

R x (R + 1) = 2R + R + 2,

which is equivalent to:

R2 - 2R -2 = 0.

This works out to:

R = 1 ± (3)1/2 = 2.735 ohms. The negative solution is an entirely different story, and not applicable to our present problem.

 

 

Posted June 4, 2024

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