What's Your EQ?
March 1963 Radio-Electronics

March 1963 Radio-Electronics [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 your thinking caps once again. No person is credited for these three "What's Your EQ?" challenges from the March 1963 issues of Radio−Electronics magazine. In other instances (see below) it is E. D. Clark. BTW, EQ stands for Electronics Quotient. My solution for the "Four−Bulb Puzzler" was different from the one proposed. Given that semiconductor diodes were available at the time, I put a 6.2−V zener diode (a common value at the time) across each bulb, figuring if the bulb burnt out (open-circuited), the zener would drop the voltage normally dropped by the 6 V bulb. While the bulb is operating, the voltage across the zener would not turn it on. Of course there could be a problem with the current through the diode being too high when turned on, without a limiting resistor. "Simple Circuit?" would take more time than I am willing to expend right now. With "Ohms, Sweet Ohms" don't let the wording of the proposition bias (pun intended) your analysis. Bon chance!

What's Your EQ?

Three 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.

Answers for this month's puzzlers are at bottom of page.

Four-Bulb Puzzler

Each black box in the diagram contains a single electronic component, connected in parallel with a 6-volt bulb. If one bulb burns out, the others remain lighted - the black box across the extinguished bulb carries the load and the total current is reduced slightly. What is in each black box? (The black boxes do not contain any of the following: fixed resistor; capacitor; relay or other electromechanical device.)

- Kendall Collins

Simple Circuit?

In the circuit shown, an alternating voltage E of 100 rms, 60 cycles, is connected to two capacitors, C1 and C2, in series. First, let C1 and C2 be equal, each having a capacitance of 2 μf. The series combination has then a value of 1 μf, which at 60 cycles will have a reactance of 2,650 ohms. The alternating current flowing in the circuit will be approximately 38 ma.

Closing switches S1 and S2 places diodes D1 and D2 across capacitors C1 and C2, respectively. Note that the diodes are connected back to back!

1. If S1 and S2 are both closed, will this change the reactance of the circuit, compared to the case when the switches are open? In other words, will the alternating current change?

2. What will a high-resistance DC voltmeter read, when placed across C1 and then across C2?

3. If only one of the switches, say S1, is closed, what will be the effect on the reactance - that is, on the alternating current?

4. And what will the DC voltages be across C1 and C2 with only S1 closed?

Better not be too sure that you know all about such a simple circuit! It might be smart to set up the circuit and measure the values. Chances are you won't believe what you find!

- Walter Richter

Ohms, Sweet Ohms

Didn't know resistors could amplify, eh? Well, watch. With the switch in position A (Fig. 1), 10 volts will be applied across the 10-ohm resistor, giving us a current of 1 ampere. The power by P = EI will be 10 x 1 = 10 watts. When we switch to position B, 1 volt across 10 ohms will give us 0.1 amp, and a power of 1 x 0.1 = 0.1 watt. Now let's find the power when both batteries are used (Fig. 2).

Eleven volts across 10 ohms gives us a current of 1.1 amps, with the resulting power of 11 x 1.1 = 12.1 watts. But hold it! The 10-volt battery delivers 10 watts, and the 1-volt battery 0.1 watts, making for a total of 10.1 watts. Yet the combination gave us 12.1 watts! Where did the 2 watts come from?

- John Timar, Jr.

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.

These are the answers.

Four-Bulb Puzzler

Each black box contains a thermistor. The thermistors are not appreciably heated by the voltage drop across a lighted bulb. When a bulb burns out, the voltage drop across the thermistor is increased and its temperature increases. This brings its resistance to a value suitable for replacing the open bulb filament.

Simple circuit?

1. Reactance of two capacitors, each of 2 μf, in series, is 2,650 ohms. This value will not be affected by the presence of the diodes. The alternating current will be 38 ma, with or without the diodes.

2. 71 volts, with C positive with respect to A or B.

3. and 4. One diode can be left out of the circuit without changing the statements under 1 and 2.

Ohms, sweet ohms

But wattage = E x I, and in the second situation both E and I are changed! The 10-volt battery that delivered 1 amp per cell is delivering 1.1 amp, or 11 watts, and the current from the 1-volt cell is also 1.1 instead of 0.1 amp, giving a wattage of 11 x 1.1. Total: 12.1 watts.

Posted May 11, 2023