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Seeing a reader name of Qutaiba Bassim El-Dhuwaib would not seem unusual in one of today's technical magazines, but having it appear in a 1964 issue of Radio-Electronics was definitely a rarity. Such was the case with this first "Black Box" circuit challenge submitted by the aforementioned subscriber. I took a WAG at the answer and got it right, but without considering the theory behind it. Fortunately, Mr. El-Dhuwaib provides that gory detail for us. It is a pretty clever scheme for effecting a phase shift while not affecting the magnitude. Such circuits were probably more intuitive to designers in the days of analog. The second mystery circuit requires being familiar with types of meter movements commonly used in DC and AC voltmeters. Having begun in the craft prior to the advent of digital multimeters (that includes me) helps. If you have never dealt with vacuum tube circuits, you probably won't have much luck with the last circuit. I worked on a couple tube-based radars whilst in the USAF in the late 1970s to early 1980s, and managed to remember enough about the operation to guess the answer correctly.

Always Half

An ac generator is connected to a network consisting of two resistors, R and RV, and the black box shown in the diagram. It is found that the voltage between A or B is independent of the value of RV. Whether RV is shorted or open, the voltage across AB is E/2 where E is the voltage of the generator. It is also independent of the frequency. Why?

- Qutaiba Bassim El-Dhuwaib

What Voltage?

Three voltmeters are connected as shown. V1 is an electrostatic type, calibrated to read peak voltage values. It indicates that the output from the rectifier is 1,000 peak volts. V2 is a dc voltmeter of the d'Arsonval type. V3 is an ac voltmeter of the electrodynamometer or iron-vane type. What will be the voltage indicated by voltmeters V2 and V3?

- Kendall Collins

A Distorted Puzzler

This could be any voltage amplifier stage - most likely audio. Could also be a triode. Signal is getting through, but with less than usual volume and with considerable distortion. The tube is good, the voltages are as shown. The components? Well, better check 'em. Won't have to disconnect anything to find any bad ones, though. Just look.

- Jack Darr

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.

These are the answers. This month's puzzles are on, page 37. If you have an interesting or unusual puzzle (with an answer) send it to us. We will pay \$10 for each one accepted. We're especially interested in service stinkers or engineering stumpers on actual electronic equipment. We get so many letters we can't answer individual ones, but we'll print the more interesting solutions - ones the original authors never thought of. Write EQ Editor, Radio-Electronics, 154 West 14th Street, New York, N. Y. - 10011.

Fig. 1 - Black Box contents.

Fig. 2 - Black Box vector voltages

Always Half

The black box contains a capacitor and a resistor equal to R as in Fig. 1. The voltage ECA = EAD = E/2 having zero angle with E; the voltage EBC is 90° out of phase with EBD, and the vector sum is equal to E: consider the modified vector diagram (Fig. 2). EBC is always perpendicular to EBD, thus point B will fall on the circumference of a circle having a diameter of E or a radius of E/2. Likewise, as point A falls in the center of this circle, the distance between A and B will become E/2. Whatever the resistance, only the phase shift will change.

This circuit is used often as a phase shifter to change phase without changing amplitude.

Note: This puzzle, in slightly different form, was published as "Output Voltage" by Cameron McCulloch in the October 1963 issue and evoked a flood of disbelieving letters as well as criticisms of an attempted abridged mathematical proof (Mr. McCulloch's proof was much longer than the one printed).

Readers may be assured that the circuit is "bench-tested," that it works, and is indeed so ruggedly based in theory that an unfortunate change of sign in one of the terms of the October explanation did not change the final result.

What Voltage?

The output from the full-wave rectifier consists of negative half-cycles at the rate of 800 per second. Therefore the output is negative with respect to the center tap on the transformer secondary.

Unless otherwise stated, dc voltmeter readings should be taken as average values, and ac voltmeter readings should be taken as rms values. Observing polarity, the dc voltmeter (V2) reading is 636 volts, determined by multiplying 0.636 times the peak value of 1,000 volts. The ac voltmeter (V3) reading is 707 volts, determined by multiplying 0.707 times the peak value of 1,000 volts.

If ac voltmeter V3 is a basic d'Arsonval type, calibrated with and using a half-wave rectifier, the indicated voltage will be entirely different. This method is used in some vom instruments. Again observing polarity, the meter would indicate 1,414 volts, or double the rms value, in the forward position and zero volts with the test leads reversed. The output waveform of the circuit shown in the puzzler, along with the relative voltages, is shown in the diagram. The 0.636 factor applies to the average value of a sine wave and also to the output of an unfiltered full-wave rectifier.

A Distorted Puzzler

A simple one, and not uncommon. The key is, of course, that 6 volts positive at the grid, which causes a high plate current and a low plate voltage. Pretty obviously, the 0.01-μf input coupling capacitor is leaking badly.

Posted July 5, 2024