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"OK, class, we're having a pop quiz today." Remember those frightening words? There's always at least one person who is prepared for a test at any time, but I'm not one of them. My blood ran cold when I heard that, even if I thought I had a good grip on the material. Stupid has a way of manifesting itself at such time. Anyway, here are three more circuit puzzlers from the September 1962 issue of Radio-Electronics magazine. The "Audio Service Stinker" will likely be lost on modern audiences. You will need more information than what is provided in the illustration, like the full pinout of the 12AU7 vacuum. Technicians of the era would know what those standard pins would be without having to look it up. I managed a lucky guess at what caused the heating issue in the relay circuit. The capacitor cube is a variation of the old resistor cube challenge. As always, when you see a circuit like this, the first thing to do is determine whether there is a way to re-draw it to make combining values easier. Remember that parallel capacitors combine like series resistors, and vice versa.

Industrial Problem

This one actually happened to me. The diagram shows part of a sequencing circuit. When the relay operated, the resistor dissipated excessive power - more than 10 times its 1.5-watt rating. The capacitor, resistor and line voltage were all checked and found OK. Meters were connected to measure the current through and the voltage across the 30-ohm resistor. These read close to the expected values of 2.20 ma and 6.6 volts. What explains the excessive heating?

- David T. Smith

Audio Service Stinker

A Heathkit WA-P2 preamp came in with the complaint of lower than normal gain. Measurements showed no voltage on pins 7 and 8 of V1 (see schematic). All resistors and capacitors tested (and were later proved) OK; all voltages on other elements in the preamplifier were within tolerances; the tube was good. What was wrong?

- Eric Leslie

Capacitive Cube

As shown, 12 capacitors are connected in a cubical framework. Each has a capacitance of 5 μf. Determine the equivalent capacitance between opposite corners of the network - say between A and G.

- Yak Chiang Yuen

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.

Solutions

Industrial Problem

The relay contacts were chattering at 120 cycles, and making contact at instances when the supply voltage was away from zero. This produced short pulses of extremely high peak current. These peaks were not measured on the meter, because of their short duration, but contributed to the total power, which was more truthfully indicated by the resistor than by the more sophisticated instruments.

Audio Service Stinker

Since the WA-P2 preamplifier has a 6-volt heater supply, the heaters for the two triode sections are in parallel (pins 4-5-9). There was a cold solder joint on pin 5, with the result that the heater in that triode section did not light.

Capacitive Cube

Junctions B, D and E are all at the same potential, as are F, C and H. Hence the network can be redrawn as shown. We now have, in effect, three capacitors, one of 15 μf, one of 30 μf and another 15 μf, in series, calculating the total capacitance by the easiest formula to use in this case:

1/C = 1/15 + 1/30 + 1/15, we find that C = 30/5, or 6 μf

Posted August 14, 2024