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This newest "What's Your EQ?" (Electronics Quotient) challenge appeared in the April 1967 issue of Radio-Electronics magazine. It only has two circuits to work out. The first is a fairly basic analysis problem where a voltage source and some resistors are connected. You need to solve for the value of one resistor, given the current through it. As it often the case, re-drawing the schematic to remove cross-connections clarifies the situation and makes the task much simpler. The other problem is a black box type, and it should not pose much of a problem. Not stated but implied is that the diodes within are ideal and do not have a junction voltage drop (sorry for the spoiler, but it really should have been included in the statement). Bon chance, viel Glück, 祝你好运, buena suerte, Удачи, καλή τύχη, がんばって, powodzenia, buona fortuna, and good luck.

Conducted by E. D. Clark

Two puzzlers for the student, theoretician and practical man. Simple? Double-check your answers before you say you've solved them. 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.

Network Problem

Given the above values, what's the value of RL? - Joseph P. Hallisey

Polarity Straightener

Within a certain voltage range, output volts equals input volts in the above box. No matter which way the input is polarized, however, output remains polarized as shown. Do you know what's in the box? - David A. Hinton

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.

Network Problem

First, redraw the circuit as a bridge, with RL temporarily removed. Now determine the Thévénin equivalent circuit.

I = Eoc / (Req + RL)

Req + RL = 2 / 0.1 = 20

Therefore, RL = 20 - 12 = 8 ohms.

Polarity Straightener

Here's the circuit.

Assume input is: 1-, 2+. D1 and D4 are forward-biased and allow conduction through the load so output is: 3 + and 4-. Reverse the input, so input is 1 + and 2-. Now D2 and D3 are forward-biased and allow conduction through the load for 3+ and 4-. When D1 and D4 are forward-biased, D2 and D3 are reverse-biased and therefore don't conduct. The same is true the other way around.

Posted April 23, 2024