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What's Your EQ?
October 1964 Radio-Electronics

October 1964 Radio-Electronics

October 1964 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.

What's Your EQ?, June 1963 Radio-Electronics - RF Cafe

Conducted by E. D. Clark

Three puzzlers for the students, 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.

Answers to this month's puzzles are on page 99.

Step Up to the Counter

In a step counter with a square pulse input of 100 volts peak from a square pulse generator, there are two capacitors and two diodes (see schematic). C2 = 9 x C1. The square pulse generator has zero impedance when the output voltage is zero. Voltage drop across the diodes is negligible when they are conducting. When the output is taken across capacitor C2, what will the output voltage and waveform be for two input pulses? -David Levine

Bypaths and Bypasses

Problem: Find current through M, and also find voltage "e". -E. D. Clark

Phase Angle

An inductive reactance of 8 ohms is connected in parallel with a capacitive reactance of 18 ohms. This combination is then connected in series with a variable resistor. For what value of resistance will the power factor be 0.5? -Joe Penny


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.

RF Cafe Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Answers to What's Your Eq?

This month's puzzles are on page 48

Step Up to the Counter

Since the voltages across capacitors in series are inversely proportional to the values of the capacitors,

EC1 = Ein ( C2/(C1 + C2) ) 

(Assume C1 = 1μf and C2 = 9μf)

So, EC2 = 90 volts and

EC2 = 10 volts.

Since the output is taken across C2, it is 10 volts. When the input voltage drops to zero after the first pulse is applied, C1 discharges through V2 and the generator. C2 cannot discharge because the cathode of V1 is positive with respect to the plate.

When the second pulse is applied to the input, we have only 90 volts applied, because the capacitor C2 is charged 10 volts and bucking the 100 volts.

The same principle applies for the second pulse as for the first. We have 81 volts across C1 and the remaining 9 volts across C2. So the output will now be 19 volts, since the 9 volts is added to the initial 10.

When the input drops to zero again, C1 discharges while C2 keeps its full charge. As more pulses are applied, the output keeps increasing in smaller steps.

Bypaths and Bypasses

IMAGES HERE

Phase Angle

The trick here is to assume a voltage, say 10, across the parallel branch to find the equivalent current and reactance.

IL = 10/8 = 1.25 amp,

and IC = 10/18 = 0.556 amp

As the currents are 180° out of phase, the equivalent current equals their difference, or 0.694 amp. Again using our assumed voltage, the equivalent reactance is found.

XL = 10/0.694 = 14.4 ohms

P.F. = 0.5 = cos θ;  θ= 60°

and tan θ = 1.732 = XL/R

R = 14.4/1.732 = 8.3 ohms

 

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