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What's Your EQ?
July 1962 Radio-Electronics

July 1962 Radio-Electronics

July 1962 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.

Another triplet of electronics circuit challenges here for you from a 1962 issue of Radio-Electronics magazine - a resistor network, a voltage divider, and a delayed switching circuit. As is usually the case with the resistor network, rearranging the circuit branches into a more familiar configuration simplifies its analysis. That is a good general rule when analyzing any circuit - rearrange and break down into recognizable subcircuits. The series circuit is about as simple as it gets, but knowing about how an ammeter is constructed helps. The Delayed Switching challenge is a little misleading in its description since the proposed solutions does use a vacuum tube duo diode (aka dual diode or full-wave diode) as part of the power supply circuit, although it is not part of the actual indicator circuit requested. This type of circuit is not required for a solid state power supply.

What's Your EQ?

What's Your EQ?, February 1962 Radio-Electronics - RF CafeIt's stumper time again. Here are three little beauties that will give you a run for the money. They may look simple, but double-check your answers before you say you've solved them. For those that get stuck, or think that it just can't be done, see the answers next month. If you've got an interesting or unusual answer send it to us. 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). Also, we're in the market for puzzlers and will pay $10 and up for each one accepted. Write to EQ Editor, Radio-Electronics, 154 West 14 St., New York, N. Y.

 

What does the voltmeter read? - RF CafeResistance Network

What does the voltmeter read?

- John Capobianco

 

Delayed Switching

A high-power audio amplifier has separate power and standby switches. The problem is to devise a circuit using two pilot lights - no tubes, relays or other moving parts - to indicate when the filaments are hot enough to switch on the plus-B. (The pilot lights used were neons.)

- Juan L. Guthman

 

A Series Circuit - RF CafeA Series Circuit

In the simple series circuit shown, a vtvm connected from ground to point A measures 66 volts. An ammeter with an internal resistance of 10 ohms reads 3 amperes when connected from ground to the same point. What is the ohmage of each resistor?

- Sgt. R. M. Rasch


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

Solutions

The voltmeter reads zero - RF CafeResistance Network

The voltmeter reads zero - the circuit is actually a balanced bridge, as can be seen in the drawing. The terminals of the voltmeter are at equal voltage points, and the 12-and 20-ohm resistors are simply shunted across the meter.

 

VTVM does not load the circuit - RF CafeDelayed Switching

When S1 is switched on, PL1 lights. S2 is shown in standby position. R2, R3 and C (a high-quality paper capacitor) are chosen so that the time taken to charge C to the firing voltage of PL2 is that necessary to warm the amplifier's filaments. After that, PL2 will keep on flashing, indicating standby ready. When S2 is switched on, PL2 is placed across the B-plus supply and will stay lit continuously, indicating "amplifier on." (If expedient, the auxiliary rectifier network can be used as negative bias source.)

 

A Series Circuit

Since the vtvm does not load the circuit, the rating of 66 volts indicates that R1 is four times larger than R2. The 10-ohm resistance of the ammeter loads the circuit so a series parallel circuit is formed.

Then the 3 amperes through the 10-ohm meter indicates a drop of 30 volts across the meter and the parallel resistor R2. The current through R2 could be written as 30/R2. The 300-volt drop across R1 can be written 300/R1 or 300/4R2, since R1 equals 4R2. The current through R1 is also equal to the sum of the current through R2 and the meter, so we have the equation:

300/4R2 = 30R2 + 3. Putting the right Side over a common denominator, we have:

300/4R2 = (30 + 3R2)/R2

Reducing the left side gives us:

75/R2 =(30 + 3R2)/R2

Multiplying both sides by R2, we have 75 = 30 + 3R2, or 3R2 = 45. R2 = 15 ohms, and R1 =60 ohms.

 

Axiom Test Equipment - RF Cafe


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Copper Mountain Technologies (VNA) - RF Cafe