What's Your EQ?
August 1964 Radio-Electronics

August 1964 Radio-Electronics [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.

Here are a couple electronics circuit analysis problems to prime you for the week ahead. They are from the August 1964 "What's Your EQ?" challenge in Radio-Electronics magazine. EQ, by the way, is for Electronics Quotient, as in IQ (Intelligence Quotient). Some others are in the list below. In the Autotransformer problem, I made the assumption that the secondary tap was actually in the middle (as drawn), so that there was an equal number of turns above and below the tap. That proved to be a good assumption since it validated my answer (not difficult if you know the basics of autotransformers). The other problem, "Case of the Lost Energy," is a variation on similar ones containing a potentially (pun intended) non-intuitive missing energy being stored in capacitors. I'll admit to not having worked through it yet. You'll probably figure it out in no time.

What's Your EQ?

Conducted by E. D. Clark

Two puzzlers for the students, theoretician and practical man. Simple? Double-cheek 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 60.

Autotransformer

Can you determine the reading on ammeter A1 and ammeter A3?

- Kendall Collins

Case of the Lost Energy

Here's a capacitor problem to test your ability in analyzing "simple" circuits:

Two one-mike capacitors, and a switch connecting them. The left capacitor has been charged up to 100 volts; the right one is unchanged. The switch is then closed.  

Since we have not changed the total amount of charge, and since the theory book tells us that:

Charge = Farads x Volts (Q = CE) the left capacitor will lose half of its charge to the right one. The voltage will now be 50, since the capacitance has doubled. The circuit now consists of a 2-μf capacitor charged to a potential of 50 volts.

So far, so good. Now once again, we refer to our theory book and find the formula for the energy stored in a capacitor:

Energy = ½ x Farads x Volts²

Let's try some numbers:

Originally,

Energy = ½ x 1 μf x 100² + ½ x 1 μf x 0²

= 2,500 μJoules

= 0.005 Joule

Finally,

Energy = ½ x 2 μf x 50²

= 2,500 μJoule

= 0.0025 Joule!

We lost half the energy when we threw the switch! But a capacitor can only store energy, not dissipate it. Also, our theory book is a firm believer in conservation of energy.

Now then, simply, the problem:

Where did the energy go?

((Assume that the switch is perfect.)

- Donald E. Lancaster

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.

Answers to This month's puzzles are on page 47

Autotransformer

Ammeters A1 and A3 each read 10 amperes. The primary and secondary currents in an autotransformer are 180° out of phase (in phase opposition). They tend to cancel in the part of the primary winding which includes the secondary winding.

Taking into consideration a winding ratio of 2 to 1, and a secondary current of 20 amperes, the apparent primary current (and hence the apparent reading on A1 and A3) would be 10 amperes. Since the primary and secondary are 180° out of phase, the total primary current is the algebraic sum of the two, or 10 amperes.

Note: Core and winding losses are disregarded.

Case of the Lost Energy

This is really not as simple a problem as it looks. Suppose our problem circuit looked like this instead.

Now, for a time after closing the switch, current will flow through the resistor until the capacitor charges have become equal. Here in this resistor is a good place to get rid of energy, turning it into heat. Since we must conserve energy, the heat energy produced by the resistor evidently must be equal to the energy lost by the capacitors. Now what if we change R to a different value? If R gets bigger, the current flows for a longer time, but is weaker. If R gets smaller, the current flows for a shorter time, but is stronger. The energy dissipated in the resistor is independent of the value of the resistance and exactly equal to the difference between the initial and final values of capacitor energy.

Because of this, if we have any resistance in the circuit at all, we have explained where the energy went. In any practical problem, the small circuit lead resistance would heat up and dissipate this energy. Since most energy values normally found in capacitor circuits are generally very small, this heating effect is not very noticeable. As an example, a 25-watt light bulb in 1 second dissipates or expends 25 joules of energy, or 10,000 times as much energy as that left in our capacitor problem!

This explains any practical problem. But what if there were absolutely no resistance in the circuit at all? Then there would be another way out of the problem. Near absolute zero (-460°F), we may have zero circuit resistance. But always, no matter what the temperature, we must have some lead inductance. Let's draw this into the circuit.

But this is a resonant circuit! It will oscillate. If it oscillates, it will radiate radio-frequency energy. And, the energy it radiates will be exactly equal to the difference between the initial and final energy in the circuit.

Any reasonable value of lead resistance will damp this circuit and it will not oscillate, so the resistance "wins" if it has half a chance.

Doodles in May 1964

The scope trace in the May 1964 issue can also be produced by quickly moving the HORIZONTAL POSITION knob when the same frequency is put into both horizontal and vertical inputs, out of phase so as to produce a circular Lissajous figure. By noting whether the cusps are up or down, you can figure out whether the spot is moving clockwise or counterclockwise. Thus you can tell which input, horizontal or vertical, is leading and which is lagging. I have generated this pattern for the purpose many times. -Paul Penfield, Jr.

Posted November 13, 2023