May 1959 Popular Electronics
Here are a few more electronics conundrums with which to exercise the old noodle. These are puzzlers from a 1959 Popular Electronics magazine, but at least one of them (#4) will likely prove to be a real stickler unless you have seen a similar resistor mesh problem before. There are no tube circuits to use as an excuse for not getting them - just resistors, batteries, switches, meters, a motor, and a couple light bulbs. All four would be fair game to present to an interviewee to see where he/she stands on basic circuit analysis. (see April Electronic Sticklers)
May 1959 Popular Electronics
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Electronic SticklersThese four thought twisters are arranged in order of increasing difficulty
1 Happy Snap, having only an s.p.s.t. switch, wanted to turn on a floodlight and a pilot light on the control board at the same time. Expecting no trouble, he wired his setup as shown. After double-checking his connections, he held his breath and inserted the wall plug. Things didn't quite work out. Do you have any idea why?
-Robert L. Noland
2 Dewey Dubblecheck, who believes in making all measurements twice, connected a voltmeter and an ammeter to measure the power drawn in this circuit. Using the formula: W = E x I, he found that the motor drew 40 watts. He made the measurement again, this time using a standard wattmeter, and read only 30 watts. Dewey is puzzled - are you?
-Donald R. Wesson
3 Sam Addit made this simple computer to add any numbers from 1 to 6. The resistors were adjusted so that if battery #1 were switched in, the 0-15 voltmeter would read 1, switching in battery #2 would give a reading of 2, and so on. He figured that if a combination like 2 and 5 were used, the meter would read 7. What did it actually read ?
4 Ned Work has a mesh of 1-ohm resistors connected as shown and extending across his living room floor. Some day he hopes to extend the mesh all the way to infinity - and maybe even beyond. Can you calculate what the resistance will be between points "A" and "B" when his "tangled web" is finished?
-Roy S. Reichert & Gene Harris
Answers to Electronic Sticklers
1. With the switch off, the lights will be in series across 117 volts ±6 volts. Pilot light will pop first, then the floodlight will go out.
2. Dewey failed to consider power factor when he made his original measurement with a voltmeter and ammeter. The wattmeter automatically took power factor into consideration. In this circuit the power factor is 0.75.
3. The reading would be 5 because the batteries are connected in parallel. Actually, unless the resistors are very large in value, the meter will read some value between 2 and 5 due to the loop current set up in the parallel circuit.
4. Although it is not practical to construct an infinite mesh, you can solve this problem by using a variation of the constant current method for solving network problems.
Assume that a battery is connected to the mesh in such a way that one terminal of the battery is connected to point "A" and the other terminal is connected at infinity. The size and polarity of the battery is such that 1 ampere of current flows "into the paper" at point "A". Since the three resistors connected to point "A" are all equal (1 ohm) and the surrounding mesh is symmetrical, the current divides equally in the three branches. Hence, the current in the resistor between "A" and "B" is 1/3 ampere (ia).
Now connect a second battery in a similar fashion, only in this case, while one terminal again connects at infinity, the other terminal is connected to point "B". The size and polarity of this battery is such that 1 ampere of current flows "out of the paper" at "B". Again, for the same reason, the current divides equally. Hence, an additional 1/3 ampere (ib) flows through the resistor between "A" and "B" in the same direction as the current from the first battery. Since one terminal of each battery is connected at infinity, the two currents at this point are equal and opposite; therefore, they cancel. The infinite extremes of the mesh may be neglected.
It can be seen that a total current through the resistor (ia + ib) is 2/3 ampere. Since this resistor equals 1 ohm, the voltage drop across it will be 2/3 volt. It follows then that since 1 ampere of current flows into point "A" and out of point "B," and the voltage drop from "A" to "B" is 2/3 volt, the total mesh resistance is: R = E/1, or 2/3 volt/1 ampere, or 2/3 of an ohm.
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