May 1959 Popular Electronics
[Table of Contents]
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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)
Electronic Sticklers
These 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 doublechecking 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 015 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
?
Hal Carlson
4 Ned Work has a mesh of 1ohm
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 (i_{a}).
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 (i_{b})
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
(i_{a} + i_{b}) 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.
If you know of a tricky Electronic Stickler, send it in with
the solution to the editors of POPULAR ELECTRONICS. If it is
accepted, we will send you a $5 check. Write each Stickler you
would like to submit on the back of a postcard. Submit as many
postcards as you like but, please, just one Stickler per postcard.
Send to: POPULAR ELECTRONICS STICKLERS, One Park Ave., New York
16, N. Y. Sorry, but we will not be able to return unused Sticklers.
Posted February 7, 2013