May 1959 Popular Electronics

## May 1959 Popular ElectronicsTable of ContentsPeople old and young enjoy waxing nostalgic about and learning some of the history of early electronics. Popular
Electronics was published from October 1954 through April 1985. All copyrights are hereby acknowledged. See all articles from |

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)

These four thought twisters are arranged in order of increasing difficulty.

-Robert L. Noland

-Donald R. Wesson

-Hal Carlson

-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