April 1959 Popular Electronics
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Here is another of the Electronic Sticklers challenges from
Popular Electronics (see May
Electronic Sticklers). These are fairly basic circuit analysis
problems that often can be solved by inspection, but sometimes
a pencil and paper are necessary. Re-drawing the circuit in
a different configuration to make the connections more obvious
often helps when solving total resistance, capacitance, inductance,
etc., as in question #1. In this case, though, you need to be
able to recognize a common configuration to be able to simplify
the circuit; otherwise, you'll be writing and solving mesh equations.
#2 has a simple answer and a more elaborate possibility. #3
and #4 are simple inspection problems.
These four thought-twisters are arranged in order of increasing
1) Harvey Matrix discovered this network covered with solder,
in his junk box (above). Rather than trust his ancient ohmmeter,
he tried unsuccessfully to compute the resistance mathematically.
Show Harvey you are slicker and quicker by solving the problem
in one minute.
2) Joe Whatsit had a black box with only two terminals showing
(above). To find out what was in the box, Joe connected a 1
1/2-volt dry cell to the two terminals and noted the current
flow. He then connected a second identical dry cell in series
with the first cell and repeated the experiment. The same current
was noted. With this information, Joe figured out what was in
the box. Can you?
3) Mr. Pennypincher, in order to save money on batteries
for his portable radio, built this little voltage supply (to
the right) to substitute for the batteries. When he plugged
in the unit, he was running a risk of a blown-out component.
Any idea why?
4) With the setup shown (to the left), a diode in series
with a capacitor, Harold Tinkertoy applied 100 volts r.m.s.
across the circuit. Then he used his vacuum-tube voltmeter to
measure the peak voltage across the diode. How much did he measure?
To make the problem easy, assume that a sine wave is applied
to the circuit.
--Louis E. Garner, Jr.
Answers to Electronic: Sticklers
1. Two ohms. Redraw the network schematic in the form of
a bridge circuit. You will find that the resistance values of
the bridge legs result in a balanced bridge. Hence the 3-ohm
resistor is an inactive component and can be omitted from the
circuit. All that remains are two series resistance circuits
2. A short circuit. The current remains the same since the
two dry cells provide not only twice the voltage but also twice
the internal resistance. (Kirt note: It could also contain a
constant current circuit)
3. Since there is no surge resistor, the charging current
of the capacitor would pull too much current through the rectifier.
Without the surge resistor (20 cents), there's a good chance
the rectifier ($1.00) would burn out.
4. About 282 volts! How come? On one half-cycle, when the
diode's plate is positive, the capacitor charges to peak line
voltage... or about 141 volts (1.41 multiplied by line voltage).
On the next half cycle, the capacitor's voltage is in series
with the peak line voltage and thus adds to it... and 141 plus
141 equals 282! This arrangement, incidentally, is basic to
voltage-doubler power supply design.
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Posted November 12, 2013