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April 1959 Popular Electronics[Table of Contents]People 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 Popular Electronics. 
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. Redrawing 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 thoughttwisters are arranged in order of increasing difficulty.
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.
Dennis Wroblewski
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 1/2volt 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?
David Borenstein
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 blownout component. Any idea why?
Ronald Wilensky
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 vacuumtube 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 3ohm resistor is an inactive component and can be omitted from the circuit. All that remains are two series resistance circuits in parallel.
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 halfcycle, 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 voltagedoubler power supply design.
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Posted November 12, 2013