April 1960 Popular Electronics
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Here is a nifty little
exercise that appeared in the April 1960 edition of Popular Electronics. It has 10 different light bulb circuits
and challenges you to figure out which bulb would burn the brightest. All are intuitively obvious to most of us
who have been in the field for decades, but do you remember how to do a circuit mesh analysis to prove your "gut,"
as the Donald would say? If you resort to building any of the circuits as a lab exercise, be sure to not use
incandescent bulbs greater than 60 watts or the green police could could be paying you a visit.
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ANSWERS: (explanations by Kirt Blattenberger)
When analyzing the circuits,
note that for many of the branches one side of the bulb returns directly to the battery, even though at first look
it might appear to share with another branch. In circuit 3, for example, you can redraw the circuit connecting the
left sides of bulbs A and D directly to the positive battery terminal (anode). Similarly, the right sides of bulbs
B and D and the bottom of bulb C directly to the negative battery terminal (cathode).
#1: C.
Bulb C
is the only one that has the full battery voltage across it. Bulbs A, B, and D are shorted across their terminals
and thus have no voltage across them.
#2: C.
Bulb C is the only one that has the full battery
voltage across it. Bulbs A and B share the battery voltage with bulb D.
#3: D.
Bulb D is the
only one that has the full battery voltage across it. Bulbs B and C share the battery voltage with bulb A.
#4: A.
Bulb A is the only one that has the full battery voltage across it. Bulbs B and C are shorted across its
terminals and therefore have zero volts across them.
#5: B.
This one is a bit tricky, as with
circuit #5. Bulbs A and B are in series and therefore share the battery voltage equally Bulb A is in series with
the parallel combination of bulbs D and E. The intent of the author is likely that you treat the bulbs as
resistors and therefore the parallel combination of bulbs D and E result in half the total resistance of a single
bulb. Therefore, the voltage across those two bulbs would be Vbatt * (0.5Rbulb / 1.5Rbulb) = 1/3*Rbulb, and the
voltage across bulb B would be 2/3Rbulb. For standard incandescent bulbs, the operational resistance is dependent
on the voltage across the bulb, so the voltage division would not be as clean-cut as that, but still bulb A would
be brighter. If the bulbs were neon types that maintain a fairly constant voltage across their terminals and
regulate current to maintain the voltage, then provided the battery voltage is high enough, all three bulbs would
burn with equal brightness.
#6: A.
This is essentially the same circuit as #4.
#7: E.
Bulb E is the only one that has the full battery voltage across it. Bulb B's terminals are shorted and thus 0
volts across it. Bulbs A and C share the battery voltage with Bulb D.
#8: A.
This one is a bit
tricky, as with circuit #5. Bulb B is shorted so it plays no role. Bulb A is in series with the parallel
combination of bulbs C and D. The intent of the author is likely that you treat the bulbs as resistors and
therefore the parallel combination of bulbs C and D result in half the total resistance of a single bulb.
Therefore, the voltage across those two bulbs would be Vbatt * (0.5Rbulb / 1.5Rbulb) = 1/3*Rbulb, and the voltage
across bulb A would be 2/3Rbulb.
#9: C.
Bulb C is the only one that has the full battery voltage
across it. Bulbs A and B are in series with each other and therefore share the battery voltage. The same goes for
bulbs D and E.
#10: E.
A quick mesh analysis, assuming the resistive model for the bulbs, shows
that the current flowing through bulb E is one unit of current while the current flowing through the other bulbs
is 1/2 unit of current. That means bulb E would be the brightest.
Posted 7/4/2012