April 1960 Popular Electronics
[Table
of Contents]
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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.
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 cleancut 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 July 4, 2012