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When we connect 2 light bulbs to a voltage source in parallel, the equivalent Resistance of the bulbs is R/2 (1/Req = 1/R + 1/R). If we connect these lights in parallel we have Req = 2R;

I= V/Req is valid for both circuits, and P=I*V (with V being the same for circuit 1 and 2) This means the parallel circuit produces more light.

How does this change for a circuit with a current source instead of a voltage source? Which of these 2 is brightest with a current source?

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The quick answer is: The series connection produces more light. This can be determined with simple reasoning as follows:

(1) The current is fixed and

(2) Series connected elements have the same current through them

For the parallel connection, each bulb gets half the source current while, for the series connection, each bulb gets the full source current. Since power is proportional to the square of the current, each bulb in the parallel combination produces 1/4 the light of the bulbs in the series combination.

Also, if you understand that current is dual to voltage, resistance is dual to conductance, and parallel is dual to series, you can immediately get the answer by duality.

For example, consider the following:

When connected to a voltage source, two light bulbs in parallel produce 4 times as much light than two light bulbs in series.

Here's the dual:

When connected to a current source, two light bulbs in series produce 4 times as much light than two light bulb in parallel.

Since the first sentence is true, and since the second sentence is its dual, the second sentence is also true.

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+1 for invoking duality. Being alert to the idea of duality from math made learning circuits so much easier. –  Jamie Banks Jun 24 '12 at 19:23
Thanks Katie, I've found that it makes teaching circuits easier too! –  Alfred Centauri Jun 24 '12 at 19:42
Thank you very much! This really made things clear to me! –  user10088 Jun 24 '12 at 21:57
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