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The greatest resistance is two bulbs in series. The equivalent resistance is series is R1+R2 while in parallel the resistance is R1R2/(R1 +R2). R1 +R2 is greater than R1R2/(R1+R2); e.g. if R1=R2 = R the series resistance is 2R and the parallel resistance is R/2.

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14y ago
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12y ago

RSeries = SummationI=1toN (RI)

Which means that if you have two bulbs in series of resistance x, then the effective resistance is 3x, while two bulbs would be 2x.

Yes, it is true that x would be slightly different, because of the temperature coefficient of the bulbs, but it is monotonically so, meaning that the resistance of 3 bulbs, while slightly less than 3x, would still be greater than 2x.

Practically, you can tell by placing an ammeter in series with the two or three bulbs, and you will see that the current is less with three bulbs. Assuming that the voltage is the same in both cases, you can simply calculate the resistance with Ohm's law as R = E / I.

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13y ago

If you had three bulbs in series with resistances of R1, R2 and R3 then RTotal = R1 + R2 + R3.

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Q: In a series circuit how the total resistance of several bulbs compare to the individual resistances?
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