In a lightbulb circuit with a constant voltage, what is the effect on current if the no. of lightbulb is doubl

Answer 1

Depends how the new lightbulbs are hooked up (and the originals, for that matter).

If we assume all the lightbulbs are in series, the current will be halved because the resistance is doubled. This is most likely the answer you are looking for.

If the new lightbulbs are wired together the same way as the first set, and in parallel with the first set, then the resistance will be halved and the current will be doubled. There are more compliacted scenarios possible - for instance, if we had four lightbulbs in series, and we add another four wired as two sets in parallel, each of two bulbs in series - and even then we can wire this in series with the original set or in parallel with it.

For that matter, I'm assuming all the lightbulbs have the same resistance, which may not be true.

Answer 2

no longer understanding no matter if the bulb is further in series with yet another bulb or in parallel that is not person-friendly. If the bulb is further in parallel, extra skill may be ate up, extra cutting-edge further that could want to diminish the voltage via the bulbs truly lowering their resistance some. the hotter a filament receives, the better its resistance. If a bulb is further in series with yet another bulb, the voltage on both may shrink, lowering their resistance some yet increasing the series resistance for this reason lowering the skill ate up and the amps dropped on the circuit. no longer understanding the output resistance of the voltage source makes it not person-friendly to foretell how a lot the voltage may upward push given a help in cutting-edge.

Answer 3

The current is divided by the same factor. Is the half than the original.

Answer 4

V=IR, and a light bulb is just an R. The two Rs are additive, so V=I*(R+R).

Answer 5

well depend whether they are in series or parallel,
if series, not change in current but parallel, current is halved