maha.4646
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can someone plz tell me why adding more bulbs in parallel circuit decreases the voltage?
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username4444284
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More bulbs in parallel would mean adding more paths with components. The charges have equal chance to reach the components hence energy carried is divided equally among the components added in parallel.
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meek-boy
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maha.4646
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thansk for heplping but I still dont undertand why the voltage decreases as more bulbs are added in parallel.
I have my exam on monday and i`ve already done my practical on this but the results I got showed that with 1 bulb the voltage was about 2.98 whereas with 6 bulbs it decreased to 2.25?
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esrever
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(Original post by maha.4646)
thansk for heplping but I still dont undertand why the voltage decreases as more bulbs are added in parallel.
I have my exam on monday and i`ve already done my practical on this but the results I got showed that with 1 bulb the voltage was about 2.98 whereas with 6 bulbs it decreased to 2.25?
Adding bulbs in parallel doesn't decrease p.d. across them. In fact p.d. across them remains constant.

EDIT: what you are observing is due to internal resistance of the power source.
Last edited by esrever; 1 year ago
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maha.4646
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thank you, so in the exam how am i suppose to explain this?
can I say:
The battery has its own internal impedence so extra current causes voltage loss across its impedence therefore, adding more bulbs decreases voltage but it is not that significant?
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esrever
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(Original post by maha.4646)
thank you, so in the exam how am i suppose to explain this?
can I say:
The battery has its own internal impedence so extra current causes voltage loss across its impedence therefore, adding more bulbs decreases voltage but it is not that significant?
That sounds like a good explanation but I would suggest that you take advice from your teacher or someone else who is studying the same exam board as you.
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maha.4646
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Ok thanks alot for helping
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DFranklin
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(Original post by maha.4646)
thank you, so in the exam how am i suppose to explain this?
can I say:
The battery has its own internal impedence so extra current causes voltage loss across its impedence therefore, adding more bulbs decreases voltage but it is not that significant?
It's "resistance" not impedence (and the difference matters), but otherwise, yes this sounds OK, although depending on the battery it might be more significant than you claim.
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maha.4646
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(Original post by DFranklin)
It's "resistance" not impedence (and the difference matters), but otherwise, yes this sounds OK, although depending on the battery it might be more significant than you claim.
Thank you
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maha.4646
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Does this sound ok:
As I added more bulbs to the parallel circuit, the brightness of the bulbs remained the same this is bcz the current goes through seperate branches in a parallel circuit and if another branch is added with an additional bulb, the current has an additional path to take. Therefore, the battery produces a constant voltage so the current through the origional bulbs doesnt change and not does their brightness, as current is shared between all the bulbs. However, the battery has its own internal resistance therefore, extra current drawn from the battery causes voltage loss across its resistance thus adding more bulbs decreases the voltage which may or may not be that significant depending on the battery.
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