# When do batteries drain more quickly?

Will a battery (for eg AA 1.5V) drain out faster when connected to 5 similar bulbs in parallel as compared to only 2 bulbs in parallel?

What I really want to know is that when we connect MORE bulbs parallel to a battery, does the battery have to do some extra work for managing more number of bulbs OR is it that the battery does the same work and that energy is simply distributed among the bulbs?

Related Other Physics Topics News on Phys.org
phinds
Gold Member
2019 Award
as you put more bulbs in parallel, the battery has to supply more current, thus drains faster

But isnt it true that current increases simply because resistance becomes less? Battery doesn't increase it's voltage, it's the resistance that decreases in parallel connection ?
I don't disagree with your answer (infact I thought that is whats true).
It's just another thought.

phinds
Gold Member
2019 Award
But isnt it true that current increases simply because resistance becomes less? Battery doesn't increase it's voltage, it's the resistance that decreases in parallel connection ?
I don't disagree with your answer (infact I thought that is whats true).
It's just another thought.
Yes, that's correct. That's why the battery drains more quickly. Do you have some belief that it is the voltage that causes the battery to drain? If that were the case, it would drain just sitting there with no load.

Yep, that's what I thought. Can you elaborate more on this topic?

Last edited:
phinds
Gold Member
2019 Award
Yep, that's what I thought. Can you elaborate more on this topic?
just study Ohm's Law

So are these right?
"Adding more bulbs/resistors to a battery, makes the battery do extra work. Hence, becoming prone to drain faster."
"Voltage doesn't cause the battery to drain."
"And that's why electricity-gen companies charge us more if we keep adding more bulbs and appliances in our homes even though voltage is always 220V. (Country wise)"

You can consider that the battery voltage remains about constant while the power delivered and consumed is about P = IE.....that's one way to start. If the voltage E is about constant, then anything you do that increases the current I increases the power consumed and requires more power from the supply. This ignores the resistance of the source.

Some finer points are discussed here:
http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

In a practical situation you can help match load to resistance of a battery source depending on what you want to accomplish....see the above article: An AGM battery, for example, has a lot less internal resistance than a wet cell lead acid battery.