# Lightbulb connected to the battery circuit helo

## Homework Statement

You have a large supply of lightbulbs and a single battery. You start with one lightbulb connected to the battery and notice its brightness. You then add one lightbulb at a time, each new lightbulb being added in parallel to the previous lightbulbs. As the lightbulbs are added, what happens to (a) the terminal voltage of the battery, (b) the brightness of each lightbulb, (c) the current in each lightbulb, (d) the power delivered by the battery, (e) to the lifetime of the battery.
Assume the lightbulbs to be ohmic and do not ignore internal resistance of the battery.

Thanks for any help!!

V = IR
P = IV

## The Attempt at a Solution

(a) The terminal voltage of the battery would stay the same because as more and more lightbulbs are added the Req decreases so the total current would increase.

(b) The brightness of each lightbulb would increase because the total current increased. P = IV so an incresed current means and increased power. The larger the power, the brighter the bulb.

(c) The current in each lightbulb increses because the total current increased. That means that there was more current to enter each junction.

(d) The power delivered by the battery would also increase. If the total current increased and the total voltage stayed constant, the power delivered by the battery would have to increase.

(e) The lifetime of the battery would decrease since it delivers a greater power.

Thank you for any help! I'm not sure if any of my explanations make sense.

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
I appriciate any help! If anyone can let me know if my answers are right or help me with this problem, it would be great! Thanks.

(a) Your answer does not really explain why the battery would continue you have the same voltage. You are talking about the overall circuit, and the question is asking about just the battery element.

(b) Are you positive that the current increases as you add more resistance? Think about ohm's law.

(c) Again, you may want to make sure that it would make sense for the current to increase.

(d) Where does the battery get the extra power or extra energy from?

(e) Sure if the battery delivered more power per unit time then the battery would have less life.

Mindscrape, the bulbs are added *in parallel* - this would decrease the resistance of the circuit. The total current would increase, but not necessarily the current in each branch of the circuit (which is to say that pink needs to think about her answer to (b) again).

Note: when there's no load on a battery, it's terminal voltage is the same as its emf. However, once the battery starts discharging, its internal resistance is going to use some of that energy. The greater the current, the more energy that's lost due to the internal resistance. Note too, that as the battery gets older, the internal resistance increases.

Pinkpolkadots, consider this: Walk into a room and turn on a lamp. Then, turn on another lamp. Does the first lamp get brighter? They're both (assuming both outlets are on the same circuit) on the same circuit, wired in parallel.

I would think that when you turn on another lamp the first one would become dimmer, but I don't know how to explain that in physics terms.

Actually, could you say that the brightness of each bulb decreases because the voltage from the battery has to be shared by more bulbs?

(a) Well, I would think that you can't change the terminal voltage of the battery, that is why I said it would stay constant. And for (d) I guess you can't change the the power delivered by the battery either. But, again, I can't explain my answer.

(c) A parallel circuit is just like parallel loops of single-bulb circuits, with each bulb directly connected to both ends of the battery, so each circuit only has the resistance of one bulb, no matter how many parallel bulbs you add. Also, by adding more lightbulbs in parallel, you aren't changing the voltage in any of the other bulbs. So would that mean the current in each bulb stays constant?

I don't know, I keep changing my mind! Thanks for any more help :)

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I looked at the sites and found this helpful for part (a):

As the number of parallel lamps increases, the terminal voltage drops and the current increases.
The terminal voltage of the cell is equal to only its emf when no current is drawn (no lamps attached).

Therefore, it seems to me that the terminal voltage of the battery decreases as more lightbulbs are added.

Here is what I have for answers now:
Please let me know if I still have the wrong idea.

(a) The terminal voltage of the battery decreases as more lightbulbs are added. The terminal voltage of the battery is only equal to its emf when no current is drawn. Also, this voltage is lost over the internal resistance of the battery. And the more current the more that is lost to internal resistance. I know the current will increase because the Req decreased as more lightbulbs were added.

(b) I know the voltage for bulbs in parallel is constant. But the brightness of each bulb decreases as the number of them increases. Is this because the current through each bulb decreases?

(c) Using Kirchoff's rule (Itotal = I1 + I2 + I3 + ...) I am thinking that the current through each bulb decreases. Even though the total current increased, because the total number of bulbs increased the current flowing to each bulb is less.

(d) Does the power delivered by the battery decrease because of internal resistance? But then I thought the total current increased, so the power would increase...which doesn't make sense.

(e) The life of the battery would decrease because as the total current continues to increase it puts a greater strain on the battery and the internal resistance increases.

Thanks again.

Pinkpolkadots,

I checked my Introductory To Circuit Analysis textbook. It says as the load resistance approaches 0 ohms, all the generated voltage will appear across the internal resistance and none at the output terminals.

You are correct in saying the resistance drops as you add bulbs to the parallel circuit. This means a drop in voltage at the output terminals of the battery.

Less voltage available would decrease the current through each individual bulb making it dimmer.

This would also decrease the power delivered by the battery but since the rest of the power is lost due to the internal resistance I think the battery life will stay the same.

I hope I wasn't to late to help but I had things to do and needed to read the section of the book that deals with batteries twice because its been 25 years since I studied electricity.

Thanks so much!

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Thanks so much!
You are welcome, I enjoyed going back and refreshing my memory. Don't hesitate to ask further questions as I'm glad I could help.

Can anyone explain to me why the battery life would stay the same?

The total resistance drops due to more paths for the current to flow through.

The resistance for each bulb stays the same.

When the voltage goes down and the resistance stays the same. The current for each bulb must go down also.

As far as the total current and therefore the life of the battery. I worked the numbers for a single circuit and the current went up. This would cause a decrease in the battery life. However I figure this depends on the specifics of the battery (the voltage and internal resistance) as well as the resistance of the bulbs in question. So my first answer about the life of the battery staying the same is not always the case. (This was an assumption due to the power consumed by the internal resistance.)

This is not covered in my books or on the websites I visited. So as I said the total current and life of the battery may depend on the specifics.

Hopefully there's someone with more knowledge than I that can help you.

Can anyone explain to me why the battery life would stay the same?
It would?
Seems to me that when I leave the headlights on in my car (2 bulbs, in parallel) that the battery dies before I'm out of work. But, if I turn off the car and the only thing on is the little LED lights that display the time, my battery hardly notices.