# lightbulbs in parallel and one burns out....

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#### sv3

##### Full Member
10+ Year Member

So here is a question that seems like a good MCAT question and since I had issues with it I thought I would post about it:

Three identical lightbulbs are connected to a battery and the lightbulbs are parallel. If the middle bulb burns out, what happens?

So the answer is: the light intensity of the other two bulbs remain the same (of course when I think about real life this makes perfect sense). However, I thought the intensity of the other two would increase. I thought so because if one bulb burns out, then total Resistance in the circuit decreases, and the total current for the circuit would then increase (I=V/R). This led me to believe that each of the remaining batteries in the circuit would get more current and have more light intensity. To my surprise, the explanation in TPR was (to paraphrase) that since there are only two bulbs left and not three, the current the battery provides decreases as their individual currents stay the same. To me it sounds like the battery can control how much voltage it supplies and I don't get that. Can anyone help me see where my thinking is incorrect or help explain it in a different way? I hate getting these concept type of questions wrong!

steve

#### funsmith

##### Full Member
I think the answer to that is the Kirchoff's second law. Where it states that the battery would only supply enough voltage to suffice for the amount that is being used in the circuit.

I hope it helps.

#### joshinjosh

##### Full Member
10+ Year Member
5+ Year Member
The bulb's brightness will remain the same.

Yes, you are correct that the overall current will decrease because the overall resistance decreases. But amount of bulbs in the circuit also decreases.

I think the easiest way to look at bulb brightness is the voltage on either side of the bulb. In circuit with a single set of parallel bulbs, all the bulbs in parallel will always have the same brightness despite the ammount of bulbs because the voltage across each one will stay the same.

Hope this helps,

Josh

#### sv3

##### Full Member
10+ Year Member
The bulb's brightness will remain the same.

Yes, you are correct that the overall current will decrease because the overall resistance decreases. But amount of bulbs in the circuit also decreases.

I think the easiest way to look at bulb brightness is the voltage on either side of the bulb. In circuit with a single set of parallel bulbs, all the bulbs in parallel will always have the same brightness despite the ammount of bulbs because the voltage across each one will stay the same.

Hope this helps,

Josh

Thanks Josh. That helped a ton. Not sure what the heck I was thinking.....

steve

#### TheBoondocks

##### StreetFighter 4 Virtuoso
15+ Year Member
So here is a question that seems like a good MCAT question and since I had issues with it I thought I would post about it:

Three identical lightbulbs are connected to a battery and the lightbulbs are parallel. If the middle bulb burns out, what happens?

So the answer is: the light intensity of the other two bulbs remain the same (of course when I think about real life this makes perfect sense). However, I thought the intensity of the other two would increase. I thought so because if one bulb burns out, then total Resistance in the circuit decreases, and the total current for the circuit would then increase (I=V/R). This led me to believe that each of the remaining batteries in the circuit would get more current and have more light intensity. To my surprise, the explanation in TPR was (to paraphrase) that since there are only two bulbs left and not three, the current the battery provides decreases as their individual currents stay the same. To me it sounds like the battery can control how much voltage it supplies and I don't get that. Can anyone help me see where my thinking is incorrect or help explain it in a different way? I hate getting these concept type of questions wrong!

steve

You were thinking and that was good. The reason for your confusion is that the previous problems in the TPR physical science section required you to calcuate Req and then use that for something else. If say, the three bulbs were also connected in series with another bulb. Then since the Req would be increased (assuming middle bulb blows), the Req of the the two bulbs along with the other bulb (4th and functional) would cause a decrease in current since the overall Req(series) would be greater.

However, in parallel, remember voltage is always constant and the same for all the resistors. The resistance remains the same. So, if V=IR, R is for the particular resistor and not Req. So, they must remain constant. Finally, this makes intuitive sense that's the advantage of parallel. If it increased then that would be almost as bad as series which causes the whole circuit to break.

In closing, you got Req confused with R. If you hadn't, you would have seen that constant V and R necessitates a constant I.

#### sv3

##### Full Member
10+ Year Member
You were thinking and that was good. The reason for your confusion is that the previous problems in the TPR physical science section required you to calcuate Req and then use that for something else. If say, the three bulbs were also connected in series with another bulb. Then since the Req would be increased (assuming middle bulb blows), the Req of the the two bulbs along with the other bulb (4th and functional) would cause a decrease in current since the overall Req(series) would be greater.

However, in parallel, remember voltage is always constant and the same for all the resistors. The resistance remains the same. So, if V=IR, R is for the particular resistor and not Req. So, they must remain constant. Finally, this makes intuitive sense that's the advantage of parallel. If it increased then that would be almost as bad as series which causes the whole circuit to break.

In closing, you got Req confused with R. If you hadn't, you would have seen that constant V and R necessitates a constant I.

You either have a crystal ball or have used TPR before! That's exactly the way it went down. Thanks very much. Just so I know, when you say Req, you mean total circuit resistance? And when you say R, you mean resistance for one resistor only right? TPR doesn't use Req, just R all the time.

Thanks again
steve

#### TheBoondocks

##### StreetFighter 4 Virtuoso
15+ Year Member
You either have a crystal ball or have used TPR before! That's exactly the way it went down. Thanks very much. Just so I know, when you say Req, you mean total circuit resistance? And when you say R, you mean resistance for one resistor only right? TPR doesn't use Req, just R all the time.

Thanks again
steve

Yep. That is correct. This is why you were confused. I also missed this question for the same reasons you did #### sv3

##### Full Member
10+ Year Member
Yep. That is correct. This is why you were confused. I also missed this question for the same reasons you did yeah it helped to go back and see what led me to that mistake. I kept thinking that the current that went into the burnt out bulb now had to go somewhere right? where in reality no it didn't for the reasons you stated above. i was actually enjoying this section but hit capacitators and dielectrics just now.....and now not so much....

#### sshah92

##### Full Member
5+ Year Member
Hey guys,

I know this is an old thread but I'm reviewing this right now and wanted some clarification on this topic!

As I understand it, the resistance of each light bulb stays the same when one bulb in parallel blows out. The voltage drop stays the same too because that's the nature of a parallel arrangement. And that leads to why current stays the same in each "route" -- which means that P (= I*V) stays the same. Power is directly related to intensity, so if power in the other two light bulbs stays the same, then so too does intensity.

However, what about for the entire circuit itself? I noticed that if the resistance of each light bulb stays the same, and we eliminate one light bulb, then Req actually increases! (I just derived this mathematically from the 1/Req = 1/R1 + 1/R2 equation.)

So if Req increases, what happens to V and I overall? Since V = IR, does the battery supply the same amount of voltage (which would make sense, considering the electric potential difference between both terminals isn't affected by anything in between) and less current overall (which would also make sense, because when you add the individual Is from each divergent path, less paths means less overall current)? Or does the battery supply the same current overall and less voltage?

Moreover, how does the battery KNOW how much current to supply if that first option is correct?

#### Socrates25

##### Full Member
10+ Year Member
Hey guys,

I know this is an old thread but I'm reviewing this right now and wanted some clarification on this topic!

As I understand it, the resistance of each light bulb stays the same when one bulb in parallel blows out. The voltage drop stays the same too because that's the nature of a parallel arrangement. And that leads to why current stays the same in each "route" -- which means that P (= I*V) stays the same. Power is directly related to intensity, so if power in the other two light bulbs stays the same, then so too does intensity.

However, what about for the entire circuit itself? I noticed that if the resistance of each light bulb stays the same, and we eliminate one light bulb, then Req actually increases! (I just derived this mathematically from the 1/Req = 1/R1 + 1/R2 equation.)

So if Req increases, what happens to V and I overall? Since V = IR, does the battery supply the same amount of voltage (which would make sense, considering the electric potential difference between both terminals isn't affected by anything in between) and less current overall (which would also make sense, because when you add the individual Is from each divergent path, less paths means less overall current)? Or does the battery supply the same current overall and less voltage?

Moreover, how does the battery KNOW how much current to supply if that first option is correct?

Battery voltage in a "theoretical" closed circuit is always the same, regardless of whether it faces 1 ohm of resistance or 1 million ohms of resistance. The current changes.

#### Socrates25

##### Full Member
10+ Year Member
You were thinking and that was good. The reason for your confusion is that the previous problems in the TPR physical science section required you to calcuate Req and then use that for something else. If say, the three bulbs were also connected in series with another bulb. Then since the Req would be increased (assuming middle bulb blows), the Req of the the two bulbs along with the other bulb (4th and functional) would cause a decrease in current since the overall Req(series) would be greater.
.

If any one of the bulbs in series blows, then resistance goes to infinity (in a theoretical circuit) and drives current to zero. None of the light bulbs will light.

#### sshah92

##### Full Member
5+ Year Member
Thanks dude!

As for your other reply, he's not talking about the bulb in series. He was saying if there was a fourth bulb in series with the three that are in parallel, when the middle parallel bulb blows out -- not the fourth one, which is in series, but one of the three that is in parallel -- then the overall Req will increase. This would cause the overall I to decrease to maintain the same voltage.