# Lightbulbs and a battery circuit

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1. Aug 26, 2016

### RoboNerd

1. The problem statement, all variables and given/known data

2. Relevant equations

no equations
3. The attempt at a solution
I know that we will have an equal current flowing through bulbs 1 and 2 when we connect them in parallel in accordance with kirchoff's voltage law which says that the voltage drops around both paths must be equal to 0, i.e., THE SAME.

Thus, bulb three will have an increased current through it, as connecting 1 and 2 in parallel will decrease the circuits overall resistance, increasing the current drawn from the battery, and increasing the current through bulb 3 which will increase its power and make it glow brighter.

OK, I get that.

Why bulb 2 does not glow brighter also? An increased overall current could mean that we have greater current through the branch of the current where bulb 2 is, increasing its power, and brightness.

Thanks in advance for the input!

2. Aug 26, 2016

### Staff: Mentor

As an exercise, you could give each bulb a resistance value, and call the battery 10V, then analyse the circuit and find the currents.

3. Aug 26, 2016

### RoboNerd

I tried doing that and I was not able to figure out how I could use it to prove it.

Besides, this is a multiple choice question for a high school students. It should be easy to solve this question quickly using a different method.

Would you be able to solve it using a fast method? If so, how?

4. Aug 26, 2016

### TomHart

If you did that exercise, what were your results? Could you show that work?

5. Aug 26, 2016

### David Lewis

When you close the switch, does voltage across bulb 2 change?

6. Aug 27, 2016

### CWatters

+1

Perhaps think about the alternative ways of calculating power dissipated in a bulb or resistor eg V^2/R What happens to the voltage across each lamp?

7. Aug 29, 2016

### RoboNerd

OK. let's look at the voltage drops across bulbs 1 and 2.

Both of them are connected in parallel, so voltage drop must be the same. Thus P = V^2/R, so the power consumed must be the same.
I get why we can eliminate answer B -> bulb 2 now. Thanks for that.

Now, let's look at why bulb 3 increases in brightness.
If its power P = V^2/R, then we would also have the same voltage drop across it as before (because regardless of the addition of another branch, it would be the same) and its resistance would be the same because it is that same bulb with the same intrinsic characteristics. That means that the power is not changing.

However, the right answer is (C) -> bulb 3 gets brighter. What is going on here?

8. Aug 29, 2016

### Staff: Mentor

⏩ Is bulb 3 connected in parallel or in series with the other pair?

9. Aug 30, 2016

### CWatters

That's not correct. Earlier you said...

and that is correct.

10. Aug 30, 2016

### RoboNerd

It is in series.

I understand why that logic may hold, but looking at this circuit by kirchoff's voltage law gives me a unique perspective. If I examine the circuit loop starting from the electromotive force, then I have either a voltage drop at bulb 1 or at bulb 2 and then the same voltage drop at bulb three. Both paths (with bulb 1 and bulb 2) must result in a net zero voltage, so the drop in voltage across bulbs 1 and 2 must be the same and the drop in voltage across bulb 3 is consistent.

Because bulb's three power is V^2/R, then its voltage will remain the same and its intrinsic resistance must remain the same, so its power is going to be the same.

Why is it not the case that the bulb is going to glow brighter?

11. Aug 30, 2016

### Staff: Mentor

No, it isn't. Now, I don't know what you exactly mean by "V", but whatever you do mean, there is no fixed V that you can be referring to. The only fixed potential in the circuit is $\Large\textrm ε$.

Is it becoming clearer that your best course is to follow my advice back in post #2?

12. Aug 31, 2016

### CWatters

The thing you are missing is that you have two different circuits here. One with the switch open and the other with the switch closed. It is true that in both cases the voltage around the circuit must sum to zero but KVL does not imply the individual voltage drops are the same in both circuits. For example..

Those voltages might be..

Switch open... +12-6-6=0

These example voltages were arrived at by adopting the strategy NascentOxygen outlined in #2

13. Sep 1, 2016

### RoboNerd

Aha. I see now why this is the case. I did a simple current analysis here, and doing it again somehow made me realize what was going on this time.

Thank you everyone for the help!!!