Calculating Resistance & Current for Two Lightbulbs in Series

[MIA6]

Two light bulbs, one rated at 120 V and another rated 40 W at 120 V, are arranged in series with each other. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in this circuit.
I first found the current, using P=IV. But here is the total power 30W+40W=70W? Then I did:70W=I*120V, I=7/12, then I found Resistance by using R=P/I^2, R=30W/(7/12)^2=88ohms. But somehow the answer seems to be wrong. Its resistance is smaller than the resistance in a parallel circuit, and it's impossible. Hope you can point out my mistake. Thanks.

 

[Doc Al]

The power ratings of the bulbs only apply when they are separately connected to a 120 V source (not when they are hooked in series). The first thing to do is to use the power ratings to find the resistance of each bulb. Then you can use that to find the current when they are put in series.

 

[MIA6]

The power ratings of the bulbs only apply when they are separately connected to a 120 V source (not when they are hooked in series). The first thing to do is to use the power ratings to find the resistance of each bulb. Then you can use that to find the current when they are put in series.

I kinda don't understand what do you mean by power ratings of the bulbs only apply when they are separately connected to the source? Btw,power ratings mean power? Also, if I find the resistance first, I used P=V^2/R? So it's 30W=(120V)^2/R? Is 120V the voltage for 30W? BUt if they connect in series, will it be the total voltage instead of one?

 

[Doc Al]

I kinda don't understand what do you mean by power ratings of the bulbs only apply when they are separately connected to the source?

A 30W/120V bulb means that when the bulb is connected (all by itself) to a 120 V source it will draw 30 W of power.

Btw,power ratings mean power? Also, if I find the resistance first, I used P=V^2/R? So it's 30W=(120V)^2/R? Is 120V the voltage for 30W?

Yes, yes, and yes.

BUt if they connect in series, will it be the total voltage instead of one?

If you connect the two bulbs in series and then apply 120V across both of them, the total voltage will be 120V. (The power used by each bulb depends on the voltage across it and the current through it, but the resistance of each bulb is assumed to remain constant no matter what.)

 

[MIA6]

If 30 W has 120 V and 40 W has 120 V, too, then in seires the total voltage should be V1+V2=240V?

 

[Doc Al]

If 30 W has 120 V and 40 W has 120 V, too, then in seires the total voltage should be V1+V2=240V?

No. I assume both bulbs are hooked in series across a single 120V source, so that voltage is divided across both bulbs. (Your first post did not describe the circuit.)

A "30 W" bulb only draws 30 W when it's hooked up to 120 V.

Hint: Find the resistance of each bulb, then use that to find the current through the bulbs when they are hooked up in series across 120 V.

 

[MIA6]

Then, I think the voltage for 30 W is 120V/2=60V? Also, btw, how do we know the brightness of a lightbulb, if they already told us their power, 30 W and 40 W?

 

[Doc Al]

Then, I think the voltage for 30 W is 120V/2=60V?

The voltage doesn't divide evenly. Find the resistances!

Also, btw, how do we know the brightness of a lightbulb, if they already told us their power, 30 W and 40 W?

The brightness will be proportional to the power drawn by the bulb, which depends on how you use it.

 

[MIA6]

ok, if according to what you said, find resistance first, then it's P=V^2/R, 30W=(120V)^2/R. However, Why I kept emphasizing on asking the single voltage for each bulb because what if these two bulbs were connected in parallel, then the Resistance and the voltage of each bulb will have no difference with the two bulbs connected in series, which will be impossible in my opinion? Since the question gave me the power for each bulb, then I also need to find the voltage for each bulb, then find the resistance. I don't think I should use the total voltage to find the resistance of a single bulb!

 

[Doc Al]

To avoid any misunderstandings, please post the problem exactly as it was given, word for word. (A diagram or scan would be a good idea.)

 

[MIA6]

Okay. There is no diagram for the question.
Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two different circuits.
a) The two bulbs are first connected in parallel to a 120 V source
i. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in this circuit.
ii. ..........rated 40 W.......
b) The bulbs are now connected in series with each other and a 120 V source.
i, ii, same questions as above.

 

[Doc Al]

Okay. There is no diagram for the question.

That's OK, the problem is clear.

Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two different circuits.

Use the method I discussed to find the resistance of each bulb. You need that before going any further. Note that this is just what you need to do to answer question a. (When the bulbs are in parallel, they both receive the full 120 V.)

 

[MIA6]

The thing is I am very clear at when they were connected in parallel because I know each of them receives 120 V, however, i don't know b) when they are connected in series because i think 120 V is then the total voltage, so How can I find the voltage for the single lightbulb. Because I need the single voltage to find the single resistance.

 

[Defennder]

The resistance of the bulbs don't change when you set them up in a different circuit. You can assume they are the same. Which is why you have to find their resistances first before doing anything else. Once you have their resistances, you can then set them up in series then add up their resistances to find the current through both of them.

 

[MIA6]

Ok. I got it. BUt then how can I figure out which lightbulb is the brightest, second brightest and so on??

 

[Defennder]

You have found the resistance of the bulbs, have you not? Once you do so, you can set them up in series. Now remember that the equivalent resistance of two circuit components (in this case, bulbs) connected in series is just the same as adding them up. With this, can you find the current through both of them? Remember than since they are both connected in series, the current through them is the same.

Once you've got the current through both of them, you can find the power dissipated in each of them. Doc AI said earlier that the brightness of the bulb is proportionate to the power dissipated in them. Can you take it from here?

 

[Redbelly98]

Ok. I got it. BUt then how can I figure out which lightbulb is the brightest, second brightest and so on??

I think it best if you could say what resistances you have calculated for each bulb. I've lost track of how many times you have been asked to do this.

 

[MIA6]

You have found the resistance of the bulbs, have you not? Once you do so, you can set them up in series. Now remember that the equivalent resistance of two circuit components (in this case, bulbs) connected in series is just the same as adding them up. With this, can you find the current through both of them? Remember than since they are both connected in series, the current through them is the same.

Once you've got the current through both of them, you can find the power dissipated in each of them. Doc AI said earlier that the brightness of the bulb is proportionate to the power dissipated in them. Can you take it from here?

Okay, I got the resistances for two bulbs in parallel circuit, which will be the same in series circuit, too.Then I found the total current in the series circuit. I also found the power for each bulb. Thanks.