# Lightbulb Brightness in a circuit.

## Homework Statement

Rank the lightbulbs in order of brighness.

http://img641.imageshack.us/img641/5871/lightbulb.png [Broken]

## Homework Equations

All lightbulbs are identical.

## The Attempt at a Solution

A is brightest, then B=C, then D=E

A>B=C>D=E

B and C are the same brightness level because they are in parallel, D and E are the least bright because they are in series.

I even downloaded a circuit creator to verify my solution but apprently I am still not correct and I am not sure why. (program from PhET Interactive Simulations Copyright © 2004-2010 University of Colorado)

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tiny-tim
Homework Helper
Welcome to PF!

Hi 1988Greens! Welcome to PF!
A is brightest, then B=C, then D=E

A>B=C>D=E

B and C are the same brightness level because they are in parallel, D and E are the least bright because they are in series.

Sorry, but that's just waffle.

Physics is equations.

So what equation are you using to find the power through each bulb?

Hi 1988Greens! Welcome to PF!

Sorry, but that's just waffle.

Heh Waffle? I like waffles they are delicious.

Physics is equations.

So what equation are you using to find the power through each bulb?

I would use V= IxR They give no numbers for the problem, so I would just have R or 1/R for the resistances in any given bulb?

B and C Resistance = 1/R + 1/R
= 2/R

D and E Resistance = R + R
= 2R

I don't really know I am pretty lost at this point.

Say if they gave you some numbers i.e. If it was a 12V battery with 1 Amp current Volts for bulb A would be V= 1xRx2/R
= 2R/R
= 2 Volts for bulb A? I feel like I am way off.

ideasrule
Homework Helper
You don't need to use any numbers. Can you figure out the current through each bulb? For example, the potential difference across D and E combined is V (since the two bulbs straddle the battery), so the current through each is V/2R. The larger the current through a bulb, the brighter it glows.

Yea, I can't seem to figure out the current for bulbs B and C. I mean I get the voltage for D and E V/2R, add the 2 resistances together so the current for bulb B would be VR? I can't figure out what it would be for A? would it be:

V/((1/R + 1/R) + R)?

I have difficulty understanding this.

collinsmark
Homework Helper
Gold Member
Yea, I can't seem to figure out the current for bulbs B and C. I mean I get the voltage for D and E V/2R, add the 2 resistances together so the current for bulb B would be VR? I can't figure out what it would be for A? would it be:

V/((1/R + 1/R) + R)?

I have difficulty understanding this.

Not quite. Remember resistors is parallel are not Req = 1/R1 + 1/R2.

You need to divide the whole thing by 1:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

which is

$$R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$

which is

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

Try that first for bulbs B and C, then add that to the resistance of bulb A in series to get the overall resistance of the path.

tiny-tim
Homework Helper
Heh Waffle? I like waffles they are delicious.

:tongue2: mmm … waffles! :tongue2:​
tiny-tim said:
So what equation are you using to find the power through each bulb?
I would use V= IxR

ah, now that's why it's not working out for you.

Do it logically …

the question asked for the power, so you need an equation for power, which V = IR isn't.

Moreover, you know that all the bulbs have the same R, but not the same I, so preferably you want an equation involving P V and R but not I.

Which is … ?