# Lightbulb Brightness in a circuit.

1. Mar 12, 2010

### 1988Greens

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

Rank the lightbulbs in order of brighness.

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

2. Relevant equations

All lightbulbs are identical.

3. 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)

Last edited by a moderator: May 4, 2017
2. Mar 12, 2010

### tiny-tim

Welcome to PF!

Hi 1988Greens! Welcome to PF!
Sorry, but that's just waffle.

Physics is equations.

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

3. Mar 12, 2010

### 1988Greens

Re: Welcome to PF!

Heh Waffle? I like waffles they are delicious.

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.

4. Mar 12, 2010

### ideasrule

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.

5. Mar 12, 2010

### 1988Greens

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.

6. Mar 12, 2010

### collinsmark

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.

7. Mar 13, 2010

### tiny-tim

:tongue2: mmm … waffles! :tongue2:​
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 … ?