Lightbulb Brightness in a circuit.

[1988Greens]

Homework Statement

Rank the lightbulbs in order of brighness.

http://img641.imageshack.us/img641/5871/lightbulb.png

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)

 

[tiny-tim]

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?

 

[1988Greens]

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]

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.

 

[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.

 

[collinsmark]

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 R_eq = 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]

Heh Waffle? I like waffles they are delicious.

mmm … waffles! ​

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 … ? ​