Relative Brightness of A-E, C+D

[Empire]

There are five light bulbs. Light bulb C and D are in a series connection. C and D are in a parallel connection with E. Light Bulb A and B are in a parallel connection. The two parallel connections are wired in a series connection. What would be their relative brightness? I'm thinking that A=B=E=C+D

********************
|----power---------------|
|*******************|
|*******************|
|**|--A--|***|-C-D-|**|
|---|****|----|****|---|
***|--B--|***|--E--|***
********************

 

[turin]

First you have to make at least one assumption. I would assume that the light bulbs are identical, and that "brightness" means "total power emited". Then brightness ~ I².

I disagree with your result (assuming I did it correctly). You need to use the voltage and current divider to examine this circuit.

 

[M98Ranger]

The relative brightness of A, B, E, C, and D would depend on the individual wattage or power rating of each light bulb. However, in general, A, B, and E would have the same brightness because they are in a parallel connection. C and D, being in a series connection, would have the same brightness as each other but potentially different from A, B, and E. Therefore, the relative brightness of A, B, E, C, and D would be A=B=E=C+D. However, this could change if the individual wattage of each light bulb is different.