# Simple DC Circuit

## Homework Statement

Rank the brightness of five bulbs (3 different circuits) considering all the batteries and light bulbs are identical.

1st circuit - 1 battery hooked to bulb A.
2nd circuit - 1 battery with positive end hooks to bulb B then bulb B connects to bulb C then comes back to negative side of battery. (series)
3rd circuit - 2 batteries (series) with positive end of first battery hooks to bulb D then bulb D connects to bulb E then comes back to negative side of other battery.

V=IR

## The Attempt at a Solution

I think it's this...A=D>E>B>C...

I think I am missing something really important...question seems to be so...simple!

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Have you just solved it intuitively or have you also set up some equations?
And why is A=D=E > B=C?

Update, D>A=E>B>C.

I can't relate this to anything. I am clueless of this kind of stuff.
I simply reordered based on what I experimented previously.
:(

On what does the brightness depend?
After you answered that question, can you relate it to your relevant equation?

Brightness depends on current, I, right?
so D should be the brightest due to number of batteries then A=E>B>C...(?)

ranger
Gold Member
The brightness of each bulb depends on the power across the bulb. Do you know how to find power?

The brightness of each bulb depends on the power across the bulb. Do you know how to find power?

Hm...is this necessary steps?
I have not learned it...
It seems like it can be solved really easily but...  By the way, this isn't suppose to be workout problems...
It is prelab that I am suppose to do...

ranger
Gold Member
Hm...is this necessary steps?
I have not learned it...
It seems like it can be solved really easily but...  By the way, this isn't suppose to be workout problems...
It is prelab that I am suppose to do...

Well, yea I guess you can solve it by making educated guesses. But that is not the way the how these things should be done. Like I said before, brightness is related to power. The general equation for power is:
P = IV

Since all of your circuits are series circuits and each bulb has equal resistance. What do you know about voltage and current in a series circuit?

indirectly proportional(?)
But that means currents gets smaller as voltage increase...

ranger
Gold Member
Voltage and current are directly proportional (V=IR). What I was asking you is how is the voltage across a bulb related to the current through it? You just need to remember the simple rules of a series circuit.

The question is not that hard. Its just relating several concepts that may cause confusion - power, ohms law, and series circuits.

ok so...back to the top...

assuming that each battery is 9 volt and since all resistor inside of bulbs is same...
current must be same as long as there are same amount of batteries, which tells me that A>B>C.

Now the circuit 3...since there are two batteries, do i simply add 9+9=18 so that D>E?

I have no idea what's keeping me from understanding this simple concept.

D=E > A > B=C...