# Simple DC Circuit

1. Feb 26, 2007

### krnhseya

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

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.

2. Relevant equations

V=IR

3. 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!

Last edited: Feb 26, 2007
2. Feb 26, 2007

### Edgardo

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

3. Feb 26, 2007

### krnhseya

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.
:(

4. Feb 26, 2007

### Edgardo

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

5. Feb 26, 2007

### krnhseya

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

6. Feb 26, 2007

### ranger

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

7. Feb 26, 2007

### krnhseya

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...

8. Feb 26, 2007

### ranger

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?

9. Feb 26, 2007

### krnhseya

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

10. Feb 26, 2007

### ranger

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.

11. Feb 26, 2007

### krnhseya

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.

12. Feb 26, 2007

### krnhseya

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