Simple DC Circuit

  • Thread starter krnhseya
  • Start date
  • #1
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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.

Homework Equations



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!
 
Last edited:

Answers and Replies

  • #2
703
13
Have you just solved it intuitively or have you also set up some equations?
And why is A=D=E > B=C?
 
  • #3
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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
703
13
Your problem asks for the brightness.
On what does the brightness depend?
After you answered that question, can you relate it to your relevant equation?
 
  • #5
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Brightness depends on current, I, right?
so D should be the brightest due to number of batteries then A=E>B>C...(?)
 
  • #6
ranger
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The brightness of each bulb depends on the power across the bulb. Do you know how to find power?
 
  • #7
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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...:confused: :frown:

By the way, this isn't suppose to be workout problems...
It is prelab that I am suppose to do...
 
  • #8
ranger
Gold Member
1,676
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Hm...is this necessary steps?
I have not learned it...
It seems like it can be solved really easily but...:confused: :frown:

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?
 
  • #9
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indirectly proportional(?)
But that means currents gets smaller as voltage increase...
 
  • #10
ranger
Gold Member
1,676
1
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
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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
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D=E > A > B=C...
My final answer :)
 
  • #13
103
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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?
I am leaning towards on that one now...
 

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