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Silly questionby shibme
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#1
Apr2107, 04:32 PM

P: 7

hi,
I am having trouble figuring this out mathematically... I have no idea why but I can't figure it out using V=IR that the circuit on the left has 1/3 the current of the circuit on the right... can anyone plz give a mathematical explaination? Left : 3Vb = I1 * RL Right : Vb = I2 * RL then if Vb = 2 and RL = 1, wouldnt that make I1 > I2, by 3 times.. anyone suggestions? 


#2
Apr2107, 05:49 PM

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P: 22,281

Why don't you trust your conclusion?



#3
Apr2107, 06:08 PM

P: 7

well based on my calculations, it says that the current through RL on the left circuit is greater than the current through RL on right... I know it should be the opposite but I can't show it mathematically... any suggestions? thanks..



#4
Apr2107, 08:36 PM

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P: 22,281

Silly question
Well that's what I'm trying to ask: why do you "know" it should be the opposite?



#5
Apr2107, 08:39 PM

P: 7

ohh... because theres an explaination in the book in words... and im trying to figure it out mathematically....



#6
Apr2107, 09:40 PM

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#7
Apr2107, 10:00 PM

P: 7

I've scoped through other posts on this forum and I came about this reply by Cliff_J...
 Ok, in series the voltages would add, in parallel the current will add. Make sure you keep the terms straight, its an important distinction between the two. Parallel is like how a home is wired, one bulb burns out and the others are fine, series is where everything is connected to the next peice in a row and if one bulb burns out like in a cheap christmas light all the lights in series with it go out too because no current will flow because the circuit is no longer complete. Lets say you have two 1.5V AA batteries that can deliver 1A of current. In series they can deliver 3V and 1A of current. In parallel they can deliver only 1.5V but 2A of current.  I dont know what is correct and what isnt... im still trying to interpret this correctly.... based on Cliff_J's answer..... it seems like the same situation as mine but with 2 batteries.... in my case it is 3..... can anyone help me on this? 


#8
Apr2107, 11:04 PM

P: 2,508

Cliff is definitely correct. But you are misinterpreting it.
  Think of it this way. You have a 120 volt circuit in your house fused at 15 amps. You plug a light bulb into it that is 120 watts. The bulb will have a current of 1 amp passing through it. It is on a circuit that is capable of supplying up to 15 amps. That is not to say that the bulb is drawing 15 amps. 


#9
Apr2107, 11:12 PM

P: 7

ahhh... thank you!
Whew... never will I doubt ohms law again... 


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