Solving for I1 and I2: Math Problem with V=IR

[shibme]

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 please give a mathematical explanation?

Left : 3Vb = I1 * RL
Right : Vb = I2 * RL

then if Vb = 2 and RL = 1, wouldn't that make I1 > I2, by 3 times..

anyone suggestions?

 

[russ_watters]

Why don't you trust your conclusion?

 

[shibme]

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..

 

[russ_watters]

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

 

[shibme]

ohh... because there's an explanation in the book in words... and I am trying to figure it out mathematically...

 

[Averagesupernova]

ohh... because there's an explanation in the book in words... and I am trying to figure it out mathematically...

Well Russ has this covered pretty well so there isn't much I can say but I will say this: If the book 'tells' you that the current on the right circuit should be more than the left then perhaps you are misinterpreting or the book explains it very poorly. Ohms law never lies.

 

[shibme]

I've scoped through other posts on this forum and I came about this reply by Cliff_J...

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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 piece 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.

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I don't know what is correct and what isnt... I am 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?

 

[Averagesupernova]

Cliff is definitely correct. But you are misinterpreting it.
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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.

The key word is can. The circuit with the two batteries in series will have a voltage of 3 volts and deliver the current of 3/R. The circuit where the 2 batteries are in parallel the voltage is 1.5 volts and it can deliver twice the amount of current compared to one battery. However, ohms law tells us that volts = ohms x amps and the batteries as a unit will deliver the current of 1.5/R. It doesn't matter how many batteries we add in parallel. All this accomplishes is more capability to supply the current if more resistors are added or a smaller ohms value resistor is installed in place of the original.
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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.

 

[shibme]

ahhh... thank you!

Whew... never will I doubt ohms law again...