(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The battery in the figure below has negligible internal resistance. Find

(a) the current in each resistor:

(b) the power delivered by the battery:

2. Relevant equations

V = IR

Resistor Connected in series

R_{eq}= R_{1}+ R_{2}+ R_{3}

Resistor Connected in Parallel

1/R_{eq}= 1/R_{1}+ 1/R_{1}+ 1/R_{3}

3. The attempt at a solution

I think that the resistors 3omega and vertical-2omega are connected in series, and the other 2 are connected in parallel.

So the ones in Parallel..

1/R_{eq1}= 1/2 + 1/4

1/R_{eq1}= 3/4

R_{eq1}= 1.333

The ones in Series

R_{eq2}= 3 + 2 = 5

so...

R_{eq}= R_{eq1}+R_{eq2}=1.333 + 5 = 6.333

But to find the Current, I you just use Ohm's law

I = V/r

so the 3.0-omega resistor would be

I = 6/3 = 2

...but thats wrong and Im at a loss to know why?? thanks for the help!

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# A multi loop curcuit- Resistors in parallel and series

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