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## Homework Statement

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

(a) the current in each resistor:

(b) the power delivered by the battery:

## Homework 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}

## 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 = 6.333But 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 that's wrong and I am at a loss to know why?? thanks for the help!