(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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: A multi loop curcuit- Resistors in parallel and series

**Physics Forums | Science Articles, Homework Help, Discussion**