A multi loop curcuit- Resistors in parallel and series

[clickcaptain]

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 = R₁ + R₂ + R₃

Resistor Connected in Parallel

1/R_eq = 1/R₁ + 1/R₁ + 1/R₃

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!

 

[LowlyPion]

Unfortunately the 2Ω, the 2Ω and the 4Ω are all in parallel. They share a common node on each side. That's what || means.

Find the equivalent of those 3 together and then you can add in series.

 

[Marthius]

The problem that you are having comes from the diagram you are using. When you see something like this you should redraw it in an easier way to see it. Start with the battery, and connect it to the first resister. Then notice that the connection splits in three, so this means you can redraw those three resisters in parallel. Finally they reconnect at the end, and the single wire returns to the battery.

When you draw it this way, you will see that all the 2 ohm resistors and the 4 ohm are in parallel, and the 3 ohm resistor is in series with them.

 

[clickcaptain]

So would the current for the first resistor be I = V/R = 6 V/ 2 = 3 ?

 

[LowlyPion]

You have to determine the R_eq for the network, in order to determine I.

 

[clickcaptain]

I got Req= 3.8

parallel = 1/Req = 1/2 + 1/2 + 1/4 = 5/4
Req-p = 0.8

so Req = 0.8 + 3 = 3.8 A

Then do I use this in V = IR?

 

[LowlyPion]

That would be the way to do it.

Except it's 3.8Ω

 

[Kruum]

Then do I use this in V = IR?

But be careful, since the current isn't nesseceraly the same in every resistor.