# Finding values in a (very) multiloop circuit (with multiple batteries as well)

1. Oct 3, 2009

### Oijl

1. The problem statement, all variables and given/known data
In the figure, R1 = R2 = 2.0, R3 = 4.0, R4 = 3.0, R5 = 1.0, R6 = R7 = R8 = 8.0, and the ideal batteries have emfs script e1 = 8.0 V and script e2 = 4.0 V.

The question asks to find the currents i1 and i2, and the energy rate of transfer in battery 1 and battery 2.

2. Relevant equations

3. The attempt at a solution
First of all, is the "energy rate of transfer" Pemf = iE?

But mostly, I was wondering if there was a good way to do this problem other than using the loop rule and junction rule so much that you could call the rules Scooby snacks and me Scooby Doo.

It's that if I use those rules to find the currents, I'd have four equations and four unknowns, which isn't really that bad except it looks like they'd be pretty long equations. And if there's a more efficient way, and I don't see it, that's because I don't understand something about circuits like this and I would like to learn.

So thanks!

2. Oct 3, 2009

### Delphi51

I would use the resistors in parallel and series formulas to find the combined resistance of the 5 resistors on the left side. That way only 2 loops to work with.

3. Oct 3, 2009

### mplayer

Yes, I think 'energy rate of transfer' is the same as asking how much power is being supplied by the two sources, battery 1 and battery 2.

Think about what resistors you could combine to simplify the circuit so as not to lose either of the two currents you are looking to solve for. Picking the most efficient node to ground and combining a few resistors, you can apply the junction rule and only have two equations and two unknowns to deal with. Hope that helps!

4. Oct 4, 2009

### Phrak

I've forgotten the Scoobie Do rules long ago.

Like Delphi is saying. Lump all the reististors on the left of E1 into one resistor and call it RA. Also, lump R4 and R5 together and call it RB. A much simpler circuit, don't you think?

5. Oct 4, 2009

### Oijl

Okay, but the only resistors I see that I can easily combine are 1 and 2. Being in series, I can add their resistances to create an equivalent resistor.

But then I have trouble combining 3, 6, 7, and (1+2). (1+2) and 6 aren't in parallel, are they? Doesn't 2 mess up simply combining (1+2) and 6 like you would any resistors in parallel? How would I combine the others?

EDIT: Got it. Thanks.

Last edited: Oct 4, 2009