# Using the Loop rule to find potential Difference

1. Oct 13, 2012

### MasterVivi

1. The problem statement, all variables and given/known data

Find the potential difference between points a and b.

Picture of Circuit involved.

2. Relevant equations

V=IR
ΔV=Vb-Va

3. The attempt at a solution

Pretty much completely lost, due to the only one current shown I'm not sure which way the current flows through the rest of the circuit so I do not know how to go about the loop rule.
It started off kinda like

-I1(30Ω)=2.118 V

But I know that cannot be correct cause if you left from the 24V battery and went down you'd have something about 23.3V across there, like I said, completely lost, any help would be great.

2. Oct 13, 2012

### Spinnor

3. Oct 13, 2012

### MasterVivi

This is useful, in a case like mine you wouldn't want multiple I notations of current cause then your introducing variables you don't know right?
should I go opposite the current from b→a?
like
ΔV=-(0.0706A)(10Ω)+12V=12.706V

Since that is the only value of I given?

4. Oct 14, 2012

### MasterVivi

Using the one know current, the current would be the same at a correct?
If so you couls just do
Va-Vb=-24V+(I)(10)
=>-24+(0.0706)(10)=-23.294

Which implies
Vb-Va=+23.294V

Or is this still the wrong path?
Do I need to use loop rules and find the unknown values of the current when it splits?