# Homework Help: Finding potential difference of Va-Vb, circuits.

1. Oct 26, 2005

### mr_coffee

Hello everyone, i'm confused, i know the loop rule and the sum of the Potential differences is 0 in a closed circuit but I need to find the Va-Vb of the circuit. Here is the picture: http://www.webassign.net/hrw/hrw7_27-40.gif and here is the info:
the resistances are R1 = 1 , R2 = 2.2 , and the ideal batteries have emfs E1 = 2.0 V, and E2 = 3 = 3 V.
Now i'm confused on how far I calculate, like I tried just going from Va to Vb, and it was wrong i did the following:
Va + R2E2 - E2 = Vb;
Va-Vb = -R2E2+E2 = -3.6;

Also i had another question, I know the direction of the current because E2 > E1, so the direction should flow counter clockwise, now when I run into any kind of resistor do I always use the current of the stronger battery? Like for instance:
If i go from a counter clockwise to b, When i run into the Resistors R1 and R1 again at the bottom, do i still use E2*R1, and never use E1*R1 because E1 is weaker then E2? Thanks.

I also tried to go all the way around a to b, and then back to a counter clockwise and got the following:
Va - R1E2 - E1 - R1E2 + Vb + E2 - E2R2 = 0;
but i ended up with Va + Vb = R1E2 + E1 + R1E2 - E2 + E2R2 = 11.6V, also wrong!

Last edited: Oct 26, 2005
2. Oct 28, 2005

### ehild

Your equations are entirely wrong. What do you mean on R*E? The potential difference across a resistor is resistance (R) multiplied current (I).

ehild

3. Oct 29, 2005

### Ouabache

You may want to find Va by making an equation in terms of Va. (using Kirchoff's Rules) Example: [Vb-(-2)]/R1 - 3 + [Vb-(-3)]/R1 = 0
You can then solve for Vb.. Try the same thing for Va. (That is easy. I showed you the harder of the two equations). Then you can find Va-Vb

Your other question has some problems. You could go all the way around the left loop and then around the right loop, you would have 2 equations and 2 unknown variables (I1 and I2) which are left and right mesh currents respectively. I wouldn't use Va or Vb in those equations, it will only confuse things and add too many variables. It is useful to assume current direction when generating your equations. The current thru R2 will have two components I1 and I2. They will be a sum or difference depending on which directions you choose for I1 and I2. The direction you choose does not matter, as long as you are consistent in your math analysis. If you get a negative current, you will know it is flowing opposite to the direction you chose.

Ex. I chose I1 and I2 to both be clockwise, my left mesh equation would be : 2 - (1)I1 - 2.2(I1-I2) - 3 - (1)I1 = 0