Potential difference between points a and b in resistor network with batteries

[mba444]

Homework Statement

A network below consists of with three bat-
teries, each having an internal resistance, and
five resistors.

http://images.upload2world.com/get-6-2009-upload2world_com_vypln.jpg

Find the magnitude of the potential differ-
ence between points a and b. Answer in units
of V.

Homework Equations

i solved it using the loop equation

The Attempt at a Solution

VA-VB= 4*i1+35-i1-7*i1-6*i2-i2-28
VA-VB=7*i3+18+i3+5*i3-6*i2-i2-28
i1=i2+i3


but I am stuck her i don't knowhat to do next

 

[quantoshake11]

following what you've done, you need one extra equation. If you take the first two and substract them, you get 0 on one side, and a linear combination of i1 i2 and i3 on the other, so with the other equation you have in total two linear equations on three variables (i1, i2, i3), so you have infinitely many solutions. you're going to need one extra equation before solving for the currents (see http://en.wikipedia.org/wiki/System_of_linear_equations#General_behavior).
for that,

Spoiler

write down the "loop" equation for the big loop in the circuit (the big loop being, well, the biggest loop you can make in that circuit, it does not go through a and b so the total voltage difference will be 0.)

Once you solve for the currents, you can get VA - VB from one of the equations you already wrote.

 

[mba444]

ahaaa

ok thanks for your help i got it now

 

[vk6kro]

There seems to be a simpler way of doing this problem.

The centre battery is open circuited isn't it?

So take the current in the outer loop only. 35 volts minus 18 volts and a total of 25 ohms. So the total current is...

Then work out the voltage at point A relative to the right vertical connector.

The voltage at B is available because the centre battery isn't supplying any current. So the resistors in series with it are irrelevant.

The difference between A and B is available by subtraction.