Joe03 said:
I actually don't know where to begin...
I tried to use V=E-Ir so did 24-2I1=4I3, 27-6I2=4I3 so 6I2-2I1=3?
Can you start me off on the correct path, bit derailed.
Those equations are correct, but you've got only two that originate from the circuit. The other is derived from those two, so doesn't provide any new information. Since you have three unknowns (I1, I2, and I3) you need three equations that originate in the circuit. These can come from KVL loop equations or KCL node equations or a combination of both.
Note that once the loops you've chosen have incorporated each component at least once, there's no new information that you can extract by writing more KVL loop equations. Then you must turn to KCL to supply any "new" relationships between the variables.
To work consistently you should set up a basic procedure to follow:
1. Identify the nodes and loops.
2. Label each component on the diagram with the potential drop expected from given current directions.
3. Write KVL for appropriate loops.
4. Write KCL at appropriate nodes.
5. Solve the simultaneous equations by whatever method you're comfortable with.
After that is done you can start to solve the simultaneous equations.
When writing KVL and KCL equations, it's often less prone to making sign errors if you write them as a sum of terms that sums to zero. That is, place all the terms on the left hand side and make the right hand size zero. When done this way you can write your KVL equations simply by taking a "KVL walk" around the loop from any starting point, ending back at the starting point. Just write down the potential rises and drops as you encounter them on the walk.
So your first equation, for example could be obtained by "walking" counterclockwise around the outer loop of the circuit starting from the negative terminal of the 24 V battery:
##+24 - 4 I_3 - 2 I_1 = 0##
