Steven W. answered 04/07/17
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Hi Amy!
These multi-loop problems pretty much always boil down to getting enough equations to solve for all the unknowns that show up as you try to get to your solution.
Kirchhoff's laws are heavily employed here. What I will do is set up the equations that I would use to try to solve for the desired quantity, and then see what questions you have about how to get them or how to work with them:
First, around the outermost loop, starting at the positive terminal of the 7.6 V battery (since we are told which way the currents flow, I will use that to determine my direction around the loop):
[1] -I1(0.7 Ω) + 1.8 V - I1(3 Ω) - I3(9.7 Ω) + 7.6 V = 0
This equation has two unknowns in it (I1 and I2), so I need at least one more equation. I will use the upper loop, starting at the positive terminal of the 3 V battery:
[2] -I2(6.4 Ω) - I3(9.7 Ω) + 3 V = 0
This equation, though, brings in yet another unknown (I3), so now I have three unknowns total in my two equations, and thus need at least one more equation. I will use Kirchhoff's junction rule (at either junction of the wires) to get:
[3] I1 = I2+I3
I did not add any new unknowns here, so I now have three equations connecting my three unknowns. You can solve them just like any system of multiple equations with multiple unknowns. I would suggest
-- using Equation 3 to get I3 in terms of I2 and I1, then
-- substitute that for I3 in Equation 2, to get Equation 2 in terms of just I1 and I2. Then
-- rearrange Equation 2 to get I2 in terms of I1, and
-- substitute this for I2 in Equation 1 to convert Equation 1 into an equation only in terms of I1
-- then you can solve for I1
If you want to talk more about any part of this, or check an answer, just let me know. I hope this helps!