A multiloop circuit is shown above. There are three branches, each drawn in a different color. The two junctions are indicated with red circles.
The resistor values are:
R1 = 1 Ω
R2 = 2 Ω
R3 = 3 Ω
R4 = 4 Ω
The battery emf's are:
ε1 = 12 V
ε2 = 3 V
ε3 = 10 V
Step 1 - Label the currents in each branch, indicating direction.
Step 2 - Label the + and - ends of the resistors. Current goes through a resistor from the + end to the - end.
Step 3 - Write down one loop equation. I chose the inside loop on the left and went clockwise. It doesn't matter where on the loop you start or which direction you go.
ε1 - I1 R1 - ε2 + I2 R2 - I1 R4 = 0
[Eq. 1] 9 - 5I1 + 2I2 = 0
Step 4 - Write down a second loop equation that includes the branch on the right. You can go around the outside of the circuit or do what I did, go around the inner loop on the right.
ε2 - ε3 + I3 R3 - I2 R2 = 0
[Eq. 2] -7 + 3I3 - 2I2 = 0
Step 5 - Apply the junction rule at either one of the junctions.
I1 + I2 + I3 = 0
Solve this for I1:
[Eq. 3] I1 = -I2 -I3
Substitute this into equation 1: 9 - 5I1 + 2I2 = 0
[Eq. 4] 9 + 7I2 + 5I3 = 0
Compare this to equation 2, -7 -2I2 + 3I3 = 0
Multiply equation 4 by 2 and equation 2 by 7:
[2 * Eq. 4] 18 + 14I2 + 10I3 = 0
[7 * Eq. 2] -49 - 14I2 + 21I3 = 0
Add to get -31 + 31I3 = 0.
This gives I3 = 1 A.
Substitute this back into equation 2:
-4 -2I2 = 0
This gives I2 = -2A
Substitute this into equation 3, I1 = -I2 -I3:
I1 = 1A
I1 = 1 A; I2 = -2 A; I3 = 1 A
There are a couple of ways to check whether the result is correct. One is to label the potential at different points in the circuit and make sure everything is consistent. Choose an arbitrary point to be zero (I chose the negative side of battery 1) and label potentials relative to that.
Going across a battery from - to + raises the potential by the battery voltage. Going from - to + drops it.
The potential difference across a resistor is IR. The current flows through a resistor from higher potential to lower potential.
The fact that I2 is negative means that the direction we chose for I2 was wrong. Make sure you remember that that current really goes the other way.
Another way to check your answer is to make sure that the power everywhere in the circuit is consistent with the Law of Conservation of Energy.
Using I2R, the power dissipated in each resistor is:
For R1: P = (1)21 = 1 W
For R2: P = (2)22 = 8 W
For R3: P = (1)23 = 3 W
For R4: P = (1)24 = 4 W
That adds up to 16 W dissipated in the resistors.
The power associated with the batteries can be determined using P = eI.
For ε1: P1 = 12*1 = 12 W
For ε2: P2 = 3*2 = 6 W
For ε3: P3 = 10*1 = 10 W
At first glance it looks like the batteries are putting more power into the circuit than is being dissipated. However, battery ε2 has current running through it from + to -, so it is being re-charged and is taking energy out of the circuit.
Batteries ε1 and ε3 supply 22 W of power to the circuit.
The resistors plus battery ε2 use 22 W of power.
Everything adds up, as it should.