A multiloop circuit

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 W
R2 = 2 W
R3 = 3 W
R4 = 4 W

The battery emf's are:

e1 = 12 V
e2 = 3 V
e3 = 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.

Which loop equation is correct for the inside loop on the left?

  1. e1 + I1 R1 - e2 - I2 R2 + I1 R4 = 0
  2. e1 + I1 R1 - e2 + I2 R2 - I1 R4 = 0
  3. e1 - I1 R1 + e2 + I2 R2 - I1 R4 = 0
  4. e1 - I1 R1 - e2 + I2 R2 - I1 R4 = 0
  5. e1 - I1 R1 - e2 + I2 R2 - I2 R4 = 0








The correct one is the fourth choice, which simplifies to:

[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.

Which is the correct loop equation here?

  1. e2 - e3 - I3 R3 - I2 R2 = 0
  2. e2 - e3 + I3 R3 - I2 R2 = 0
  3. e2 + e3 + I3 R3 - I2 R2 = 0
  4. e2 + e3 - I3 R3 - I2 R2 = 0
  5. e2 + e3 - I3 R3 + I2 R2 = 0








The second one is correct, and it simplifies to:

[Eq. 2] -7 + 3I3 - 2I2 = 0

Step 5 - Apply the junction rule at either one of the junctions.

Which equation represents a correct junction equation for this circuit?

  1. I1 + I2 = I3
  2. I1 + I3 = I2
  3. I1 = I2 + I3
  4. I1 + I2 + I3 = 0
  5. At least one of the current directions shown is incorrect so none of the above are valid equations.








The fourth one is correct. At least one of the current directions is incorrect but it doesn't matter. If you set up the equations consistently based on the directions shown your equations are perfectly valid.

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