Resistor Circuit

• 3 resistors, R1, R2, R3
• 3 voltages at terminals, V1, V2, V3
• Find currents, I1, I2, I3, (taken as directed towards the central node) and central voltage VC.

Equations

1. Current balance at central node:
   I1 + I2 + I3 = 0
2. First voltage equation (from V = IR, Ohm's Law):
   R1 * I1 = V1 - VC
3. Second voltage equation:
   R2 * I2 = V2 - VC
4. Third voltage equation:
   R3 * I3 = V3 - VC

With a little rearrangement, (moving VC to the other side of the equations), this can be put in standard matrix form:

  [ 1   1   1   0 ] [I1]   [ 0]
  [ R1  0   0   1 ] [I2] = [V1]
  [ 0   R2  0   1 ] [I3]   [V2]
  [ 0   0   R3  1 ] [VC]   [V3]

Here we have 4 equations and 4 unknowns (the three currents and the central voltage). This is a fine system, written directly from observation and simple electrical laws, and can be solved as it stands given the Rs and Vs.

The central voltage, VC, seems a bit superfluous. We introduced it as an auxiliary variable to make the equations easy to write, but we have a voltage as an unknown as well as three currents, which does not seem as elegant as it might be, given the symmetry of the original system. If we want, we can eliminate VC from the system by subtracting the 3rd and 4th equations from the 2nd. This yields a system in the three current variables:

1. Current balance at central node:
   I1 + I2 + I3 = 0
2. First voltage equation minus the second:
   R1 * I1 - R2 * I2 = V1 - V2
3. First voltage equation minus the third:
   R1 * I1 - R3 * I3 = V1 - V3

In matrix form:

  [ 1    1    1  ] [I1]   [  0  ]
  [ R1  -R2   0  ] [I2] = [V1-V2]
  [ R1   0   -R3 ] [I3]   [V1-V3]

If you have taken an EE course, you may recognize these last two equations as what could be obtained by applying Kirchhoff's Laws to (virtual) loops in the circuit.

Adding or subtracting (multiples of) one equation from another is a standard way of simplifying systems of linear equations and, in organized form, the basis for Gaussian elimination, a basic approach for solving such systems.