Q: Determine the current Io through the 12K ohm resistor.

Solution:

At node (1)

        (V₁+6)/12 + V₁/12 + (V₁ - V₂)/12 = 0

                            V₁ + 6 +V₁ +V₁ -V₂ = 0

                                              3V₁ -V₂ = -6 ..........................A

By multiplying both sides with 2

                                            6V₁ - 2V₂ = -12 ................. A1

At node (2)

                                    (V₂ - V₁)/12 + V₂/12 = 2mA

                                                    V₂ -V₁ +V₂ = 24

                                                     2V₂ -V₁ =24 ....................B

                        By adding equation A1 and B we get

                                                     2V₂ - V₁   = 24

                                                - 2V₂ + 6V₁= -12

                                                           5V₁ = 12 => V₁=12/5

            By putting the value of V₁ in A we get

                                                           V₂ =3V₁ +6           // V₁=12/5

                                                           =3 x 12/5 + 6

                                                            = (36 + 30)/5 =66/5

We know now the voltage at both of the nodes therefore now we can calculate the value of current through 12K resistance. Therefore for output current Io

                                                            Io = (V₁ - V₂)/12

                                                                = (12/5 - 66/5)/12            // putting value of V₁ &V₂

                                                          = -54/5 x 1/12

                                                          = - 54/60 =-0.9mA                           Ans.