Q. Determine current passing through 6k ohm resistor.

Solution:

For node 1

       (V₁/3k) + ((V₁- V₂)/6k) =2mA   //      V=IR

         V₁ - V₂ +2V₁ =6k x 2mA

            -V₂ + 3V₁ =12

             9V₁ -3V₂ = 36                                             (1)  

Now for node 2

      V₂/12k + (V₂-V₁)/6k +4mA =0       //   V=IR

          V₂+2V₂-2V_{1 +} 48=0

              3V₂ - 2V₁ = -48                                            (2)

By addition of equation (A) & (B)

3V₂ - 2V₁ = -48

-3V₂+ 9V₁ = 3

    V₁= -12/7

Now put the value of V₁ in equation (1)

                                 V₂= 3V₁-12

                              = 3 x (-12/7) -12

                                 = (-36/7) -12

                        = (-36-84)/7=-120/7        

Now we can say    V₀ =V₂ - V1

                                         = (-120/7) + 12/7= -108/7

                          I₀ = -120/7   x (1/6k)

                                     =-120/42= 2.85 mA