Q.   Compute the current I_o.

Solution:

            In the circuit at node 1

                                     (V₁/12k)  + ((V₁ - V₂)/10k) = 6mA          // V=IR

                                                       5V₁ +6V₁ - 6V₂ = (60k) *(6mA)

                                                                11V₁- 6V₂ = 360

Now at node 2

        (V₂/3k) +(V₂/6k) + ((V₂- V₁)/10k) =0              // V=IR

                               10V₂ +5V₂+3V₂-3V₁=0

                         18V₂ -3V₁ =0

   Equating  of node1 & node 2

   33V₁ -18V₂ =1080

  -3V₁ + 18V₂ = 0    

   30V₁ =1080 

   V₁ = 36 volts

18V₂ -3V₁ =0

               6V₂ - V₁=0

                 6V₂ -36 =0

V₂ = 36/6

V₂    = 6volts

  I_o =V₂/6k

     = 6/6k

      = 1mA