Q:  Find current passing through given circuit.

 Solution:          Through the circuit 3k Ω is in parallel with 6k Ω,

                                    Therefore,        3k||6k  = (3 x 6)/ (3 + 6)

                                                            = 18/9 =2k Ω

Through the circuit 4kΩ is series with 2kΩso their joint effect = 2k + 4k = 6kΩ.

                         Now through the circuit 6k Ω is in parallel along with 12k Ω therefore

                                                 6k||12k = (6 x 12)/ (12+6)

                                                  = 72/18

                                                  = 4k Ω

The resistors 4kΩ, 2kΩ, 4kΩ are in series therefore the joint effect will be                                                                                                                           = 2k+4k+4 =10kΩ

            Hence, the current passing through all of the Resistors by Ohm's law will be

                                                              I=V/R

                                                                =20/10k =2mA   Ans.