Q:   Calculate the value of current passing through all of the resistors in the given circuit.

Solution:

                        To simplify the circuit first

                                                3k||6k= (3 x6)/ (3+6) =18/9=2kΩ

Through the circuit 2k Ω is in series with 2k Ω so their joint effect    =2k+2k=4k Ω

Through the circuit 3k is in series along 1k so their joint effect= 3k+1k=4k

Now 4k is in parallel along 4k, therefore

                                                 4k||4k=16/8=2k

 Now through the circuit the resistors 2kΩ, 6kΩ & 4kΩare in series

                    So they will be joint as=6k+2k+4k=12k Ω.

Therefore the current passing through all of resistors by Ohm's law

                                            =12/12k

                                            = 1k^-1 A = 1mA   Ans.