Q. Shorten the circuit given below.

Solution:

                        In the circuit 4µF is in parallel combination with 12mF

                                                Therefore 12||4=12+4=16µF

                        And 16mµF is in series with 3µF therefore    

                                                          = (3x16)/(3+16) = 48/19

                                                                                     = 2.5µF

            Now 12µF is parallels connected with 2.5mF so their joint effect will be

                                                                                     = 12 +2.5

                                                                                      =14.5µF

                        Both of the capacitors are in series combination

                                    Therefore C_AB = (4 x14.5)/18.5

                                                        =3.13µF               Ans.