Q:    Supposing the diodes to be ideal .Discover the value of I & V in the circuit indicated in diagram below.

Solution :

            Here we are using ideal diode. In diagram (a) positive 5 V battery is attached to the anode of the diode via a resistance of 2.5 k ohm. The cathode terminal in a diode is attached along the neutral terminal. The diode will be in forward biased condition. From the diagram we can see that all of components are in series therefore the same current will pass.

                                    Thus by using Ohm's Law we get

                                                                        I = 5/2.5          

                                                                        I = 2mA           

                        '           Diode behave like short circuit thus,

                                                                         V=0 V

            In diagram (b) the circuit is same just the diode has reversed. 5 V battery is attached to the cathode of the diode via a resistance of 2.5 k ohm. The anode terminal in a diode is attached along the neutral terminal. The diode is now in reversed biased condition therefore, it will behave as an open circuit so no current will pass.

                                                            I = 0mA

                                    Overall voltage of 5 volts shall appear at +ve terminal.

                                                            V= 5V

            In diagram (c) -ve 5 Volt battery is connected to the anode in a diode through a resistance of      2.5 k ohm. The cathode terminal in a diode is attached with the neutral terminal. The diode will in reversed biased condition and it will behave as an open circuit so, no current will pass.

                                                                  I = 0mA

                                    overall voltage of 5 V will appear

                                                            V= - 5V

            In diagram (d) the circuit is similar only the diode has reversed. -ve 5 Volt battery is connected to the cathode in a diode through a resistance of 2.5 k ohm. The anode terminal in diode is linked with the neutral terminal. The diode now is in forward biased condition therefore, it will behave like short circuit so no current will pass.

                        Therefore by Ohm's Law we have

                                                                        I = 5/2.5          

                                                                        I = 2mA           

                                    Diode acting likes short circuit thus,

                                                                         V=0 volts