Consider the circuit given in Fig 515 The circuit is in the

Consider the circuit given in Fig. 5.15. The circuit is in the ON state for times 0 lessthanequal t lessthanequal 2 and in the OFF state otherwise. For the time range t greaterthanequal 0, please find the time dependant equation of: the voltage across a capacitor with capacitance 0. IF if it is connected to the output port of this circuit. the current through an inductor of inductance 3H if it was connected to the output port of this circuit.

Solution

a) we can solve the above problem by SUPERPOSITION theorem i.e. considering one source at a time.

Let us consider the source voltage source at first, calculate the net resistance.

       R= 60+(45*30/45+30)

                 = 78 ohm

By considering only the first source the current source will be open circuit.Therefore by applying KVL we get the current across the load.

                 6 - 60 I - 45 I - 30 I = 0

                     I = 0.044 mA

By considering the source i.e. 75 mA current source we get the

                     net current I = 0.044+75 = 75.044 mA

Therefore the voltage across capacitor is given by

              V = 1/C integral ( i dt)

                V = 1/0.1 integral (75.04 dt)

                 V = 750.4t volts

b) To find the current across the 3H inductor load we can conside the NORTON theorem i.e converting the voltage sorce into current source and finding the norton cirrent and norton resistance.

To find the norton equivelant resistance open circuit the current source and short circuit the voltage source and the net equivalent resistance when see through the load we get

               R_eq = (45 + 60) II 30

                         = 23.33 ohm

similarly considering one source at atime the norten current is

                           I_N = 0.044 + 75

                                 = 75.044 mA

The current at the load is given by the Norton equation

               I = I_N + Req / (R_eq + R_L)

               I = 75.044 + 23.33 / (23.33+ 3H)

                      = 75.99 mA

Voltage across inductor = 0.075 * 45 = 3.75 mA

current across inductor is given by the equation

     I = L integral( v dt)

      I = 3 * 3.75 t

       I = 10.125t mA