The switch below has been closed for a long period of time and is opened att=0.a) Find the ENERGY stored in theSwitch_open_t=0capacitor just before theswitch is opened6ka1kQS1ka= 12V=1µFThe switch opens at t=0. Thecapacitor begins to discharge.Q(t)cQ(0). · eb) At what time 't' will the charge on the capacitor be equal to 20% of thecharge at t=0?c) How much charge remains on the capacitor at this time?d) Find the total energy dissipated by one of the 1k2 resistors usingE, = p(t)dt where p(t)is the instantaneous power%3D

Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please re-post the complete question and mention the sub-parts to be solved. The stored energy in capacitor (EC) is given by: Where, The capacitance of capacitor is “C”, The voltage across capacitor is “VC”. A: In the circuit, when switch is closed for a long period of time, the capacitor will act as open circuit. The equivalent circuit for this condition is shown below: In the above circuit, both 1 kΩ resistor, 6 kΩ resistor and 12 V voltage source are connected in parallel. To calculate the voltage across the capacitor terminal (VC), apply voltage division rule in above circuit across both 1 kΩ resistor: To calculate the stored energy in the 1 μF capacitor before opening the switch (EC), use the energy expression and substitute all known values:The equivalent circuit after the switch is open is shown below: In the above circuit, both 1 kΩ resistor are connected in series and the series combination is connected to capacitor. In the expression of charge (Q) given in question, the term “τ” represent the time constant of the RC circuit. The time constant of the RC circuit is the product of the total resistance of the RC circuit and capacitance of RC circuit. So, for the circuit after switch is open, the value of time constant (τ) is given by: To calculate the time after which charge on capacitor is 20% of initial charge, use the charge expression and substitute all known values:The charge initial stored across the capacitor is given by: Where, The capacitance of capacitor is “C”, The voltage across the capacitor before opening the switch is “VC”. So, the charge initial stored across the capacitor is: C: Thus, 20% of the charge initial stored across the capacitor is given by: Hence, answer to part A, part B and part C are 4.5 μJ, 3.2188 msec and 0.75 μC respectively.