Consider the circuit below and the followingevents. ε = 12V;C1 =C2 = 47 μF and R1 = 15MΩ.Switch S1 is closed, C1 andC2 areallowed to charge up completely and thenS1 is opened.Now switch S2 isclosed.
A) What is the total energy of the circuit afterS1is opened and before S2 is closed?
B) What is the time constant of the circuit afterS2is closed?
C) At what time is the potential the same across thebothcapacitors and the resistor?
D) What value capacitor if added between S2andR1 decreases the time constant by 30 percent?
For Part A, I got PE = (1/2)CeqV2=(1/2)(2*47x10-6)(12) = 6.768 mJ
For Part B, Ï = RCeq =(15)(2*47x10-6) =1410 s
For Part C, I know the answer is supposed to be "Always", butIdo not understand why. Is potential always the constant in anRCcircuit?
Part D, I am not sure of the equation to use here.Originally,I thought I would just find a new Ceq2 =C1 +C2 + C3 and plug that in toRCeq2= 0.70RCeq1 and solve for C3,but Icalculated a negative value (-2.82x10-5), whichisn'tpossible. Please help me find the right equation tomanipulate.Thanks!