Consider the circuit below and the followingevents. = 12V; C 1 =C 2 = 47 F and R

Question

Consider the circuit below and the followingevents. = 12V; C1 =C2 = 47 F and R1 = 15M. Switch S1 is closed, C1 andC2 are allowed to charge up completely and thenS1 is opened. Now switch S2 isclosed.
A) What is the total energy of the circuit after S1is opened and before S2 is closed?
B) What is the time constant of the circuit afterS2 is closed?
C) At what time is the potential the same across the bothcapacitors and the resistor?
D) What value capacitor if added between S2 andR1 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, = RCeq =(15)(2*47x10-6) = 1410 s
For Part C, I know the answer is supposed to be "Always", butI do not understand why. Is potential always the constant in an RCcircuit?
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 to RCeq2= 0.70RCeq1 and solve for C3, but Icalculated a negative value (-2.82x10-5), which isn'tpossible. Please help me find the right equation to manipulate.Thanks!

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