RC Circuit capacitor charge Help

[cp51]

[SOLVED] RC Circuit Help

Homework Statement

What is the potential of point a with respect to point b in the circuit shown when switch S is open? Which point, a or b is at a higher potential? b.) What is the potential (after a long time) of point a with respect to ground when switch S is closed? How much does the charge in each capacitor change when S is closed?
R1 = 6 Ohms, C1 = 6microFarads, C2 = 3 microFarads, R2 = 3 ohms, C3 = 6 microFarads
V = 18V

Im sorry for the poor drawing

Homework Equations

V = IR
C = Q/V

The Attempt at a Solution

Ok, I know the solution to this question, I just don't know why...
A.) Apparently this circuit simplifies to a circuit of just the capacitors, the resistors drop out and you end up with C1 and C2 in parallel, connected to C3. Why do the resistors drop out?
From this you can combine all 3 resistors into 1 resistor Ctotal = 3.6 microFarad, and so
Q = CV = (3.6microF)(18V) = 6.48x10^-5
then combining C1 and C2 to get C12 = 9microF, i can get the voltage drop across the capacitors = Q/9microF = 7.2V

B.) When the switch is closed, aparently all the voltage drop is across the bottom capacitor, C3... why?

Thank you very much for any help

 

[CEL]

What is the current in each resistor when the switch is open? What is the voltage drop across each resistor?

 

[cp51]

well, as I understand it:

Once C1 and C2 are fully charged no current flows through the two branches because the capacitors act as an open circuit. So that would mean that the current through the reisitors is 0 and therefore no voltage drop.

However, because the two capacitors are now charged, isn't there a volage difference between the two capacitors? So does that cause current to flow through o C3?? If so, doesn't that also require current to flow through R2? That is how I see it, which is apparently wrong cause it isn't the answer... heh...

 

[CEL]

well, as I understand it:

Once C1 and C2 are fully charged no current flows through the two branches because the capacitors act as an open circuit. So that would mean that the current through the reisitors is 0 and therefore no voltage drop.

However, because the two capacitors are now charged, isn't there a volage difference between the two capacitors? So does that cause current to flow through o C3?? If so, doesn't that also require current to flow through R2? That is how I see it, which is apparently wrong cause it isn't the answer... heh...

After the switch is closed, there is a direct path from the source V to C3 through the resistors R1 and R2, so C3 will be charged until its voltage reaches V. During the transient, capacitors C1 and C2 discharge trough R2 into C3.

 

[cp51]

ok, I am still not getting why when the switch is open, the circuit simplifies to a circuit without resistors...

 

[CEL]

ok, I am still not getting why when the switch is open, the circuit simplifies to a circuit without resistors...

If there is no current, there is no voltage drop on the resistor.

 

[cp51]

omg, I am so dumb... i totally didnt think of it that way... Thanks.

I started to look at it as charge and stuff, and I kinda thought i figured it out, but man...

Thanks a lot.