1. The problem statement, all variables and given/known data Two resistors and two uncharged capacitors are arranged as shown above. Then a potential difference of V0 = 12 V is applied across the combination as shown. a. What is the potential at point a with S open? (Let V = 0 at the negative terminal of the battery.) b. What is the potential at point b with the switch open? c. When the switch is closed, what is the final potential of point b? d. How much charge flows through the switch S after it is closed? 2. Relevant equations Voltage of Capacitor = Q/C V=IR Kirchhoff's Laws (Junction Rule and Loop Rule) 3. The attempt at a solution I combined the two resistors so I would have one equivalent resistor and I combined the capacitors to have one equivalent capacitor. Now, the circuit is just the voltage source, a single resistor (13.2 Ohms) and a single capacitor (0.16 microFarads) I found the current through the resistor to be .909 Amps. Applying Kirchhoff's Laws, I got three equations: i1-i2-i3=0 12-13.2(i2)=0 13.2(i2)-Q/.16microF(i3)=0 --> not sure how to solve for Q since there's two unknowns here (i2 and Q) From here, I'm not sure what I need to do.