1. The problem statement, all variables and given/known data Two identical bulbs (bulb A and bulb B), a capacitor, and a switch are connected to a battery as shown at right. The switch has been open for a long time. The battery is ideal and has a voltage (Vbat) of 4.0 volts. The switch is now closed. Just after the switch is closed, rank the magnitudes of the four voltages Vbat, VA, VB, and Vcap. A) Vbat=VA=VB= Vcap B) Vbat=VA=VB>Vcap C) Vbat=VA=Vcap>VB D) Vbat>VA=VB>Vcap E) VA=VB>Vcap>Vbat After the switch has been closed for a long time, rank the magnitudes of the currents ibat, iA, and iB. (Assume that the battery is ideal.) A) ibat = iA = iB B) ibat > iA = iB C) ibat = iA > iB D) iA > iB > ibat Just after the switch is opened, the voltage across bulb A (in V) is: 2 2. Relevant equations V=IR; C=Q/V; Kirchhoff Loop Rules 3. The attempt at a solution For the first, I know that in parallel, voltages will be equal. The capacitance is at zero at this moment, thus less than. Second, after a long time, the circuit will act as a loop w/o capacitor since the capacitor will now be fully charged. I bolded my answers. Wanted a check to see if my thought processes were indeed correct. If not, I would love to learn from my mistake. Thank you!