Ranking Voltages and Currents in a Circuit with 1 Capacitor

[orangefruit]

Homework Statement

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

Homework Equations

V=IR; C=Q/V; Kirchhoff Loop Rules

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!

 

[CWatters]

All your bold answers look ok to me.

However...

For the first, I know that in parallel, voltages will be equal. The capacitance is at zero at this moment, thus less than.

You mean the voltage on the capacitor is zero. The capacitance of the capacitor is unchanged.

Second, after a long time, the circuit will act as a loop w/o capacitor since the capacitor will now be fully charged.

I wouldn't use the words "w/o capacitor" (because of question 3). I think it's better to say...

After the switch has been closed for a long time the capacitor becomes fully charged and Ib falls to zero. Hence Ia>Ib.