# Circuit with 1 capacitor

1. Feb 17, 2013

### orangefruit

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!

2. Feb 18, 2013

### CWatters

All your bold answers look ok to me.

However...

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

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.